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A. a sure win of $30 B. 80% chance to win $45Your choice must be made before the game starts, i.e., before the outcome of the first stage is known. Please indicate the option you prefer.
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37, 41, 45, 50, 52, 58, 60, 66, 70, 74, 76, 84, 87, 91, 95, 101, 103, 111, ...
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a team that has a .500 winning percentage has won 50% of their matches.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A baseball team won 45 percent of the first 80 games it played. How many of the remaining 85 games will the team have to win in order to have won exactly 40 percent of all the games it played?
A)30
B)45
C)50
D)55
E)81
Answer:
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A)30
Justification: Games won out of first 80 questions = (45/100)*80 = 36
Total Games = 80+85 = 165
Games to be won = (40/100)*165 = 66
Remaining games to be won = 66-36 = 30
Answer: Option A
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The perimeter of a square whose four sides have length ℓ {\displaystyle \ell } is P = 4 ℓ {\displaystyle P=4\ell } and the area A is A = ℓ 2 . {\displaystyle A=\ell ^{2}.} Since four squared equals sixteen, a four by four square has an area equal to its perimeter.
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the shape must be 'a rectangle' and must have 'four sides of equal length'.
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If a rectangle has length ℓ {\displaystyle \ell } and width w {\displaystyle w} it has area A = ℓ w {\displaystyle A=\ell w\,} , it has perimeter P = 2 ℓ + 2 w = 2 ( ℓ + w ) {\displaystyle P=2\ell +2w=2(\ell +w)\,} , each diagonal has length d = ℓ 2 + w 2 {\displaystyle d={\sqrt {\ell ^{2}+w^{2}}}} , and when ℓ = w {\displaystyle \ell =w\,} , the rectangle is a square.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
If a rectangular billboard has an area of 91 square feet and a perimeter of 40 feet, what is the length of each of the shorter sides?
A)4
B)7
C)8
D)13
E)26
Answer:
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B)7
Justification: This question can be solved algebraically or by TESTing THE ANSWERS.
We're told that a rectangle has an area of 91 and a perimeter of 40. We're asked for the length of one of the SHORTER sides of the rectangle.
Since the answers are all INTEGERS, and the area is 91, the shorter side will almost certainly be less than 10 (since 10x10 = 100, but we're NOT dealing with a square).
Let's TEST Answer b: 7
IF...
The shorter side = 7...
The area = 91....91/7 = 13 = the longer side
Perimeter = 7+7+13+13 = 40
B
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This method takes the list 1, 7, 3, 13 and orders it to read 1, 3, 7, 13. Then the 1 and 13 are removed to obtain the list 3, 7. Since there are two elements in this remaining list, the median is their arithmetic mean, (3 + 7)/2 = 5.
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, XN, and add up their values. By Wald's equation, the resulting value on average is E E = 1 + 2 + 3 + 4 + 5 + 6 6 ⋅ 1 + 2 + 3 + 4 + 5 + 6 6 = 441 36 = 49 4 = 12.25 . {\displaystyle \operatorname {E} \operatorname {E} ={\frac {1+2+3+4+5+6}{6}}\cdot {\frac {1+2+3+4+5+6}{6}}={\frac {441}{36}}={\frac {49}{4}}=12.25\,.}
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This is rearranged to 2,3,4,5. Second Column 4,1,4,2 is rearranged to 1,2,4,4, and column 3 consisting of 3,4,6,8 stays the same because it is already in order from lowest to highest value.) The result is: A 5 4 3 becomes A 2 1 3 B 2 1 4 becomes B 3 2 4 C 3 4 6 becomes C 4 4 6 D 4 2 8 becomes D 5 4 8 Now find the mean for each row to determine the ranks A (2 + 1 + 3)/3 = 2.00 = rank i B (3 + 2 + 4)/3 = 3.00 = rank ii C (4 + 4 + 6)/3 = 4.67 = rank iii D (5 + 4 + 8)/3 = 5.67 = rank iv Now take the ranking
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
find the average of first 20 multiples of 7?
A)73.5
B)70.5
C)63.5
D)54.5
E)40.5
Answer:
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A)73.5
Justification: Required average =7(1+2+3+…….+20)/20
(7*20*21)/(20*2)
(147/2)=73.5.
ANSWER A 73.5
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stomach, prostate, breast, thyroid, non-Hodgkin lymphoma, chondrosarcoma and osteosarcoma cancers (see WRN).
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Cancerous bone or soft tissue tumors (e.g. osteosarcoma, chondrosarcoma, fibrosarcoma, epithelioid sarcoma, Ewing's sarcoma, synovial sarcoma, sacrococcygeal teratoma, liposarcoma), melanoma
Document 2:::
include the lung, pancreas, breast, prostate, stomach, liver, and colon.The remaining 10 percent are either poorly or undifferentiated malignant neoplasms (5%), or squamous cell carcinomas (5%). Rarely, CUP may appear as neuroendocrine tumors, or mixed tumors, such as sarcomatoid, basaloid, or adenosquamous carcinomas.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Most radiosensitive tumor is
A. Renal cell carcinoma
B. Carcinoma colon
C. Hepatocellular carcinoma
D. Testicular seminoma
Answer:
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D. Testicular seminoma
Justification: Ref Robbins 8/e p989 In general, radiation therapy is mainly used for patients with seminoma, which is very sensitive to radiation. Sometimes it's used after orchiectomy (the operation to remove the testicle) and is directed at the lymph nodes at the back of the abdomen (the retroperitoneal lymph nodes). This is to kill any tiny bits of cancer in those lymph nodes that can't be seen. It can also be used to treat small amounts of seminoma that have spread to the nodes (based on changes seen on CT and PET scans).
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: 36–37 Consider the following thought experiment: The red arrow illustrates a train that is moving at 0.4 c with respect to the platform. Within the train, a passenger shoots a bullet with a speed of 0.4 c in the frame of the train. The blue arrow illustrates that a person standing on the train tracks measures the bullet as traveling at 0.8 c. This is in accordance with our naive expectations.
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will give the total time t the projectile must travel for to reach the target's horizontal displacement, neglecting air resistance.
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The points at which the two light flashes hit the ends of the train are at the same level in the diagram. This means that the events are simultaneous. In the second diagram, the two ends of the train moving to the right, are shown by parallel lines. The flash of light is given off at a point exactly halfway between the two ends of the train, and again form two 45° lines, expressing the constancy of the speed of light. In this picture, however, the points at which the light flashes hit the ends of the train
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A bullet train 220 m long is running with a speed of 59 kmph. In what time will it pass a man who is running at 7 kmph in the direction opposite to that in which the bullet train is going?
A)23 sec
B)15 sec
C)12 sec
D)11 sec
E)16 sec
Answer:
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C)12 sec
Justification: C
12 sec
Speed of the bullet train relative to man = (59 + 7) kmph
= 66 *5/18 m/sec = 55/3 m/sec.
Time taken by the bullet train to cross the man = Time taken by it to cover 220 m at (55/3) m / sec
= (220 *3/55) sec = 12 sec
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With work schedules, employees will often use a tradeoff of "9/80" where an 80-hour work period is compressed into a narrow group of 9 nearly-9 hour working days over the traditional 10 8-hour working days, allowing the employee to take every second Friday off.
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in the ranges of 5 to 7 hours and 14 to 17 hours, respectively.
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would be present after one hour? The question implies a = 1, b = 2 and τ = 10 min.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A can do a half of certain work in 70 days and B one third of the same in 35 days. They together will do the whole work in.
A)17 days
B)97 days
C)19 days
D)60 days
E)15 days
Answer:
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D)60 days
Justification: A = 140 days
B = 105 days
1/140 + 1/105 = 7/420 = 1/60
=>60 days
Answer: D
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Leucyl aminopeptidases (EC 3.4.11.1, leucine aminopeptidase, LAPs, leucyl peptidase, peptidase S, cytosol aminopeptidase, cathepsin III, L-leucine aminopeptidase, leucinaminopeptidase, leucinamide aminopeptidase, FTBL proteins, proteinates FTBL, aminopeptidase II, aminopeptidase III, aminopeptidase I) are enzymes that preferentially catalyze the hydrolysis of leucine residues at the N-terminus of peptides and proteins. Other N-terminal residues can also be cleaved, however. LAPs have been found across
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Aminopeptidase is active in the anterior and posterior midgut regions before and after feeding. In the whole midgut, activity rises from a baseline of approximately three enzyme units (EU) per midgut to a maximum of 12 EU at 30 hours after the blood meal, subsequently falling to baseline levels by 60 hours. A similar cycle of activity occurs in the posterior midgut and posterior midgut lumen, whereas aminopeptidase in the posterior midgut epithelium decreases in activity during digestion.
Document 2:::
If a stroke is suspected, magnetic resonance imaging is indicated.The activities of the enzymes creatine kinase (CK) and aspartate aminotransferase (AST) are elevated due to the death of muscle cells in the blood. If cardiac disease is present, which is often the case, brain natriuretic peptide is above the reference range. The "kidney values" (creatinine, urea, SDMA) may also be elevated due to the shock-induced reduced renal function (prerenal azotemia).
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Leucine aminopeptidase is elevated in obstruction of
A. Ureter
B. Urethra
C. Common bile duct
D. Spermatic cord
Answer:
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C. Common bile duct
Justification: Leucine Aminopeptidase Increased leucine aminopeptidase (Lap) activity is seen in : Carcinoma of the pancreas, choledocholithiasis, acute pancreatitis Viral hepatitis, cirrhosis, carcinoma with liver metastases In common bile duct obstruction, whether due to carcinoma pancreas or choledocholithiasis, the elevated serum Lap levels returned to normal following relief of the obstruction. This is in agreement with the hypothesis that the increased serum LAP activity in these conditions is the result of bile duct obstruction Ref: American journal of Gastroenterology, Dec 1963, vol 41 issue 6 Pgno : 620
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the second law of thermodynamics, are innumerable: When we reflect how generally physical phenomena are connected with thermal changes and relations, it at once becomes obvious that there are few, if any, branches of natural science which are not more or less dependent upon the great truths under consideration. Nor should it, therefore, be a matter of surprise that already, in the short space of time, not yet one generation, elapsed since the mechanical theory of heat has been freely adopted, whole
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and latent heats are observed in many processes of transfer of energy in nature.
Document 2:::
The second law of thermodynamics conceptualizes that the entropy of a closed system can never decrease. As the law relates to power plants, it dictates that heat is to flow from a body at high temperature to a body at low temperature (the device in which electricity is being generated). This law is particularly pertinent to thermal power plants which derive their energy from the combustion of a fuel source.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Energy is lost as heat between each trophic level due to the second law of what?
A. particles
B. reactions
C. thermodynamics
D. chemistry
Answer:
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C. thermodynamics
Justification: One major factor that limits the length of food chains is energy. Energy is lost as heat between each trophic level due to the second law of thermodynamics. Thus, after a limited number of trophic energy transfers, the amount of energy remaining in the food chain may not be great enough to support viable populations at yet a higher trophic level. The loss of energy between trophic levels is illustrated by the pioneering studies of Howard T. Odum in the Silver Springs, Florida, ecosystem in the 1940s (Figure 46.5). The primary producers generated 20,819 kcal/m2/yr (kilocalories per square meter per year), the primary consumers generated 3368 kcal/m2/yr, the secondary consumers generated 383 kcal/m2/yr, and the tertiary consumers only generated 21 kcal/m2/yr. Thus, there is little energy remaining for another level of consumers in this ecosystem.
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including olefins, acetic acid, formaldehyde, ammonia, urea and others.
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health and disease is a subject of research. : 104 Some phenols are germicidal and are used in formulating disinfectants.
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disinfectants, and sunscreen agents. These products may be found in cosmetics, perfumes, menstrual care products, lotions, shampoos, soaps, toothpastes, and sunscreen. These products typically enter the environment when passed through or washed off the body and into the ground or sewer lines, or when disposed of in the trash, septic tank, or sewage system. Traces of illicit drugs can be found in waterways and may even be carried by money.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Best virucidal disinfectant is?
A. Phenol
B. Hypochloe
C. BPL
D. Formaldehyde
Answer:
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D. Formaldehyde
Justification: Ans. is 'd' i.e., Formaldehyde Among the options, formaldehyde is high level disinfectant. Categories of effective potency Sterilants :- are capable of completely eliminating or destroying in all forms of microbial life, including spores. Disinfectants :- Destroy some, but not necessarily all organisms. The category is fuher divided into subcategories:- i) High level disinfectants :- In their usual concentration and contact period, they destroy all microorganisms, with exception of high number of bacterial spores ( small number of spores can be destroyed). It is woh noting that at higher concentrations and prolonged contact period, high level disinfectant can act as steriliant, i.e. can kill high numbers of spores as well. Examples are :-2% gluteraldehyde, 8% formaldehyde, 6-10% hydrogen peroxide and ethylene oxide gas. ii) Intermediate level disinfectants :- Inactivate even resistant organisms such as mycobacterium tuberculosis as well as vegetative bacteria, most viruses and most fungi, but do not necessarily kill bacterial spores. Examples are :-0.5% iodine, 70-90% ethanol and isopropanol, chlorine compounds (hypochlorite), some phenolic compounds and iodophor based disinfectants. iii) Low level disinfectants :- kill most bacteria, some viruses and some fungi, but cannot be relied on to kill resistant microorganisms such as tubercular bacilli or bacterial spores. Examples are :- quaerly ammonium compounds, mercurials,some phenolic compounds and iodophores. Note:- the disinfectant levels of iodophors (iodines) and phenolic compounds may be classified as intermediate or low depending on the concentration employed.
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In ordеr to storе and rеtriеvе thе data nееdеd for procеssing, AI rеquirеs еffеctivе mеmory systеms. To avoid bottlеnеcks in data accеss, rapid connеctivity and largе-capacity mеmory is crucial.
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34-et reduces the combined mistuning from the theoretically ideal just thirds, fifths and sixths from 11.9 to 7.9 cents.
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N_{M_{j}}={\begin{cases}0&j=p{\text{ and }}m_{p}={\text{reject}}\\1&j=p{\text{ and }}m_{p}={\text{accept}}\\\max _{m_{j+1}}N_{M_{j+1}}&j
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
NRHM goal for IMR is to reduce it by ?
A. 10
B. 30
C. 40
D. None
Answer:
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B. 30
Justification: Ans. is 'b' i.e., 30 NRIIM goal for infant moality rate is to reduce it to 30 per 1000 live bihs.
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Melanoma is a type of neuroectodermal neoplasm. There are four main types of melanoma: Other histopathologic types are: Mucosal melanoma; When melanoma occurs on mucous membranes. Desmoplastic melanoma Melanoma with small nevus-like cells Melanoma with features of a Spitz nevus Uveal melanoma Vaginal melanoma Polypoid melanoma, a subclass of nodular melanoma.
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In children, most cases are associated with neuroblastoma and most of the others are suspected to be associated with a low-grade neuroblastoma that spontaneously regressed before detection. In adults, most cases are associated with breast carcinoma or small-cell lung carcinoma. It is one of the few paraneoplastic (meaning 'indirectly caused by cancer') syndromes that occurs in both children and adults, although the mechanism of immune dysfunction underlying the adult syndrome is probably quite different.It
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nerves that can only be seen microscopically but not radiologically and are often confined to the main tumor mass. The transition from PNI to PNS is not precisely defined, but PNS is detectable on MRI and may have clinical manifestations that correlate with the affected nerve.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Which of the following is primary neurogenictumour ?
A. Meninginoma
B. Glioblastoma
C. Acoustic neuroma
D. Neurobl astoma
Answer:
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D. Neurobl astoma
Justification: Ans. is 'd' i.e., Neuroblastoma
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one showing "heads" with 2/3 probability and one showing "heads" with 9/10 probability.
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{2T} }} , which are the probability that coins 1 and 2 come up heads or tails.
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A binomial distributed random variable Y with parameters n and p is obtained as the sum of n independent and identically Bernoulli-distributed random variables X1, X2, ..., XnExample: A coin is tossed three times. Find the probability of getting exactly two heads. This problem can be solved by looking at the sample space. There are three ways to get two heads. The answer is 3/8 (= 0.375).
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
If a coin is tossed three times, what is the probability that on the first toss the coin lands on heads and then lands on tails the next two tosses?
A)1/8
B)1/3
C)1/4
D)1/2
E)1
Answer:
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A)1/8
Justification: P(HTT) = 1/2*1/2*1/2=1/8
The answer is A.
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Increased pressure at the LES over time may result in an epiphrenic diverticulum. Further evaluation for mechanical causes of obstruction may include CT scans, MRI, or endoscopic ultrasound.Several additional tests may be used to further evaluate EGJOO. Further evaluation of esophageal motor function may be accomplished with functional lumen imaging probe (FLIP). Although not widely available, FLIP may help assess esophageal wall stiffness and compliance. FLIP may help identify individuals with EGJOO who
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EGJOO is diagnosed using esophageal manometry. High resolution esophageal manometry will show elevated pressure at the LES with normal peristalsis. The LES pressure is evaluated immediately following a swallow, when the sphincter should relax. The overall LES pressure after a swallow is represented by the integrated relaxation pressure (IRP).
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The diagnosis of nutcracker esophagus is typically made with an esophageal motility study, which shows characteristic features of the disorder. Esophageal motility studies involve pressure measurements of the esophagus after a patient takes a wet (fluid-containing) or dry (solid-containing) swallow. Measurements are usually taken at various points in the esophagus.Nutcracker esophagus is characterized by a number of criteria described in the literature. The most commonly used criteria are the Castell
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Investigation of choice in diffuse esophageal spasm is -
A. Manometry
B. Esophagoscopy
C. Barium examination showing teiary contractions
D. CT thorax
Answer:
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A. Manometry
Justification: Diffuse esophageal spasm (DES) Motility disorder of the esophagus Representing uncoordinated teiary contractions of the esophagus, Corkscrew esophagus is associated with diffuse esophageal spasm On barium swallow, DES may appear as a corkscrew or rosary bead esophagus, but this is uncommon. Manometry is the gold-standard diagnostic test / investigation of choice. Differential Diagnosis * Hypercontracting esophagus (nutcracker esophagus) has normal peristalsis but high manometric intra-esophageal pressures o Barium swallow will be normal or show dysmotility
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histopathology, genetic abnormalities, and course. These subtypes, which are now (i.e. primary gastrointestinal tract FL) or may in the future (pediatric-type FL) be considered distinctive diseases, are:
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gas pain, nutrient deficiencies, and the inability to digest certain foods.
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of complications. This disease as well as the primary one requires immediate treatment (for example, type 2 diabetes).
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
All are true about Zollinger-Ellison syndrome except:
A. Recurrent ulceration after acid reducing surgery
B. Raised gastrin levels in all cases
C. Decreased BAO/MAO
D. Diarrhea
Answer:
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C. Decreased BAO/MAO
Justification: Ans. (c) Decreased BAO/MAOBlood levels of ZES:* 100% all patients will have Se. gastrin >100pg/ml* BAO >15meq/hr in most patients* BAO will be 60% of MAO (BAO - Basal Acid output, MAO - Maximal Acid Output)* Levels of Se. Gastrin >1000pg/ml is diagnostic* If there is only borderline elevation provocative tests with secretin, Calcium etc are done* With Secretin stimulation gastrin increases >200pg/ml in 15 minutes is diagnostic
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cells, macrophages, eosinophils, fibroblasts, and Reed–Sternberg cells (HRS cells, also termed Hodgkin Reed-Sternberg cells) into lymphoid and other tissues. HRS cells are large mono- or poly-nuclear cells which: 1) derive from lymph node and/or spleen germinal center B cells; 2) may contain EBV and viral products indicative of stage II latency; and 3) are the only malignant cells in, and the mediators of, HD.
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Microscopically, the ulcers consist of lymphocytes, including EBV+ B cells, sometimes a scattering of other EBV+ lymphoid cell types, and histiocytes, plasma cells, eosinophils, and scattered large immunoblasts which may closely resemble but are not the Reed–Sternberg cells seen in Hodgkin lymphoma. These Reed-Sternberg–like cells are EBV+ B cells that express the tumor marker cell surface membrane protein, CD30, the B cell surface membrane marker, CD20, and the proteins typical of the EBV replication cycle
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It is a more aggressive disease in the elderly. Histologic features of DLBCL can be divided into three patterns based on the cell types in tissue infiltrates; the anplastic variant (~3% of cases) exhibits prominent Reed–Sternberg-like cells embedded in a background of histiocytes and lymphocytes; the immunoblastic variant (8–10% of cases) has 90% immunoblasts; and the centroblastic variant (80% of cases) is dominated by centroblasts. These histological features are typically accompanied by the invasion and
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Reed Sternberg cells are found in -
A. Hodkin's disease
B. Sickle cell anaemia
C. Thalassemia
D. CML
Answer:
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A. Hodkin's disease
Justification: Ans. is 'a' i.e., Hodgkin's disease o Pathologically, Hodgkins disease is distinguished from other lymphomas by the presence of Reed Sternberg cells.o Reed Sternberg cells are not absolutely specific for Hodgkins disease and have also been noticed in cases of infectious mononucleosis and other malignancies including lymphoma, carcinoma and sarcomas,o Therefore, Reed-Stemberg, cells are not sufficient to establish the diagnosis of Hodgkins disease,o The diagnosis of hodgkins1 lymphoma requires the presence ofboth the characteristic Reed-Stemberg cells and the characteristics cellular environment comprising of lymphocytes, histiocytes, granulocytes, eosinophils and plasma cells.o One more method which can be useful in the diagnosis is immunophenotyping for CD-I5 and CD-30. The neoplastic cells of Hodgkins disease, both classic Reed-stemberg cell and Reed-sterberg variants, tend to stain positively with these antibodies.o In addition R.S. cells of lymphocytic predominant cells are also positive for CD20.o Atypical cells can be seen in some types of Hodgkins disease, but they are not diagnostic.
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Vehicles are generally equipped with a variety of instruments mounted on the dashboard to indicate driving parameters and the state of the mechanics. The placement of the instruments can vary. While they are usually mounted behind the steering wheel, they may also be mounted centrally below the windshield, or integrated into the center stack above the climate control and audio system.
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When the outside temperature is 21 °C (70 °F), the temperature inside a car parked in direct sunlight can quickly exceed 49 °C (120 °F). Young children or elderly adults left alone in a vehicle are at particular risk of succumbing to heat stroke. "Heat stroke in children and in the elderly can occur within minutes, even if a car window is opened slightly." As these groups of individuals may not be able to open car doors or to express discomfort verbally (or audibly, inside a closed car), their plight may
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the effect of temperature changes on the monitor response, and so on.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
if the temperature gauge on a dashboard goes up while vehicle is motion, what could be responsible?
A. the snow falling outside
B. the rubbing of the mechanical parts
C. the car sales man
D. the anger of the driver
Answer:
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B. the rubbing of the mechanical parts
Justification: friction causes the temperature of an object to increase
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Receptor-mediated endocytosis is a form of pinocytosis where a cell takes in specific molecules or solutes. Proteins with receptor sites are located on the plasma membrane, binding to specific solutes. The receptor proteins that are attached to the specific solutes go inside coated pits, forming a vesicle.
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through a clathrin-mediated process, into an endosome, where acidic conditions are thought to lead to exposure of membrane-destabilizing portions of L2.
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As it promotes the acidification of lysosomes, endosomes, and secretory vesicles, V-ATPase contributes to processes including: vesicular/protein trafficking receptor recycling endocytosis protein degradation autophagy cell signalingWith its role in lysosomal acidification, V-ATPase is also crucial in driving the transport of ions and small molecules into the cytoplasm, particularly calcium and amino acids. Additionally, its acidification of endosomes is critical in receptor endocytosis as low pH tends to
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
The ligand-receptor complex dissociates in the endosome because
A. Of its large size
B. The vesicle looses its clathrin cost
C. Of the acidic pH of the vesicle
D. Of the basic pH of the vesicle
Answer:
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C. Of the acidic pH of the vesicle
Justification: Ligand-gated channels are opened by binding of effectors. The binding of a ligand to a receptor site on the channel results in the opening (or closing) of the channel (Fig. 2.11). The ligand may be an extracellular signaling molecule or an intracellular messenger.Ref: DM Vasudevan, page no: 15
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m / s 2 , R = 6500 km {\displaystyle g=10{\text{ m}}/{\text{s}}^{2},R=6500{\text{ km}}} that time is T = π ω = π g R ≈ 3.1415926 10 6500000 ≈ 2532 s {\displaystyle T={\frac {\pi }{\omega }}={\frac {\pi }{\sqrt {\frac {g}{R}}}}\approx {\frac {3.1415926}{\sqrt {\frac {10}{6500000}}}}\approx 2532{\text{ s}}}
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20 seconds, therefore, giving an average speed of 187.5 m/s.
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would be present after one hour? The question implies a = 1, b = 2 and τ = 10 min.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A train 250 m long passed a pole in 10 sec. How long will it take to pass a platform 1250 m long?
A)60 sec
B)90 sec
C)120 sec
D)150 sec
E)180 sec
Answer:
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A)60 sec
Justification: Speed = 250/10 = 25 m/sec.
Required time = (250 + 1250)/25 = 60 sec
Answer:A
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If infection occurs or is suspected, treatment is generally with the antibiotics streptomycin or gentamicin. Doxycycline was previously used. Gentamicin may be easier to obtain than streptomycin. There is also tentative evidence to support the use of quinolone antibiotics.
Document 1:::
Since then, many more antibiotics and other secondary metabolites have been isolated and manufactured by microbial fermentation on a large scale. Some important antibiotics besides penicillin are cephalosporins, azithromycin, bacitracin, gentamicin, rifamycin, streptomycin, tetracycline, and vancomycin. Cell Cultures Animal or plant cells, removed from tissues, will continue to grow if cultivated under the appropriate nutrients and conditions.
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Treatment is typically with the antibiotic streptomycin. Gentamicin, doxycycline, or ciprofloxacin may also be used.Between the 1970s and 2015, around 200 cases were reported in the United States a year. Males are affected more often than females.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Probenecid interacts with ?
A. Streptomycin
B. Ampicillin
C. Vancomycin
D. Erythromycin
Answer:
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B. Ampicillin
Justification: Ans. is 'b' i.e., Ampicillin Interactions of probenecid Probenecid inhibits urinary excretion of penicillins, cephalosporins, methotrexate, sulfonamides and indomethacin. It inhibits biliary excretion of rifampicin. Salicylates block uricosuric action of probenecid. Probenecid inhibits tubular secretion of nitrofurantoin. Pyrazinamide and ethambutol may interfere with uricosuric action of probenecid. Note : Ampicillin is a penicillin.
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may include respiratory depression (decreased breathing), sleepiness, adrenal insufficiency, QT prolongation, low blood pressure, allergic reactions, constipation, and opioid addiction. Among those with a history of seizures, a risk exists of further seizures.
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disease, attention deficit/hyperactivity disorder, Parkinson's disease, Tourette's syndrome, multiple sclerosis, head trauma, sleep disorders, neuromuscular diseases, and various infections and tumors of the nervous system.
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effects include irregular heartbeat, seizures, heart attack, stroke, and death.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Analeptics will:
A. Depress CNS
B. Stimulate CNS
C. Causes seizures
D. Control seizures
Answer:
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B. Stimulate CNS
Justification: Ans: (b) Stimulate CNSRef: Kaplan and Sadock's Comprehensive Textbook of Psychiatry, 10th editionANALEPTICSAnaleptic drugs are basically CNS stimulants*.They also stimulate respiration by neural and other mechanisms.They are grouped based on their mechanism of action:Serotonin receptor agonists - BuspironeAMPA agonists - Ampakines (under trial)Adenosine antagonists - Caffeine (Methylxanthines)Potassium channel blockers - DoxapramUses:Serotonin receptor agonists - Depression*AMPA agonists - Used for research purposesAdenosine antagonists - Bronchodilator (Respiratory distress syndrome, Reactive airway disease)Potassium channel blockers - Reversal of respiratory depression (Following anesthesia, drug overdose, neonatal apnea, respiratory failure etc.)Side Effects:Excessive CNS stimulation - Irritability, anxiety, tremors, palpitations, seizures, insomnia, elevated pulse and BP.
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can be given in a controlled environment. The most commonly used drugs for this purpose are ajmaline, flecainide, and procainamide, with some suggestions indicating that ajmaline may be the most effective. Precaution must be taken in giving these medications as there is a small risk of causing abnormal heart rhythms.
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Several classes of medications, collectively referred to as antihypertensive medications, are available for treating hypertension. First-line medications for hypertension include thiazide-diuretics, calcium channel blockers, angiotensin converting enzyme inhibitors (ACE inhibitors), and angiotensin receptor blockers (ARBs). These medications may be used alone or in combination (ACE inhibitors and ARBs are not recommended for use together); the latter option may serve to minimize counter-regulatory
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Since beta blockers are known to relax the cardiac muscle and to constrict the smooth muscle, beta-adrenergic antagonists, including propranolol, have an additive effect with other drugs which decrease blood pressure, or which decrease cardiac contractility or conductivity. Clinically significant interactions particularly occur with: Verapamil Epinephrine (adrenaline) β2-adrenergic receptor agonists Salbutamol, levosalbutamol, formoterol, salmeterol, clenbuterol etc. Clonidine Ergot alkaloids Isoprenaline
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
What is the drug of choice to control supraventricular tachycardia :
A. Adenosine
B. Propranolol
C. Verapamil
D. Digoxin
Answer:
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A. Adenosine
Justification: Answer is A (Adenosine) Drug of choice to control supraventricular tachycardia is adenosine.
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and invasin, binding to integrins on the outer membrane of the cell.
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Membrane conditions Spherocytosis Hereditary elliptocytosis Enzyme conditions Glucose-6-phosphate dehydrogenase deficiency (also called G6PD deficiency) Pyruvate kinase deficiency Congenital erythropoietic porphyria (CEP, also called Morbus Günther; Uroporphyrinogen III Synthase deficiency) Globin synthesis defect sickle cell disease Alpha-thalassemia, e.g. HbH disease
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Integrin beta-1 can exist as different isoforms via alternative splicing. Six alternatively spliced variants have been found for this gene which encode five proteins with alternate C-termini. Integrin receptors exist as heterodimers, and greater than 20 different integrin heterodimeric receptors have been described. All integrins, alpha and beta forms, have large extracellular and short intracellular domains. The cytoplasmic domain of integrin beta-1 binds to the actin cytoskeleton. Integrin beta-1 is the
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Hereditory spherocytosis is due to deficiency of-
A. Ankyrin
B. Actin
C. Selectin
D. Integrin
Answer:
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A. Ankyrin
Justification: HS is caused by diverse mutations that lead to an insufficiency of membrane skeletal components. The pathogenic mutations most commonly affects ankyrin, band 3, spectrin, or band 4.2. Ref :Robbins pathologic basis of disease ; south east asia edition ;pg:633
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Three types of cars were manufactured for these trains, which were motor, trailer, and cab-trailer. The minimal quantity of cars is four (2 motor & 2 cab-trailer cars). The maximal quantity was twelve (6 motor cars, 4 trailers, and 2 cab-trailers.) The trains could be controlled and/or operated only from cab cars.
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Suppose one wants to determine the 5-combination at position 72. The successive values of ( n 5 ) {\displaystyle {\tbinom {n}{5}}} for n = 4, 5, 6, ... are 0, 1, 6, 21, 56, 126, 252, ..., of which the largest one not exceeding 72 is 56, for n = 8. Therefore c5 = 8, and the remaining elements form the 4-combination at position 72 − 56 = 16.
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The number of ways to choose a member of {A, B, C}, and then to do so again, in effect choosing an ordered pair each of whose components are in {A, B, C}, is 3 × 3 = 9. As another example, when you decide to order pizza, you must first choose the type of crust: thin or deep dish (2 choices). Next, you choose one topping: cheese, pepperoni, or sausage (3 choices). Using the rule of product, you know that there are 2 × 3 = 6 possible combinations of ordering a pizza.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
John has 4 friends who want to ride in his new car that can accommodate only 3 people at a time (John plus 2 passengers). How many different combinations of 2 passengers can be formed from John's 4 friends?
A)6
B)8
C)10
D)15
E)18
Answer:
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A)6
Justification: 4C2 = 6
The answer is A.
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Initially, it is present in the limbic cortices; it may then spread to the adjacent frontal and temporal lobes. Damage to specific areas can result in reduced or eliminated ability to encode new explicit memories, giving rise to anterograde amnesia. Patients with anterograde amnesia may have episodic, semantic, or both types of explicit memory impaired for events after the trauma that caused the amnesia.
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The posterior cerebral artery is a main locale for the cause of this deficit because this artery is not just responsible for itself. It also supplies the anterior temporal branches, the posterior temporal branches, the calcarine branch, and the parieto-occipital branch. What is important about these arteries is their location.
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The frontal lobe contains the precentral gyrus and prefrontal cortex and, by some conventions, the orbitofrontal cortex. These three areas are represented in both the left and the right cerebral hemispheres. The precentral gyrus or primary motor cortex is concerned with the planning, initiation and control of fine motor movements dorsolateral to each hemisphere.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Which area is involved for the memory deficit in posterior cerebral aery embolism?
A. Pre frontal coex
B. Hippocarpal gyrus
C. Angular gyrus
D. Superior temporal gyrus
Answer:
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B. Hippocarpal gyrus
Justification: Memory deficit is present only if dominant hemisphere is involved. The most common long-term sequelae of PCA strokes are visual and sensory deficits. Infarction of the medial temporal lobe, fornices, or medial thalamic nuclei may result in permanent anterograde amnesia. Lesions of the lingual gyrus in the inferior occipital lobe may produce disorders of color perception. Pure alexia may result from infarction of the dominant occipital coex. Prosopagnosia refers to an inability to recognize faces. Typically, this deficit results from bilateral lesions of the lingual and fusiform gyri. The major posterior cerebral aery (PCA) stroke syndromes (many of which occur concomitantly) include the following: Paramedian thalamic infarction Visual field loss Visual agnosia Balint syndrome Prosopagnosia Palinopsia, micropsia, and macropsia Disorders of reading Disorders of color vision Memory impairment Motor dysfunction
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look directly for micro-organisms or to provide a thorough inventory of complex organic compounds.
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They are required to make proteins, DNA, membranes, organelles, and exoskeletons. The major elements that constitute >95% of organic matter mass are carbon, hydrogen, nitrogen, oxygen, sulfur, and phosphorus. Minor elements are iron, manganese, cobalt, zinc and copper.
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The particulate organic matter in sediments is about 20% of known molecules 80% of material that cannot be analysed. Detritivores consume some of the fallen organic materials. Aerobic bacteria and fungi also consume organic matter in the oxic surface parts of the sediment.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Where could you find organic matter?
A. plastic items
B. carbon dioxide
C. a rabbit
D. table salt
Answer:
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C. a rabbit
Justification: an organism is a source of organic matter
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seen in patients without pre-existing medical conditions, and with infection localized to the extremities.
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The most difficult aspect of the differential diagnosis may arise from the similarity of two other diseases: Familial exudative vitreoretinopathy which is a genetic disorder that also disrupts the retinal vascularization in full-term infants. Persistent fetal vasculature that can cause a traction retinal detachment difficult to differentiate but typically unilateral.
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They are also known to develop in the conjunctiva of the eye, anogenital (perineum) region, and esophagus. Cutaneous lesions tend to follow the onset of mucosal lesions. The blisters often erupt in waves, usually affecting the upper trunk, head, neck, and proximal extremities.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Soft exudates are found in all except-
A. Diabetic Retinopathy
B. Hypertensive retinopathy
C. CMV retinitis
D. Retinopathy of prematurity
Answer:
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D. Retinopathy of prematurity
Justification: Ans. is 'd' i.e., Retinopathy of prematurity Causes of soft exudates (cotton wool spots)* All types of severe hypertension (hypertensive retinopathy) including renal disease, Toxemia of pregnancy.* Diabetic retinopathy* Collagen vascular disordersSLE, PAN, systemic sclerosis, scleroderma, wegener granulomatosis.* Infections AIDS, septic retinitis (including SABE), rarely CMV retinitis in AIDS.* Microembolization IV drug abuse, Post cardiac surgery, Fat embolism or Purtscher retinopathy.* OtherEale's disease, Adults anemia (Hb < 6.6-8.0 g/dl), Pseudoxanthoma elasticum, Neoplasia (Leukemia, Hodgkin's disease), Sickle cell retinopathy, radiation retinopathy. Causes of Hard exudates* Sever hypertension (Hypertensive retinopathy), especially with renal disease* Diabetic retinopathy, especially with renal disease* Infections Measles, influenza, meningitis, erysipelas, psittacosis, parasitic infection, coccidioidomycosis, condidiosis, TB, syphilis.* Collagen vascular diseases Dermatomyositis, SLE, PAN, Bechet's syndrome, Systemic sclerosis, Rheumatic polyarthritis.* HematologicalPernicious anemia, other severe adult anemia (Hb < 8 g/dl), leukemia, multiple myeloma.* Eye diseasesDisciform macular degeneration, Coats disease, Eales disease.* OtherCerebral trauma, after strangulation, fat embolism, lead poisoning, sarcoidosis, old central retinal vein thrombosis, Retinal artery macroaneurysm, Choroidal neovascularization, Hypercholesterolemia.
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schizophrenia that experience flat affect, can also experience difficulty perceiving the emotions of a healthy individual.
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The symptoms of Morvan's Syndrome have been noted to bear a striking similarity to limbic encephalitis (LE). These include the CNS symptoms consisting of insomnia, hallucinations, and disorientation, as well as dementia and psychosis. Both entities can be paraneoplastic and associated with thymoma.
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Abnormal activity in the PCC has been linked to schizophrenia, a mental disorder with common symptoms such as hallucinations, delusions, disorganized thinking, and a lack of emotional intelligence. What is common between symptoms is that they have to do with an inability to distinguish between internal and external events. Two PET studies on patients with schizophrenia showed abnormal metabolism in the PCC. One study reports that glucose metabolism was decreased in people with schizophrenia, while another
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A patient of Schizophrenia on neurolepts, his psychotic symptoms gets relieved but developed sadness, talks less to others, remain on bed, all of following are likely causes except:
A. Parkinsonism
B. Major depression
C. Negative symptoms are still persisting
D. He is reacting to external stimuli
Answer:
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D. He is reacting to external stimuli
Justification: D i.e. He is reacting to external stimuli Post psychotic depressive disorder of schizophreniaQ occurs only during residual phase of schizophrenia in 25% Differential diagnosis of post psychotic depressive disorder of schizophrenia are - Residual phase (negative symptoms) of schizophreniaQ - Adverse effects (eg. parkinsonism, akinesia, akathisia) of antipsychotic drugsQ - Schizoaffective disorder - depressive type
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Haemorrhage into the necrotic skin causes purpura fulminans lesions to become painful, dark and raised, sometimes with vesicle or blister (bulla) formation.The distribution of purpura fulminans lesions may be different according to the underlying pathogenesis. Purpura fulminans in severe sepsis typically develops in the distal extremities and progresses proximally or appears as a generalised or diffuse rash affecting the whole body surface. In cases of severe inheritable protein C deficiency, purpura
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Purpura fulminans is an acute, often fatal, thrombotic disorder which manifests as blood spots, bruising and discolouration of the skin resulting from coagulation in small blood vessels within the skin and rapidly leads to skin necrosis and disseminated intravascular coagulation.
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increase the blood's tendency to coagulate when treatment is first begun (many patients when starting on warfarin are given heparin in parallel to combat this), leading to massive thrombosis with skin necrosis and gangrene of limbs. Its natural counterpart, purpura fulminans, occurs in children who are homozygous for certain protein C mutations.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Henoch Schoenlein Purpura is characterized by following except:
A. Palpable purpura
B. Renal failure
C. GIT hemorrhage
D. Thrombocytopenia
Answer:
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D. Thrombocytopenia
Justification: Ans D. Thrombocytopenia. (Ref. H - 17th / pg. 2128)# Purpura in 'Henoch Scholein purpura' is due to vasculitis and not thrombocytopenia.HENOCH-SCHONLEIN PURPURA (anaphylactoid purpura)Definition:# It is a small vessel vasculitis, is a distinct systemic vasculitis syndrome characterized by:- palpable purpura, arthralgias, GI involvement and glomerulonephritis.Incidence and Prevalence# It is usually seen in children; most patients range in age from 4-7 years; however, the disease may also be seen in infants and adults.# It is not a rare disease.# The male-to-female ratio is 1.5:1.# A seasonal variation with a peak incidence in spring has been noted.Pathology and Pathogenesis# A immune-complex deposition disease.# A number of inciting antigens have been suggested including upper respiratory tract infections, various drugs, foods, insect bites, and im- munizations. IgA is the antibody class most often seen in the immune complexes and has been demonstrated in the renal biopsies of these patients.Clinical and Laboratory Manifestations# Palpable purpura (most commonly distributed over the buttocks and lower extremities and in pediatric patients, palpable purpura is seen in virtually all patients),# Arthralgias (most patients develop poly- arthralgias in the absence of frank arthritis),# GI signs and symptoms and# Glomerulonephritis (occurs in 10-50% of patients).# Myocardial involvement can occur in adults but is rare in children.# Lab studies show a mild leukocytosis, a normal platelet count, and occasionally eosinophilia. Serum complement components are normal, and IgA levels are elevated.Diagnosis# The diagnosis of Henoch-Schonlein purpura is based on clinical signs and symptoms.# Skin biopsy useful in confirming leukocytoclastic vasculitis with IgA and C3 deposition.# Renal biopsy is rarely needed for diagnosis but may provide prognostic information in some.Prognosis# The prognosis of Henoch-Schonlein purpura is excellent.# Mortality is exceedingly rare, and 1-5% of children progress to end-stage renal disease.# Most patients recover completely, and some do not require therapy.Rx:# Rx is similar for adults and children. When glucocorticoid therapy is required, prednisone, in doses of 1 mg/kg per day and tapered according to clinical response; however, it has not proven beneficial in the Rx of skin or renal disease and does not appear to shorten the duration of active disease or lessen the chance of recurrence.# Patients with rapidly progressive glomerulonephritis have been anecdotally reported to benefit from intensive plasma exchange combined with cytotoxic drugs.
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Examples of in vitro studies include: the isolation, growth and identification of cells derived from multicellular organisms (in cell or tissue culture); subcellular components (e.g. mitochondria or ribosomes); cellular or subcellular extracts (e.g. wheat germ or reticulocyte extracts); purified molecules (such as proteins, DNA, or RNA); and the commercial production of antibiotics and other pharmaceutical products. Viruses, which only replicate in living cells, are studied in the laboratory in cell or
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The number of viral plaques are counted and can be used to calculate the total number of infectious viral particles in the library. Most viral vectors also carry a marker that allows clones containing an insert to be distinguished from those that do not have an insert. This allows researchers to also determine the percentage of infectious viral particles actually carrying a fragment of the library.A similar method can be used to titer genomic libraries made with non-viral vectors, such as plasmids and
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After a genomic library is constructed with a viral vector, such as lambda phage, the titer of the library can be determined. Calculating the titer allows researchers to approximate how many infectious viral particles were successfully created in the library. To do this, dilutions of the library are used to transform cultures of E. coli of known concentrations. The cultures are then plated on agar plates and incubated overnight.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Virus quantification is done by -
A. Egg inoculation
B. Hemadsorption
C. Plaque assay
D. Electron microscopy
Answer:
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C. Plaque assay
Justification: Quantitative assays measure the actual number of infectious paicles in the inoculum. Two methods available-plaque assay and pock assay. REF:ANATHANARAYAN AND PANIKER'S TEXTBOOK OF MICROBIOLOGY 8TH EDITION PAGE NO:437
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Some environmental factors, including toxins, are also under active investigation, as they may play a role in the disease's cause.
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of complications. This disease as well as the primary one requires immediate treatment (for example, type 2 diabetes).
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and disruption of muscle and blood cells due to lack of ATP.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
True about caisson's disease -
A. O2 release from tissues
B. CO2 release from tissues
C. N, release from tissues
D. H, release from tissues
Answer:
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C. N, release from tissues
Justification: Ans. is 'c' i.e., N, release from tissues Caisson's diseaseo Caisson's disease, (also called decompression sicknessQ, Bendst Diver's Paralysis, Dysbarism) is a particular form of gas embolism, which occurs when individuals are exposed to sudden lowering of atmospheric pressure like - rapid ascend of scuba and deep sea divers0, individuals in unpressurized aircraft in rapid ascent (e.g., in pilotsQ), underwater construction workers and extra-vehicular activity from spacecraft.o When air is breathed at high pressure (e.g., during a deep sea dive), increased amounts of gas (particularly nitrogen) become dissolved in the blood and tissues. If the diver then ascends (depressurizes) too rapidly, the nitrogen expands in the tissues and bubbles out of solution in the blood to form gas emboli.o Since bubbles can form in or migrate to any part of the body, it can produce many symptoms, and its effects may vary from joint pain, ischemia in tissues to paralysis and death,o The rapid formation of gas bubbles within skeletal muscles and supporting tissues in and about joints is responsible for the painful condition called the bends. In the lungs, edema hemorrhages, and focal atelectasis or emphysema may appear, leading to respiratory distress, the so-called chokes,o Treatment of gas embolism requires placing the individual in a compression chamber where the barometric pressure may be raised, thus forcing the gas bubbles back into solution. Subsequent slow decompression theoretically permits gradual resorption and exhalation of the gases so that obstructive bubbles do not re-form.
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Extraskeletal calcification, e.g. calciphylaxis Brain, e.g. primary familial brain calcification (Fahr's syndrome) Choroid plexus usually in the lateral ventricles Tumor calcification Arthritic bone spurs Kidney stones Gall stones Heterotopic bone Tonsil stones Pulp stone
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to make a differential diagnosis of possible causative conditions. In this case, an epidemiology-based differential diagnostic procedure is used, and its first step is to find candidate conditions that can explain the finding. Hypercalcemia (usually defined as a calcium level above the reference range) is mostly caused by either primary hyperparathyroidism or malignancy, and therefore, it is reasonable to include these in the differential diagnosis.
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by the specific condition. This can come in form of Medical Nutrition.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Intranasal calcitonin is given in ?
A. Pagets's disease
B. MEN syndrome
C. Hypercalcemia
D. Post menopausal osteoporosis
Answer:
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D. Post menopausal osteoporosis
Justification: Ans. is 'd' i.e., Post menopausal osteoporosis o Salmon calcitonin is approved for clinical use. The latter product also is available as a nasal spray, introduced for once-daily treatment of postmenopausal osteoporosis. o For Paget c disease, calcitonin generally is administered by subcutaneous injection because intranasal delivery is relatively ineffective owing to limited bioavailability.
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At the same time a fly that travels at a steady 15 mph starts from the front wheel of the southbound bicycle and flies to the front wheel of the northbound one, then turns around and flies to the front wheel of the southbound one again, and continues in this manner till he is crushed between the two front wheels. Question: what total distance did the fly cover? The slow way to find the answer is to calculate what distance the fly covers on the first, southbound, leg of the trip, then on the second,
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45, 50, 75 is: ( 4 × 36 × 45 × 50 × 75 ) 1 5 = 24 300 000 5 = 30. {\displaystyle (4\times 36\times 45\times 50\times 75)^{\frac {1}{5}}={\sqrt{24\;300\;000}}=30.}
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rate of 50% over two periods gives a holding period return of 125%: ( 1 + 50 % ) 2 − 1 = 2.25 − 1 = 1.25 = 125 % {\displaystyle (1+50\%)^{2}-1=2.25-1=1.25=125\%}
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A student traveled 25 percent of the distance of the trip alone, continued another 20 miles with a friend, and then finished the last half of the trip alone. How many miles long was the trip?
A)60
B)80
C)100
D)120
E)150
Answer:
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B)80
Justification: Let x be the total length of the trip.
0.25x + 20 miles + 0.5x = x
20 miles = 0.25x
x = 80 miles
The answer is B.
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If 5% of female students are computer science majors, what percentage of computer science majors are female? We are asked to compute the ratio of female computer science majors to all computer science majors. We know that 60% of all students are female, and among these 5% are computer science majors, so we conclude that 60/100 × 5/100 = 3/100 or 3% of all students are female computer science majors.
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Whenever communicating about a percentage, it is important to specify what it is relative to (i.e., what is the total that corresponds to 100%). The following problem illustrates this point. In a certain college 60% of all students are female, and 10% of all students are computer science majors.
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test was 50% with a false positive rate of 3%. Given a patient with a positive test result, what is the probability that the patient has cancer?
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
In a sample of National college students, 30 percent are third year students and 60 percent are not second-year students. What fraction of those students who are not third-year students are second-year students?
A) 3/4
B) 2/3
C) 4/7
D) 1/2
E) 3/7
Answer:
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C) 4/7
Justification: = 40/70. = 4/7
my answer is C too
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Examples include: Hippocratic facies – eyes are sunken, temples collapsed, nose is pinched with crusts on the lips, and the forehead is clammy Moon face (also known as "Cushingoid facies") – Cushing's syndrome Elfin facies – Williams syndrome Potter facies – oligohydramnios Mask like facies – parkinsonism Leonine facies – lepromatous leprosy or craniometaphyseal dysplasia Mitral facies – mitral stenosis Amiodarone facies (deep blue discoloration around malar area and nose) Acromegalic facies – acromegaly
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Bones: Joint anomalies including abnormal position and function, altered palmar crease patterns, small distal phalanges, and small fifth fingernails. Kidneys: Horseshoe, aplastic, dysplastic, or hypoplastic kidneys. Eyes: Strabismus, optic nerve hypoplasia (which may cause light sensitivity, decreased visual acuity, or involuntary eye movements). Occasional problems: ptosis of the eyelid, microphthalmia, cleft lip with or without a cleft palate, webbed neck, short neck, radioulnar synostosis, spina bifida,
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In cephalopods and insects the optic tracts do not cross the body midline, so each side of the brain processes the ipsilateral eye.
Other neurodegenerative disorders include Rett syndrome, Prader–Willi syndrome, Angelman syndrome, and Williams-Beuren syndrome.
In cephalochordata, metamorphosis is iodothyronine-induced and it could be an ancestral feature of all chordates.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
All of the following are the classical presentation of Cranioverebral junction anomalies except
A. Pyramidal signs
B. Low hairline
C. Sho neck
D. Pupillary asymmetry
Answer:
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D. Pupillary asymmetry
Justification: . Pupillary asymmetry
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planets—Mercury, Venus, Earth, Mars, Jupiter, and Saturn.
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Centaurs appear to be grouped into two classes: very red – for example 5145 Pholus blue (or blue-grey, according to some authors) – for example 2060 Chiron or 2020 MK4There are numerous theories to explain this colour difference, but they can be broadly divided into two categories: The colour difference results from a difference in the origin and/or composition of the centaur (see origin below) The colour difference reflects a different level of space-weathering from radiation and/or cometary activity.As
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Observatory. This two-dimensional (temperature and luminosity) classification scheme is based on spectral lines sensitive to stellar temperature and surface gravity, which are related to luminosity (the Harvard classification is based on surface temperature).
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
What celestial bodies are classified by color and temperature, ranging from blue to red and hottest to coolest?
A. galaxies
B. stars
C. astroids
D. planets
Answer:
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B. stars
Justification: Stars are classified by color and temperature. The most common system uses the letters O (blue), B (blue-white), A (white), F (yellow-white), G (yellow), K (orange), and M (red), from hottest to coolest.
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is a cubic equation in y. Solve for y using any method for solving such equations (e.g. conversion to a reduced cubic and application of Cardano's formula). Any of the three possible roots will do.
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Let f ( x ) = x 4 {\displaystyle f(x)=x^{4}} . Then f ′ ( x ) = 4 x 3 {\displaystyle f'(x)=4x^{3}} and f ″ ( x ) = 12 x 2 {\displaystyle f''(x)=12x^{2}} . Therefore, the third derivative of f is, in this case, f ‴ ( x ) = 24 x {\displaystyle f'''(x)=24x} or, using Leibniz notation, d 3 d x 3 = 24 x .
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The solutions of the cubic are: y = a 6 + w − p 3 w {\displaystyle \ y={\frac {a}{6}}+w-{\frac {p}{3w}}\ } w = − q 2 + q 2 4 + p 3 27 3 {\displaystyle \ w={\sqrt{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}\ }}\ }}} using any one of the three possible cube roots. A wise strategy is to choose the sign of the square-root that makes the absolute value of w as large as possible. p = − a 2 12 − c , {\displaystyle \ p=-{\frac {a^{2}}{12}}-c\ ,} q = − a 3 108 + a c 3 − b 2 8 . {\displaystyle \
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Find the value of y from given equation: (12)^3 x 6^4 ÷ 432 = y?
A)2345
B)2790
C)3490
D)7389
E)5184
Answer:
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E)5184
Justification: Given Exp. = (12)3 x 64 = (12)3 x 64 = (12)2 x 62 = (72)2 = 5184
432 12 x 62
E
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The residual cake is usually disposed. However, it has been recently shown that waxes can be submitted to solid state fermentation, using Starmerella bombicola in combination with sources of sugar, such as beetroot molasses, and it can be exploited to produce surfactants. == References ==
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butyrate, n-amyl butyrate and hexyl acetate; mixtures of various kinds of ESTISOL; and dimethyl siloxane oils.
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Oleoresins are naturally-occurring mixtures of an oil and a resin; they can be extracted from various plants. Other resinous products in their natural condition are a mix with gum or mucilaginous substances and known as gum resins. Several natural resins are used as ingredients in perfumes, e.g., balsams of Peru and tolu, elemi, styrax, and certain turpentines.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
The main ingredient of inlay wax is:
A. Candelilla wax
B. Paraffin wax
C. Carnauba wax
D. Gum dammar
Answer:
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B. Paraffin wax
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PL/I provides several 'storage classes' to indicate how the lifetime of variables' storage is to be managed – STATIC, AUTOMATIC, CONTROLLED, and BASED. The simplest to implement is STATIC, which indicates that memory is allocated and initialized at load-time, as is done in COBOL "working-storage" and early Fortran. This is the default for EXTERNAL variables. PL/I's default storage class for INTERNAL variables is AUTOMATIC, similar to that of other block-structured languages influenced by ALGOL, like the
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of identifying peaks and storing them as a single sample, as seen in standard LC, these droplet libraries allow for the specific concentration of the analyte to be retained along with its identity.
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Megakaryocyte and platelet production is regulated by thrombopoietin, a hormone produced in the kidneys and liver. Each megakaryocyte produces between 1,000 and 3,000 platelets during its lifetime. An average of 1011 platelets are produced daily in a healthy adult. Reserve platelets are stored in the spleen and are released when needed by splenic contraction induced by the sympathetic nervous system.The average life span of circulating platelets is 8 to 9 days. Life span of individual platelets is
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Platlets can be stored at
A. 20-24degC for 5 days
B. 20-24degC for 8 days
C. 4-8degC for 5 days
D. 4-8degC for 8 days
Answer:
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A. 20-24degC for 5 days
Justification: Ans. is (a) i.e. 20-24degC for 5 days
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Using this technique he found that 1166 out of the first three million primes are divisors of Sylvester numbers, and that none of these primes has a square that divides a Sylvester number. The set of primes which can occur as factors of Sylvester numbers is of density zero in the set of all primes: indeed, the number of such primes less than x is O ( π ( x ) / log log log x ) {\displaystyle O(\pi (x)/\log \log \log x)} .The following table shows known factorizations of these numbers (except the first
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= 6 ≡ 2 (mod 4), if n is composite then (n − 1)! is congruent to 0 (mod n). The proof is divided into two cases: First, if n can be factored as the product of two unequal numbers, n = ab, where 2 ≤ a < b ≤ n − 2, then both a and b will appear in the product 1 × 2 × ... × (n − 1) = (n − 1)!
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It does not matter whether B is prime (in fact, it is not). Finally, for each prime factor p of A, use trial and error to find an ap that satisfies (6) and (7). For p = 2 {\displaystyle p=2} , try a 2 = 2 {\displaystyle a_{2}=2} .
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
If (6)(x^2) has four different prime factors, at most how many different prime factors does x have?
A)1
B)2
C)3
D)4
E)5
Answer:
|
D)4
Justification: x can have at most 4 prime factors, namely the prime factors 2 and 3, plus 2 others.
If x had more than this number of prime factors, then (6)(x^2) would have more than 4 prime factors.
The answer is D.
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in 125 μg or 250 μg doses.Digoxin elimination is mainly by renal excretion and involves P-glycoprotein, which leads to significant clinical interactions with P-glycoprotein inhibitor drugs. Examples commonly used in patients with heart problems include spironolactone, verapamil and amiodarone.
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and include diclofenac, bosentan, entacapone, tacrolimus, cimetidine, and flucloxacillin.
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cytarabine, gemcitabine, decitabine, azacitidine, fludarabine, nelarabine, cladribine, clofarabine, and pentostatin. The thiopurines include thioguanine and mercaptopurine.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Drug contraindicated in HOCM is?
A. Verapamil
B. Propranolol
C. Digoxin
D. None of the above
Answer:
|
C. Digoxin
Justification: ANSWER: (C) DigoxinREF: Harrisons Principles of Internal Medicine 17th edition chapter 231Repeat in December 2010"Digoxin is contraindicated in hypertrophic cardiomyopathy and in patients with AV conduction blocks"
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dragon's body, whose inherent brain was destroyed in a recent battle.
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Mammalian prey typically consists of rodents and leporids; the Komodo dragon can kill prey as large as water buffalo. Dragons are prolific scavengers, and a single decaying carcass can attract several from 2 km (1.2 mi) away. A 50 kg (110 lb) dragon is capable of consuming a 31 kg (68 lb) carcass in 17 minutes.Around 2 percent of lizard species, including many iguanids, are herbivores.
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this movie was screened on regular activity days. Addicted to Life (since 2007, Hebrew only) – how does the body deal with using drugs.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
"Chasing the dragon" is a slang used with
A. Cocaine
B. Heroin
C. LSD
D. Ketamine
Answer:
|
B. Heroin
Justification: "Chasing the dragon" refers to inhalation of vapours of Heroin (opioid) placed on a heated aluminium foil.
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This means that the four quarters of the year are treated as having equal length. The average capital of the cash account is: average capital = start value - time weight × outflow amount = 10,000 - 1/4 × $8,000 = 10,000 - $2,000 = $8,000The average capital of the shares over the last quarter requires no calculation, because there are no flows after the beginning of the last quarter. It is the $8,000 invested in the shares.
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weighted flows = 0so average capital = start value = $10,000 so the return is: gain or loss/average capital = 900/10,000 = 9 %This 9% portfolio return breaks down between 8 percent contribution from the $800 earned on the shares and 1 percent contribution from the $100 interest earned on the cash account, but how more generally can we calculate contributions? The first step is to calculate the average capital in each of the cash account and the shares over the full year period. These should sum to the
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, XN, and add up their values. By Wald's equation, the resulting value on average is E E = 1 + 2 + 3 + 4 + 5 + 6 6 ⋅ 1 + 2 + 3 + 4 + 5 + 6 6 = 441 36 = 49 4 = 12.25 . {\displaystyle \operatorname {E} \operatorname {E} ={\frac {1+2+3+4+5+6}{6}}\cdot {\frac {1+2+3+4+5+6}{6}}={\frac {441}{36}}={\frac {49}{4}}=12.25\,.}
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
If each of 4 subsidiaries of Corporation R has been granted a line of credit of $464,800 and each of the other 3 subsidiaries of Corporation R has been granted a line of credit of $112,000, what is the average (arithmetic mean) line of credit granted to a subsidiary of Corporation R?
A) $1,568,000
B) $448,000
C) $406,000
D) $313,600
E) $116,000
Answer:
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D) $313,600
Justification: My take is D.
(464800*4 + 112000*3)/7 = 313600
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would be present after one hour? The question implies a = 1, b = 2 and τ = 10 min.
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There is a complex project that is composed of several sub-tasks. We would like to assign tasks to workers, such that the project finishes in the shortest possible time. As an example, suppose the "project" is to feed the goats. There are three goats to feed, one child can only feed one goat at a time, and there are two children that can feed them: Shmuel feeds each goat in 10 minutes and Shifra feeds each goat in 12 minutes. Several schedules are possible: If we let Shmuel feed all goats, then the makespan
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was based on how quickly (in seconds) they would respond to each of the tasks.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Sandy can do a job in 12 days and Molly can do the same job in 24 days. If they work together, in how many days will they complete the job?
A)6
B)7
C)8
D)9
E)10
Answer:
|
C)8
Justification: Sandy can do 1/12 of the job each day.
Molly can do 1/24 of the job each day.
The combined rate is 1/12 + 1/24 = 1/8 of the job each day.
The job will take 8 days.
The answer is C.
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assets in proportion to his/her investment in the partnership.
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{\displaystyle n_{B}=0.75\ {\text{mol}}\times 1+1\ {\text{mol}}=1.75\ {\text{mol}}} n C = 0.75 mol × 3 + 0 mol = 2.25 mol {\displaystyle n_{C}=0.75\ {\text{mol}}\times 3+0\ {\text{mol}}=2.25\ {\text{mol}}} == References ==
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The end value of the portfolio is $2,100 in cash, plus shares worth $8,800, which is in total $10,900. There has been a 9 percent increase in value since the beginning of the year. There are no external flows in or out of the portfolio over the year.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A, B and C enter into a partnership. A invests 3 times as much as B invests and 2/3 of what C invests. At the end of the year, the profit earned is Rs. 11000. What is the share of C?
A)Rs. 2250
B)Rs. 3000
C)Rs. 6750
D)Rs. 5625
E)None of these
Answer:
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B)Rs. 3000
Justification: Explanation:
Let the investment of C be Rs. x.
The inverstment of B = Rs.(2x/3)
The inverstment of A = Rs. (3 × (2/3)x) = Rs. (2x)
Ratio of capitals of A, B and C = 2x : 2x/3 : x = 6 : 2 : 3
C's share = Rs. [(3/11) × 11000] = Rs. 3000
Answer: Option B
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percentage, convert both percentages to fractions of 100, or to decimals, and multiply them.
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{\displaystyle {\frac {80}{100}}=80\%} and 20 100 = 20 % {\displaystyle {\frac {20}{100}}=20\%} respectively.This gives the contributions to the overall return, which are: 50 % × 80 % = 40 % {\displaystyle 50\%\times 80\%=40\%} and 25 % × 20 % = 5 % {\displaystyle 25\%\times 20\%=5\%} respectively.The sum of these contributions is the return: 40 % + 5 % = 45 % {\displaystyle 40\%+5\%=45\%} This is equivalent to the simple return, adjusting the end value for outflows: Start value = 1 , 000 dollars
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The percent value is computed by multiplying the numeric value of the ratio by 100. For example, to find 50 apples as a percentage of 1250 apples, one first computes the ratio 50/1250 = 0.04, and then multiplies by 100 to obtain 4%. The percent value can also be found by multiplying first instead of later, so in this example, the 50 would be multiplied by 100 to give 5,000, and this result would be divided by 1250 to give 4%. To calculate a percentage of a percentage, convert both percentages to fractions
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
If 20 gallons of grape juice are added to 40 gallons of a mixture, which contains 10 percent grape juice then what percent of the resulting mixture is grape juice?
A)14%
B)25%
C)28%
D)34%
E)40%
Answer:
|
E)40%
Justification: OFFICIAL SOLUTION:
If we start with 40 gallons of a mixture that is 10% grape juice, then we have:
40 × 0.10 = 4 gallons of grape juice.
40 × 0.90 = 36 gallons of other components.
If we add 20 gallons of grape juice, we will end up with 24 gallons of grape juice and 36 gallons of other components, and we will have a total of 60 gallons of the mixture.
So 24/60 of the new mixture is grape juice. Now we convert this to a percent:Percent Grape Juice = 40/100 =40%.
The correct answer is choice (E)
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size ≥7 cm in diameter; involvement of ≥3 nodes in 3 distinct areas, each of which is ≥3 cm in diameter; organ compression; presence of ascites or pleural effusion (i.e. build-up of fluid in the abdominal or pleural cavities); poor performance status due to the disease; elevated levels of serum lactose dehydrogenase or beta-2 microglobulin; presence of localized bone lesions; kidney involvement; reduced levels of circulating blood platelets or any of the various types of white blood cells; onset of
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In medicine, diagnoses can be made with the assistance of microscopic observation of patient biopsies, such as cancer cells. Pathology and cytology reports include a microscopic description, which consists of analyses performed using microscopes, histochemical stains or flow cytometry. These methods can determine the structure of the diseased tissue and the severity of the disease, and early detection is possible through identification of microscopic indications of illness.
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and what histological type it is (small cell or non-small cell).
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
In giant cell arteritis :
A. Histological diagnosis is based on fragmentation of the internal intimal
B. C-reactive protein is always raised
C. Giant cell is needed for diagnosis
D. Anaemia is a feature
Answer:
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C. Giant cell is needed for diagnosis
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endoscopy, and a natural history that may include stricturing. Eosinophils are normally present in other parts of a healthy gastrointestinal tract, these white blood cells are not normally found in the esophagus of a healthy individual.
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Endoscopically, ridges, furrows, or rings may be seen in the esophageal wall. Sometimes, multiple rings may occur in the esophagus, leading to the term "corrugated esophagus" or "feline esophagus" due to similarity of the rings to the cat esophagus. Presence of white exudates in esophagus is also suggestive of the diagnosis. On biopsy taken at the time of endoscopy, numerous eosinophils can be seen in the superficial epithelium.
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pancreatitis, ovarian hyperstimulation, exudate pleural effusion, congestive heart failure, metastatic tumors to the peritoneal surfaces, collagen-vascular disease, and cirrhosis of the liver.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Which of the following conditions leads to the following endoscopic appearance?
A. Intestinal lymphangiectasia
B. Gastric antral vascular ectasia
C. Amoebic colitis
D. Necrotising Enterocolitis
Answer:
|
B. Gastric antral vascular ectasia
Justification: Intestinal lymphangiectasis Biopsy: vacuolated cells because fat will not be absorbed Necrotizing enterocolitis Diagnosed by Radiological finding (Pneumatosis intestinalis) Amoebic colitis Biopsy -Flask- shaped ulcer(Bleeding) Gastric antral vascular ectasia: Vascular information-Watermelon stomach
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The density of the air the aircraft moves through affects its performance as well as winds, weight, and power settings. The basic formula for DR is Distance = Speed x Time. An aircraft flying at 250 knots airspeed for 2 hours has flown 500 nautical miles through the air.
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equates to a speed of about 806 kilometres per hour (501 mph).
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for time T, and the wind is assumed to move in a constant direction with speed w. The solution of the problem is that the airplane should travel in an ellipse whose major axis is perpendicular to w, with eccentricity w/v.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
An aeroplane covers a certain distance of 420 Kmph in 6 hours. to cover the same distance in 4 2/3 hours, it Must travel at a speed of ?
A)450 Km
B)480 Km
C)500 Km
D)540 Km
E)590 Km
Answer:
|
D)540 Km
Justification: Speed of aeroplane = 420 Kmph
Distance travelled in 6 hours
= 420 * 6 = 2520 Km
Speed of aeroplane to acver 2520 Km in 14/3
= 2520*3/14 = 540 Km
Answer: Option 'D'
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inclusion (AI). The presence of both inclusions can be diagnostic for this virus.
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and parvovirus, enterovirus, adenovirus, and alphavirus infections (e.g., chikungunya, Mayaro, Ross River, Barmah Forest, O'nyong'nyong, and Sindbis viruses).
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by the presence of two glycoproteins (E1 and E2) embedded as trimeric dimers in a host-derived lipid envelope.Because RRV is transmitted by mosquitos, it is considered an arbovirus, a non-taxonomic term for viruses borne by arthropod vectors.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Slapped cheek sign is seen in -
A. Parvovirus B19
B. JC virus
C. Rota virus
D. Mumps
Answer:
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A. Parvovirus B19
Justification: parvo virus B19 causes erythema infectiosum it sta with erythema of cheeks called slapped cheek appearance REF:ANATHANARAYANAN MICROBIOLOGY NINTH EDITION PAGE.554
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) are 7, 15, 23, 28, 31, 39, 47, 55, 60, 63, 71 ... (sequence A004215 in the OEIS).
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29, 31, 37, or 47. The former are ≡ 1 or ≡ 3 (mod 8), and the latter are ≡ 5, 7 (mod 8).
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268, 15, 272, 66, 34, 28, 138, 112, 116, 179, 5, 378, 388, 18, 204, 418, 6, 219, 32, 48, 66, 239, 81, 498, ... (sequence A128858 in the OEIS)
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Which of the following numbers completes
the sequence 8, 14, 21, 29, 38___?
A)35
B)36
C)37
D)48
E)69
Answer:
|
D)48
Justification: The numbers increase at a somewhat steady rate, so you have to figure out how much
you have to add to each number to produce the next in the sequence: 8 + 6 = 14;
14 + 7 = 21; 21 + 8 = 29, 29 + 9 = 38 and so on. The rule for the sequence is to add successively larger
numbers to each number; therefore, the next number is 38 + 10 = 48
correct answer D)48
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Bones: Joint anomalies including abnormal position and function, altered palmar crease patterns, small distal phalanges, and small fifth fingernails. Kidneys: Horseshoe, aplastic, dysplastic, or hypoplastic kidneys. Eyes: Strabismus, optic nerve hypoplasia (which may cause light sensitivity, decreased visual acuity, or involuntary eye movements). Occasional problems: ptosis of the eyelid, microphthalmia, cleft lip with or without a cleft palate, webbed neck, short neck, radioulnar synostosis, spina bifida,
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onchodermatitis – hyperpigmented papules and plaques, with edema, lymphadenopathy, pruritus and common secondary bacterial infections Skin atrophy – loss of elasticity, the skin resembles tissue paper, 'lizard skin' appearance Depigmentation – 'leopard skin' appearance, usually on anterior lower leg Glaucoma effect – eyes malfunction, begin to see shadows or nothingOcular involvement provides the common name associated with onchocerciasis, river blindness, and may involve any part of the eye from
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of the limbs, the face, and/or the bulbar muscles; abnormalities of the pupils; and absent reflexes.Like some other autoimmune diseases, the condition usually follows a minor infection, such as a respiratory tract infection or gastroenteritis.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Posner-schlossman syndrome is -
A. Ipsilateral optic atrophy with contralateral papilloedema
B. Unilateral glaucomatous changes with mild anterior uveitis
C. Granulomatous uveitis with iris heterochromia
D. None of the above
Answer:
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B. Unilateral glaucomatous changes with mild anterior uveitis
Justification: It is also known as glaucomatocyclitic crisis. It shows mild uveitis with severe rise of IOP. Disease typically effects young adults, 40% of whom are positivs for HLA-BW54 REF:khurana page no. 167
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In previous stages acetyl-CoA is transferred from the mitochondria to the cytoplasm where fatty acid synthase resides. The acetyl-CoA is transported as a citrate, which has been previously formed in the mitochondrial matrix from acetyl-coA and oxaloacetate. This reaction usually initiates the citric acid cycle, but when there is no need of energy it is transported to the cytoplasm where it is broken down to cytoplasmic acetyl-CoA and oxaloacetate. Another part of the cycle requires NADPH for the synthesis
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the catalysts involved in the citric acid cycle, a central pathway in cellular metabolism, and it is located within the mitochondrial matrix of a cell.
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The majority of cellular ATP is generated by this process. Although the citric acid cycle itself does not involve molecular oxygen, it is an obligately aerobic process because O2 is used to recycle the NADH and FADH2. In the absence of oxygen, the citric acid cycle ceases.The generation of ATP by the mitochondrion from cytosolic NADH relies on the malate-aspartate shuttle (and to a lesser extent, the glycerol-phosphate shuttle) because the inner mitochondrial membrane is impermeable to NADH and NAD+.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
All the following cycles operate in mitochondria except
A. Ketogenesis
B. Beta oxidation
C. TCA cycle
D. EMP
Answer:
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D. EMP
Justification: Ans. is 'd' i.e., EMPRepeat from previous sessions. See explanation-2 of sessions-6.This was the session in which many questions of biochemistry were repeated from previous sessions.
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to most microvesicles. For example, nearly all contain the cytoplasmic proteins tubulin, actin and actin-binding proteins, as well as many proteins involved in signal transduction, cell structure and motility, and transcription.
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including relevant publications, chemical structures or protein structure via the Protein Data Bank.
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There are two types, rough ER (containing ribosomes) and smooth ER (lacking ribosomes). The Golgi apparatus consists of multiple membranous sacs, responsible for manufacturing and shipping out materials such as proteins. Lysosomes are structures that use enzymes to break down substances through phagocytosis, a process that comprises endocytosis and exocytosis. In the mitochondria, metabolic processes such as cellular respiration occur. The cytoskeleton is made of fibers that support the structure of the
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Proteins are soed by:
A. Golgi Bodies
B. Mitochondria
C. Ribosomes
D. Nuclear Membrane
Answer:
|
A. Golgi Bodies
Justification: The Golgi apparatus is involved in the glycosylation and soing of proteins. Ref: Biochemistry By Mary K. Campbell, Shawn O. Farrell, 7th Edition, Page 21; Harper's Textbook of Biochemistry, 27th Edition, Pages 506, 518
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that auditory hallucinations are connected to a malfunction in the left hemisphere of the brain.
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of depression.It may also play a role in temporal lobe seizures.
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may include respiratory depression (decreased breathing), sleepiness, adrenal insufficiency, QT prolongation, low blood pressure, allergic reactions, constipation, and opioid addiction. Among those with a history of seizures, a risk exists of further seizures.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Gustatory hallucinations are most commonly associated with -
A. Temporal lobe epilepsy
B. Grand mal epilepsy
C. Anxiety disorders
D. Tobacco dependence
Answer:
|
A. Temporal lobe epilepsy
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In 2000, the World Health Organization estimated that 25% of the units of blood transfused in Africa were not tested for HIV, and that 10% of HIV infections in Africa were transmitted via blood.Poor economic conditions (leading to the use of dirty needles in healthcare clinics) and lack of sex education contribute to high rates of infection. In some African countries, 25% or more of the working adult population is HIV-positive. Poor economic conditions caused by slow onset-emergencies, such as drought, or
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are more susceptible to infection (through still unknown routes of sexual transmission) whereas the virus is transmitted through non-sexual routes in developing countries.
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more restricted sexual behavior, which may reduce pathogen transmission.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
True about HIV:
A. Not transmitted through semen
B. More chances of transmission during LSCS than normal labour
C. More infectious than hepatitis B
D. Male to female transmitted > female to male
Answer:
|
D. Male to female transmitted > female to male
Justification: Ans. is. 'd' i.e., Male to female transmission > female to male transmission(Ref: Harrison, 18th/e, p. 1510, 1512, 1515; 17th/e, p. 1143)* Male to female transmission is approximately eight times more efficient than female to male transmission.* Seminal fluid (Semen) and vaginal fluid can transmit the infection.* Caesarean section (LSCS) results in decreased transmission from mother to infant.* After percutaneous injury Hepatitis B virus is more infectious than HIV.HBV - 6 to 30% chance of transmission.HCV - 1.8%.HIV - 0.3%.
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In production, research, retail, and accounting, a cost is the value of money that has been used up to produce something or deliver a service, and hence is not available for use anymore. In business, the cost may be one of acquisition, in which case the amount of money expended to acquire it is counted as cost. In this case, money is the input that is gone in order to acquire the thing. This acquisition cost may be the sum of the cost of production as incurred by the original producer, and further costs of
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The Simpsons episode "Dark Knight Court" has Mr. Burns asking Comic Book Guy how much he wants for his entire comic book inventory. He says "the speed of light expressed as dollars" and Mr. Burns tells Smithers to "just give him Faraday's Constant". The check is written for $96,485.34.
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His 2 to 1 bet pays 0.5 if he wins and costs 0.25 if he loses. At the new 5 to 1 payout, he could get a bet that pays 0.625 if he wins and costs 0.125 if he loses, this is 0.125 better than his original bet in both states. Therefore his original bet now has a value of -0.125.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Jim is able to sell a hand-carved statue for $670 which was a 35% profit over his cost. How much did the statue originally cost him?
A)$496.30
B)$512.40
C)$555.40
D)$574.90
E)$588.20
Answer:
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A)$496.30
Justification: 670 is the final price that is the selling price. now suppose, x is the cost price.
therefore 670 = 1.35 * x
now while doing x = 670/1.35 calculation. I saw that the answer choices begin with 5.
So i did 135 * 5 = 675. As 670 is less than 675. Therefore the quotient will be less than 5.
Which is only in the answer A.
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Breastmilk may be stored for various amounts of time depending on storage temperature and conditions. The content and quality of expressed milk changes over time as it is stored, particularly when frozen. For example, there is a decrease in the ability of breastmilk to kill bacteria when it is stored in the refrigerator for more than 48 hours. Additionally, the quantity of fat, protein, and calories in breastmilk decreases when the milk is frozen for more than 3 months. While several components of
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more than 3 months. While several components of breastmilk change over time, inflammatory factors (cytokines) and maternal antibodies, and growth factors are thought to be stable for at least 6 months when the breastmilk is frozen. Storage guidelines, according to the CDC, La Leche League International and the Academy of Breastfeeding Medicine, are noted in the table below.
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and a need to suck on a nipple, and breastfed babies nurse for both nutrition and for comfort. Breast milk provides all necessary nutrients for the first six months of life, and then remains an important source of nutrition, alongside solid foods, until at least one or two years of age.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
For how many hours can breast milk be stored in the refrigerator?
A. 48 hours
B. 24 hours
C. 12 hours
D. 6 hours
Answer:
|
B. 24 hours
Justification: Breast milk stored in the refrigerator should be used within 24 hours. Refrigerated or frozen milk should be thawed rapidly by holding under running tepid water and used completely within 24 hours after thawing. Milk should be microwaved. Expressed breast milk Stored for Room Temperature 6 hours In refrigerator 24 hours Frozen milk 3 months A difference of opinion among authors is noted regarding the refrigeration of expressed breast milk. Nelson&;s (21st edition, pgno: 324) mentions that breast milk can be refrigerated and used within 48 hours and frozen for around 6 months . However AIIMS in colloboration with WHO recommends that breast milk can be refrigerated upto 24 hours and frozen at -20oC for about 3 months . Ref: Nelson Textbook of pediatrics 20th edition Pgno: 288 Aiims Protocols in Neonatology
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Sodium thiopental is an ultra-short-acting barbiturate and has been used commonly in the induction phase of general anesthesia. Its use has been largely replaced with that of propofol, but may retain some popularity as an induction agent for rapid-sequence induction and intubation, such as in obstetrics. Following intravenous injection, the drug rapidly reaches the brain and causes unconsciousness within 30–45 seconds. At one minute, the drug attains a peak concentration of about 60% of the total dose in
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Propranolol may be contraindicated in people with: Reversible airway diseases, particularly asthma or chronic obstructive pulmonary disease (COPD) Slow heart rate (bradycardia) (<60 beats/minute) Sick sinus syndrome Atrioventricular block (second- or third-degree) Shock Severe low blood pressure
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Their early use is contraindicated if there are signs of congestive heart failure (e.g., Killip class II or above) or hypotension, along with other contraindications to beta blockers (slow heart rate, atrioventricular block); in the absence of contraindications beta blocker therapy should begin in the first 24 hours. It may be prudent to prefer oral rather than intravenous forms.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Regarding propofol which One of the following is false
A. it is used as an intravenous induction agent
B. It causes severe vomiting
C. It is painful on injecting intravenously
D. It has no muscle relaxant propey
Answer:
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B. It causes severe vomiting
Justification: Propofol also possesses significant antiemetic activity with small (sub hypnotic) doses (i.e., 10 mg in adults). The median concentration of propofol with an antiemetic effect was 343 ng/mL, which also causes a mild sedative effect. This concentration can be achieved by an initial dose of propofol infusion of 10 to 20 mg followed by 10 mg/kg/minute. Propofol used as a maintenance anesthetic during breast surgical procedures was more effective than 4 mg of ondansetron given as prophylaxis in preventing postoperative nausea and vomiting (PONV). Propofol as an infusion of 1 mg/kg/hour (17 mg/kg/minute) also has provided excellent antiemetic action after anticancer chemotherapy. Ref: Miller's anesthesia 8th edition
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Within the first 2–3 months after birth, surgery is performed to close the cleft lip. While surgery to repair a cleft lip can be performed soon after birth, often the preferred age is at approximately 10 weeks of age, following the "rule of 10s" coined by surgeons Wilhelmmesen and Musgrave in 1969 (the child is at least 10 weeks of age; weighs at least 10 pounds, and has at least 10g hemoglobin). If the cleft is bilateral and extensive, two surgeries may be required to close the cleft, one side first, and
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Measuring the outcomes of CLP treatment has been laden with difficulty due to the complexity and longitudinal nature of cleft care, which spans birth through young adulthood. Prior attempts to study the effectiveness of specific interventions or overall treatment protocols have been hindered by a lack of data standards for outcomes assessment in cleft care.The International Consortium for Health Outcome Measurement (ICHOM) has proposed the Standard Set of Outcome Measures for Cleft Lip and Palate. The
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surgery of cleft lip/or palate treatment and aesthetics.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
In cleft lip surgery rule of 10 says
A. 10 weeks of age, having 10 dL/mg Hb and weight 10 pounds
B. 10 weeks of age, having 10 dL/mg Hb and weight 10 kgs
C. 10 months of age, having 10 dL/mg Hb and weight 10 pounds
D. 10 months of age, having 10 dL/mg Hb and weight 10 kgs
Answer:
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A. 10 weeks of age, having 10 dL/mg Hb and weight 10 pounds
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When abnormal, evidence from neuroimaging may suggest the timing of the initial damage. The CT or MRI is also capable of revealing treatable conditions, such as hydrocephalus, porencephaly, arteriovenous malformation, subdural hematomas and hygromas, and a vermian tumour (which a few studies suggest are present 5–22% of the time). Furthermore, abnormalities detected by neuroimaging may indicate a high likelihood of associated conditions, such as epilepsy and intellectual disability.
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The main features of this condition are hypertrichosis, osteochondrodysplasia, and cardiomegaly. There is also a characteristic facies. Other features include patent ductus arteriosus, congenital hypertrophy of the left ventricle, and pericardial effusions.Neurodevelopmental outcome appears normal, but obsessive traits and anxiety have been reported. It may also be associated with recurrent infections with low immunoglobulin levels and gastric bleeding, and additional possible associations include
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A relatively new approach in the diagnosis is magnetic resonance imaging (MRI), which is noninvasive. MRI uses the property of nuclear magnetic resonance to image nuclei of atoms inside the body. MRI is non-radiographic and therefore can be repeated more often in short periods of time. In addition, different studies show that the MRI is better as an imaging tool than videofluoroscopy for visualizing the anatomy of the velopharynx.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A child presents with unilateral propotosis which is compressible and increases on bending forwards. It is non-pulsatile and has no thrill or bruit. MRI shows retroorbital mass with ecogenic shadows (with enhancement). The most probable diagnosis is:
A. Neurofibromatosis
B. Orbital varix
C. Orbital A-V fstula
D. Orbital encephalocele
Answer:
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B. Orbital varix
Justification: B i.e. Orbital varix
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duration of the wind and the distance over which it blows.
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The quick way is to observe that the bicycles meet exactly one hour after their start, so that the fly had just an hour for his travels; the answer must therefore be 15 miles. When the question was put to von Neumann, he solved it in an instant, and thereby disappointed the questioner: "Oh, you must have heard the trick before!" "What trick?"
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times (such as fastest mile, 3-second gust, 1-minute and mean hourly) which designers may have to take into account. To convert wind speeds from one averaging time to another, the Durst Curve was developed which defines the relation between probable maximum wind speed averaged over t seconds, Vt, and mean wind speed over one hour V3600.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
It was a beautiful sunny morning. The air was fresh and a mild wind was blowing against my wind screen. I was driving from Bangalore to Brindavan Gardens. It took me 1 hour and 30 minutes to complete the journey.
After lunch I returned to Bangalore. I drove for 90 rhinutes. How do you explain it ?
A)75 min
B)60 min
C)90 min
D)80 min
E)88 min
Answer:
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C)90 min
Justification: C
90 min
Jansi went out for shopping. She had in her handbag approximately Rs. 15/- in one rupee notes and 20 p. coins. When she returned she had as many one rupee notes as she originally had and as many 20 p. coins as she originally had one rupee notes. She actually came back with about one-third of what she had started out with.
How much did she spend and exactly how much did she have with her when she started out ?
There is nothing to explain here. The driving time there and back is absolutely the same because 90 minutes and 1 hour and 30 minutes are one and the same thing.
This problem is meant for inattentive readers who may think that there is some difference between 90 minutes and 1 hour 30 minutes.
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Morphine has the structure shown on the right. The standard InChI for morphine is InChI=1S/C17H19NO3/c1-18-7-6-17-10-3-5-13(20)16(17)21-15-12(19)4-2-9(14(15)17)8-11(10)18/h2-5,10-11,13,16,19-20H,6-8H2,1H3/t10-,11+,13-,16-,17-/m0/s1 and the standard InChIKey for morphine is BQJCRHHNABKAKU-KBQPJGBKSA-N.
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With respect to equianalgesic dosing, when used sublingually, the potency of buprenorphine is about 40 to 70 times that of morphine. When used as a transdermal patch, the potency of buprenorphine may be 100 to 115 times that of morphine.
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The smallest known dose of TAA that has killed a person is 30 mL.An overdose produces symptoms similar to alcohol poisoning and is a medical emergency due to the sedative/depressant properties which manifest in overdose as potentially lethal respiratory depression. Sudden loss of consciousness, simultaneous respiratory and metabolic acidosis, fast heartbeat, increased blood pressure, pupil constriction, coma, respiratory depression and death may follow from an overdose. The oral LD50 in rats is 1 g/kg. The
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Fatal dose morphine is: BHU 12
A. 100 mg
B. 200 mg
C. 300 mg
D. 500 mg
Answer:
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B. 200 mg
Justification: Ans. 200 mg
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A periodontal examination is a clinical examination of the periodontium (gums). It is routinely carried out in dentistry and allied specialties. Many different techniques are used around the world. A report by World Health Organization in 1978 led to the creation of the Community Periodontal Index of Treatment Needs (CPITN) and a periodontal probe termed WHO 621 ("Trintity").
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Radiographs utilized to find dental caries and bone loss laterally or at the apex. Assessment of biting on individual teeth (which sometimes helps to localize the problem) or the separate cusps (may help to detect cracked cusp syndrome).Less commonly used tests might include trans-illumination (to detect congestion of the maxillary sinus or to highlight a crack in a tooth), dyes (to help visualize a crack), a test cavity, selective anaesthesia and laser doppler flowmetry. Establishing a diagnosis of
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By formulating evidence-based best-practice clinical guidelines that practitioners can refer to with simple chairside and patient-friendly versions, this need can be addressed. Evidence-based dentistry has been defined by the American Dental Association (ADA) as "an approach to oral healthcare that requires the judicious integration of systematic assessments of clinically relevant scientific evidence, relating to the patient's oral and medical condition and history, with the dentist's clinical expertise
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
As per American Dental Association (ADA) Type III oral
examination includes
A. Screening using tongue depressor and available illumination
B. Screening using mouth mirror and available illumination
C. Inspection, using mouth mirror and adequate illumination
D. Inspection using explore and adequate illumination
Answer:
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C. Inspection, using mouth mirror and adequate illumination
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Symptoms include persistent right upper quadrant pain, jaundice, palpable mass. Biliary dyskinesia is a disease with the abnormal release of bile from the gallbladder leading to chronic biliary colic. Diagnosis is based on several studies examining the most common cause of gallstones and looking at the ejection fraction through a HIDA scan with cholecystokinin.
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A serious complication of cholecystectomy is biliary injury, or damage to the bile ducts. Laparoscopic cholecystectomy has a higher risk of bile duct injury than the open approach, with injury to bile ducts occurring in 0.3% to 0.5% of laparoscopic cases and 0.1% to 0.2% of open cases. In laparoscopic cholecystectomy, approximately 25–30% of biliary injuries are identified during the operation; the rest become apparent in the early post-operative period.Damage to the bile ducts is very serious because it
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Jaundice is an uncommon occurrence in intrahepatic (metabolic) cholestasis, but is common in obstructive cholestasis. The majority of patients with chronic cholestasis also experience fatigue. This is likely a result of defects in the corticotrophin hormone axis or other abnormalities with neurotransmission.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A 50year old with history of jaundice in the past has presented with RUQ abdominal pain. Examination and investigations reveals chronic calculous cholecystitis. The liver function tests are within normal limits and on ultrasound examination, the common bile ducts is not dilated. Which of the following will be the procedure of choice in her
A. Laparoscopic cholecystectomy
B. Open choledocholithotomy with CBD exploration
C. ERCP + choledocholithotomy followed by laparoscopic cholecystectomy
D. Laparoscopic Cholecystectomy followed by ERCP +choledocholithotomy
Answer:
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A. Laparoscopic cholecystectomy
Justification: In the given question, there was an episode of jaundice, but at present, LFT is normal and CBD is not dilated. The best option is laparoscopic cholecystectomy only Management of CBD stones associated with GB stones Pre-operatively detected stones Experienced laparoscopic surgeon - Cholecystectomy and choledochotony in same suiting Inexperienced laparoscopic surgeon - pre-op ERCP with stone removal and laparoscopic cholecystectomy later Unsuspected stones found at the time of cholecystectomy Experienced Laparoscopic surgeon - Laparoscopic CBD exploration and stone retrieval through the cystic duct. Laparoscopic choledochotomy and stone extraction Inexperienced laparoscopic surgeon - Conve to open procedure and remove CBD stones. Complete the cholecystectomy and refer the patient for ERCP. Conversion to an open procedure is preferred over ERCP because the success rate of ERCP is not 100% Ref: Sabiston 20th edition Pgno :1496
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father a female's successive offspring) than under full siblings.
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P\wedge Q} , each doctor gets their worst choice among the two hospitals they are matched to in P {\displaystyle P} and Q {\displaystyle Q} (if these differ) and each hospital gets its best choice. (The same operations can be defined in the same way for any two sets of elements, not just doctors and hospitals.
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system). Conversely, any of those three cases may or may not be indeterminate.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Which of the following potentially represents the most dangerous situation?
A. Rh+ve mother with 2nd Rh-ve child
B. Rh-ve mother with 2nd Rh+ve child
C. Rh+ve mother with 1st Rh-ve child
D. Rh-ve mother with 1st Rh+ve child
Answer:
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B. Rh-ve mother with 2nd Rh+ve child
Justification: * Rh-ve mother with 2nd Rh+ve child can result in the development of hemolytic disease of newborn or erythroblastosis fetalis. So, it is a dangerous condition. * This condition is a type II hypersensitivity reaction. * This is Not to be confused with Hemorrhagic disease of the newborn which is a coagulation disturbance in the newborns due to vitamin K deficiency. As a consequence of vitamin K deficiency there is an impaired production of coagulation factors II, VII, IX, X, C and S by the liver.
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dimensions of 50 women between the ages of 18 and 50, with a mean age of 35.6:
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fraction (PAF) of 26%. Of these, 17,000 were males and 4,400 females, 13,000 (60%) were aged between 50 and 69 years of age, and the majority of cases (15,000) were in developed regions compared to developing regions (6,400).
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For example, an 1871 census in the UK (the first of its kind, but personal data from other censuses dates back to 1841 and numerical data back to 1801) found the average male life expectancy as being 44, but if infant mortality is subtracted, males who lived to adulthood averaged 75 years. The present life expectancy in the UK is 77 years for males and 81 for females, while the United States averages 74 for males and 80 for females. Studies have shown that black American males have the shortest lifespans
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
The average age of seven persons sitting in a row facing east is 28 years. If the average age of the first three persons is 25 years and the average age of the last three persons is 36 years, then find the age of the person sitting in the middle of the row?
A)9 years
B)29 years
C)18 years
D)13 years
E)17 years
Answer:
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D)13 years
Justification: Total age seven persons = (28 * 7)years
Total age of the first three persons and the last three persons are (25 * 3) years and (36 * 3) years respectively.
Age of the person sitting in the middle of the row = 28 * 7 - 25 * 3 - 36 * 3 = 196 - 75 - 108 = 13 years.
ANSWER:D
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If a rectangle has length ℓ {\displaystyle \ell } and width w {\displaystyle w} it has area A = ℓ w {\displaystyle A=\ell w\,} , it has perimeter P = 2 ℓ + 2 w = 2 ( ℓ + w ) {\displaystyle P=2\ell +2w=2(\ell +w)\,} , each diagonal has length d = ℓ 2 + w 2 {\displaystyle d={\sqrt {\ell ^{2}+w^{2}}}} , and when ℓ = w {\displaystyle \ell =w\,} , the rectangle is a square.
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The perimeter of a square whose four sides have length ℓ {\displaystyle \ell } is P = 4 ℓ {\displaystyle P=4\ell } and the area A is A = ℓ 2 . {\displaystyle A=\ell ^{2}.} Since four squared equals sixteen, a four by four square has an area equal to its perimeter.
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The area of a rectangle is defined to be length × width of the rectangle. If a long thin rectangle is stood up on its short side then its area could also be described as its height × width. The volume of a solid rectangular box (such as a plank of wood) is often described as length × height × depth. The perimeter of a polygon is the sum of the lengths of its sides. The circumference of a circular disk is the length of the boundary (a circle) of that disk.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
If the perimeter of a rectangular garden is 600 m, its length when its breadth is 100 m is?
A)286 m
B)899 m
C)200 m
D)166 m
E)187 m
Answer:
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C)200 m
Justification: 2(l + 100) = 600 => l
= 200 m
Answer: C
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Both RNA and DNA contain two major purine bases, adenine (A) and guanine (G), and two major pyrimidines. In both DNA and RNA, one of the pyrimidines is cytosine (C). However, DNA and RNA differ in the second major pyrimidine. DNA contains thymine (T) while RNA contains uracil (U). There are some rare cases where thymine does occur in RNA and uracil in DNA.Here are the 4 major ribonucleotides (ribonucleoside 5'-monophosphate) which are the structural units of RNAs.
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Three nucleobases found in nucleic acids, cytosine (C), thymine (T), and uracil (U), are pyrimidine derivatives: In DNA and RNA, these bases form hydrogen bonds with their complementary purines. Thus, in DNA, the purines adenine (A) and guanine (G) pair up with the pyrimidines thymine (T) and cytosine (C), respectively. In RNA, the complement of adenine (A) is uracil (U) instead of thymine (T), so the pairs that form are adenine:uracil and guanine:cytosine.
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nucleic acid.In some occasions, DNA and RNA may contain some minor bases. Methylated forms of the major bases are most common in DNA.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Not present in DNA ?
A. Uracil
B. Thymine
C. Cytosine
D. Adenine
Answer:
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A. Uracil
Justification: Ans. is 'a' i.e., Uracil Two types of bases are found in nucleotides : (i) purines and (ii) pyrimidines. Purines : Two major purine bases found both in DNAs as well as RNAs are (i) adenine (A) and (ii) guanine (G). Pyrimidines : Three major pyrimidine bases are (i) cytosine (C), (ii) Uracil (U) and (iii) Thymine (T). Cytosine and uracil are found in RNAs and cytosine and thymine are found in DNAs. Uracil is not found in DNAs e and thymine is not found in RNAs. Different major bases with their corresponding nucleosides and nucleotides Base Ribonucleoside Ribonucleotide Adenine (A) Adenosine Adenosine monophosphate (AMP) Guanine (G) Guanosine Guanosine monophosphate (GMP) Uracil (U) Uridine Uridine monophosphate (UMP) Cytosine (C) Cytidine Cytidine (Monophosphate) (CMP) Base Deoxyribonucleoside Deoxyribonucleotide Adenine Deoxyadenosine Deoxyadenosine monophosphate (dAMP) Guanine Deoxyguanosine Deoxyguanosine monphosphate (dGMP) Cytosine Deoxycytidine Deoxycytidine monophosphate (dCMP) Thymine Deoxythymidine Deoxythymidine monophosphate (dCMP)
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to 10 different groups. Most of the 26 identified neurotoxin-like peptides have a different molecular structure compared to the neurotoxins found in spiders, snakes, scorpions, marine cone snails, and sea anemones. The functional mechanism of these peptides are similar to the neurotoxins of the mention venomous animals, yet their primary structures remain unique.
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The venom of H. schmidti contains a large variety of neurotoxins, which function to paralyze the spider's prey. So far, 14 of the isolated primarily neurotoxic peptide components have been characterized and investigated. In the following, two subfamilies of the HWTX are described: those targeting voltage-gated calcium channels, and those targeting voltage-gated sodium channels.
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Many smaller predators such as the box jellyfish use venom to subdue their prey, and venom can also aid in digestion (as is the case for rattlesnakes and some spiders). The marbled sea snake that has adapted to egg predation has atrophied venom glands, and the gene for its three finger toxin contains a mutation (the deletion of two nucleotides) that inactives it. These changes are explained by the fact that its prey does not need to be subdued.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Neurotoxin in which snake: NEET 13
A. Viper
B. Krait
C. Sea snake
D. None
Answer:
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B. Krait
Justification: Ans. Krait
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For a given request the service time varies little as the workload increases – to do X amount of work it always takes X amount of time. The wait time is how long the request had to wait in a queue before being serviced and it varies from zero, when no waiting is required, to a large multiple of the service time, as many requests are already in the queue and have to be serviced first. With basic queueing theory math you can calculate how the average wait time increases as the device providing the service
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The motor load factor is then 12/15 = 80%. The motor above may only be used for eight hours a day, 50 weeks a year. The hours of operation would then be 2800 hours, and the motor use factor for a base of 8760 hours per year would be 2800/8760 = 31.96%. With a base of 2800 hours per year, the motor use factor would be 100%.
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Variations in the time needed to complete the tasks can be accommodated by "buffering" (holding one or more cars in a space between the stations) and/or by "stalling" (temporarily halting the upstream stations), until the next station becomes available. Suppose that assembling one car requires three tasks that take 20, 10, and 15 minutes, respectively. Then, if all three tasks were performed by a single station, the factory would output one car every 45 minutes.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A car runs at the speed of 55 km per hour when not serviced and runs at 90 kmph when serviced. After servicing the car covers a certain distance in 3 hours. How much time will the car take to cover the same distance when not serviced?
A)8 hours
B)6 hours
C)9 hours
D)7 hours 12 minutes
E)None
Answer:
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B)6 hours
Justification: Explanation :
Time = 90*3 / 45 = 6 hours
Answer – B
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For example, 50% of 40% is: 50/100 × 40/100 = 0.50 × 0.40 = 0.20 = 20/100 = 20%.It is not correct to divide by 100 and use the percent sign at the same time; it would literally imply division by 10,000. For example, 25% = 25/100 = 0.25, not 25%/100, which actually is 25⁄100/100 = 0.0025. A term such as 100/100% would also be incorrect, since it would be read as 1 percent, even if the intent was to say 100%.
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percentage, convert both percentages to fractions of 100, or to decimals, and multiply them.
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{\tfrac {1}{5}}={\tfrac {3}{10}}} 8 1 3 = 8 × 3 1 = 24.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
What will be the fraction of 10%
A)1/10
B)1/50
C)1/75
D)1/25
E)None of these
Answer:
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A)1/10
Justification: Explanation:
10*1/100 = 1/10
Option A
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Benzodiazepines are often used to reduce anxiety symptoms, muscle tension, seizure disorders, insomnia, symptoms of alcohol withdrawal, and panic attack symptoms. Their action is primarily on specific benzodiazepine sites on the GABAA receptor. This receptor complex is thought to mediate the anxiolytic, sedative, and anticonvulsant actions of the benzodiazepines. Use of benzodiazepines carries the risk of tolerance (necessitating increased dosage), dependence, and abuse. Taking these drugs for a long period
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This disorder is usually acquired in one of four ways: One cause is benzodiazepine drugs such as midazolam, flunitrazepam, lorazepam, temazepam, nitrazepam, triazolam, clonazepam, alprazolam, diazepam, and nimetazepam; all of these are known to have powerful amnesic effects. This has also been recorded in non-benzodiazepine sedatives or "z-drugs" which act on the same set of receptors; such as zolpidem (also known as Ambien), eszopiclone (also known as Lunesta), and zopiclone (also known by brand names
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There are four principal protein targets with which drugs can interact: Enzymes- (e.g. neostigmine and acetyl cholinesterase) Inhibitors Inducers Activators Membrane carriers- (e.g. tricyclic antidepressants and catecholamine uptake-1) Enhancer (RE) Inhibitor (RI) Releaser (RA)Ion channels (e.g. nimodipine and voltage-gated Ca2+ channels) Blocker OpenerReceptor (e.g. Listed in table below) Agonists can be full, partial or inverse. Antagonists can be competitive, non-competitive, or uncompetive. Allosteric
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Action of flumazenil on benzodiazepine receptor is :
A. Agonist
B. Partial agonist
C. Inverse agonist
D. Antagonist
Answer:
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D. Antagonist
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To find the probability of drawing a red card or a club, for example, add together the probability of drawing a red card and the probability of drawing a club. In a standard 52-card deck, there are twenty-six red cards and thirteen clubs: 26/52 + 13/52 = 39/52 or 3/4. One would have to draw at least two cards in order to draw both a red card and a club.
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If the events are not (necessarily) mutually exclusive then For example, when drawing a card from a deck of cards, the chance of getting a heart or a face card (J,Q,K) (or both) is 13 52 + 12 52 − 3 52 = 11 26 , {\displaystyle {\tfrac {13}{52}}+{\tfrac {12}{52}}-{\tfrac {3}{52}}={\tfrac {11}{26}},} since among the 52 cards of a deck, 13 are hearts, 12 are face cards, and 3 are both: here the possibilities included in the "3 that are both" are included in each of the "13 hearts" and the "12 face cards", but
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The probability of doing so in two draws depends on whether the first card drawn was replaced before the second drawing since without replacement there is one fewer card after the first card was drawn. The probabilities of the individual events (red, and club) are multiplied rather than added. The probability of drawing a red and a club in two drawings without replacement is then 26/52 × 13/51 × 2 = 676/2652, or 13/51.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Six cards numbered from 1 to 6 are placed in a cardboard box . First one card is drawn and then put back into the bowl; then a second card is drawn. If the cards are drawn at random and if the sum of the numbers on the cards is 10, what is the probability that one of the two cards drawn is numbered 5 ?
A)1/2
B)1/4
C)1/5
D)1/6
E)1/3
Answer:
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E)1/3
Justification: (6,4) (4,6) (5,5) – only 3 possible scenarios for sum to be 10. One from this 5 has already happened.
From this three cases, only in one we have 5. So, the probability is 1 chances out of 3 that the one that occurred had 5: P=1/3.
Answer: E.
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Surgery (thyroidectomy to remove the whole thyroid or a part of it) is not extensively used because most common forms of hyperthyroidism are quite effectively treated by the radioactive iodine method, and because there is a risk of also removing the parathyroid glands, and of cutting the recurrent laryngeal nerve, making swallowing difficult, and even simply generalized staphylococcal infection as with any major surgery. Some people with Graves' may opt for surgical intervention. This includes those that
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surgical intervention. This includes those that cannot tolerate medicines for one reason or another, people that are allergic to iodine, or people that refuse radioiodine.A 2019 systematic review concluded that the available evidence shows no difference between visually identifying the nerve or utilizing intraoperative neuroimaging during surgery, when trying to prevent injury to recurrent laryngeal nerve during thyroid surgery.If people have toxic nodules treatments typically include either removal or
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of calcium ions in the blood plasma within narrow limits. The level of the calcium in plasma is regulated by the hormones parathyroid hormone (PTH) and calcitonin. PTH is released by the chief cells of the parathyroid glands when the plasma calcium level falls below the normal range in order to raise it; calcitonin is released by the parafollicular cells of the thyroid gland when the plasma level of calcium is above the normal range in order to lower it.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Pt. after thyroid surgery presents with perioral paresthesia. serum Ca level is 7 mg/dl. What will be the best min (management) -
A. Oral vit D3
B. Oral vit D3 with Ca
C. I.V.Ca. gluconate
D. Oral calcium
Answer:
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D. Oral calcium
Justification: Ans. is 'd' i.e. Oral Calcium Since the pt. is presenting with mild symptoms he should be treated with oral calcium supplementation.
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were shown seven letters in a row and then asked to repeat the order of the letters by pushing buttons on a screen.
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So, for example, to find the weekday of 16 June 2020: Column "20" meets row "20" at "D". Row "June" meets column "16" at "F". As F is two letters on from D, so the weekday is two days on from Sunday, i.e. Tuesday.
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In numerical calculation, the days of the week are represented as weekday numbers. If Monday is the first day of the week, the days may be coded 1 to 7, for Monday through Sunday, as is practiced in ISO 8601. The day designated with 7 may also be counted as 0, by applying the arithmetic modulo 7, which calculates the remainder of a number after division by 7. Thus, the number 7 is treated as 0, the number 8 as 1, the number 9 as 2, the number 18 as 4, and so on.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Sam wrote 7 letters on Tuesday and 3 letters on Wednesday. How many letters did he write on average per day?
A)10
B)4
C)21
D)3
E)5
Answer:
|
E)5
Justification: (7 + 3)/2 = 5
The correct answer is E
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and "contrary to the common law privilege against self-incrimination. "The Crimes Act 1914, 3LA(5) "A person commits an offence if the person fails to comply with the order. Penalty for contravention of this subsection: Imprisonment for 2 years."
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Loi no 2001-1062 du 15 novembre 2001 relative à la sécurité quotidienne, article 30 (Law #2001-1062 of 15 November 2001 on Community Safety) allows a judge or prosecutor to compel any qualified person to decrypt or surrender keys to make available any information encountered in the course of an investigation. Failure to comply incurs three years of jail time and a fine of €45,000; if the compliance would have prevented or mitigated a crime, the penalty increases to five years of jail time and €75,000.
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a criminal offence and is punished with jail time between two and seven years.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Issuing false ceificate is punishable under Sec 197 IPC for imprisonment upto
A. 4 years
B. 5 years
C. 7 years
D. 10 years
Answer:
|
C. 7 years
Justification: Issuing false ceificate is punishable under Sec 197 IPC : imprisonment upto 7 years and fine and erasure of name from medical register. Ref: FORENSIC MEDICINE AND TOXICOLOGY Dr PC IGNATIUS THIRD EDITION PAGE 11
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(sequence A226898 in the OEIS).
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(4\mid 2464;\quad 16\mid 2464;\quad 28\mid 2464)} 2821 = 7 ⋅ 13 ⋅ 31 ( 6 ∣ 2820 ; 12 ∣ 2820 ; 30 ∣ 2820 ) {\displaystyle 2821=7\cdot 13\cdot 31\qquad (6\mid 2820;\quad 12\mid 2820;\quad 30\mid 2820)} 6601 = 7 ⋅ 23 ⋅ 41 ( 6 ∣ 6600 ; 22 ∣ 6600 ; 40 ∣ 6600 ) {\displaystyle 6601=7\cdot 23\cdot 41\qquad (6\mid 6600;\quad 22\mid 6600;\quad 40\mid 6600)} 8911 = 7 ⋅ 19 ⋅ 67 ( 6 ∣ 8910 ; 18 ∣ 8910 ; 66 ∣ 8910 ) . {\displaystyle 8911=7\cdot 19\cdot 67\qquad (6\mid 8910;\quad 18\mid 8910;\quad 66\mid 8910).} These
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) and two permutations with one cycle ( 312 = ( 132 ) {\displaystyle 312=(132)} and 231 = ( 123 ) {\displaystyle 231=(123)} ). Thus, = 1 {\displaystyle \left=1} , = 3 {\displaystyle \left=3} and = 2 {\displaystyle \left=2} .
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Simplify (212 * 212 + 312 * 312 )
A)132288
B)142088
C)142188
D)142288
E)None of these
Answer:
|
D)142288
Justification: Explanation:
Trick: Above equation can be solved by using following formula
(a2+b2)=1/2((a+b)2+(a−b)2)
Option D
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Amniotic fluid is essential to normal fetal movement, fetal organ development, and cushioning of the fetus within the mother's uterus. It is also imperative in preventing umbilical cord compression so that the fetus is able to obtain nourishment and oxygenation from its mother. Premature delivery: Premature birth of the fetus occurs at less than 37 weeks of gestation, compared to full-term delivery at about 40 weeks. Miscarriage: Loss of a fetus prior to birth via miscarriage can transpire if any of the
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be given to patients with a placental abruption in attempts to increase blood pressure and minimize the effects of severe blood loss. If oligohydramnios occurs as a result of a circumvallate placenta, a treatment called amnioinfusion may be considered to replenish the amount of lost amniotic fluid within the amniotic sac.
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fluid, and amniocentesis to sample amniotic fluid.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A case of 35 week pregnancy with hydromnios and marked respiratory distress is best treated by:
A. Intravenous frusemide
B. Saline infusion
C. Amniocentesis
D. Aificial rupture of membranes
Answer:
|
C. Amniocentesis
Justification: As the patient has respiratory distress and the gestational age is 35 weeks -fetal maturity has not acquired our main aim is to relieve the distress continuing the pregnancy to atleast 37 weeks. This can be achieved by Amniocentesis - drainage of good amount of liquor. Ref:Dutta Obs 9e pg 203.
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Look back to the definition of the outer code and linear inner codes, this definition of the Justesen code makes sense because the codeword of the outer code is a vector with N {\displaystyle N} elements, and we have N {\displaystyle N} linear inner codes to apply for those N {\displaystyle N} elements. Here for the Justesen code, the outer code C o u t {\displaystyle C_{out}} is chosen to be Reed Solomon code over a field F q k {\displaystyle \mathbb {F} _{q^{k}}} evaluated over F q k − { 0 }
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For the other codes that are not linear, we can consider the complexity of the encoding algorithm. So by far, we can see that the Wonzencraft ensemble and Reed-Solomon codes are strongly explicit. Therefore, we have the following result: Corollary: The concatenated code C ∗ {\displaystyle C^{*}} is an asymptotically good code(that is, rate R {\displaystyle R} > 0 and relative distance δ {\displaystyle \delta } > 0 for small q) and has a strongly explicit construction.
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The following slightly different code is referred to as the Justesen code in MacWilliams/MacWilliams. It is the particular case of the above-considered Justesen code for a very particular Wonzencraft ensemble: Let R be a Reed-Solomon code of length N = 2m − 1, rank K and minimum weight N − K + 1. The symbols of R are elements of F = GF(2m) and the codewords are obtained by taking every polynomial ƒ over F of degree less than K and listing the values of ƒ on the non-zero elements of F in some predetermined
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Let $\mathbb F$ be a field of cardinality $q$ and let $0<k<n\leq q$ be unspecified integers. As seen in the lecture, we generate a $(n,k,d_{min})$ Reed-Solomon code with the following mapping: $$\mathbb F^k
ightarrow \mathbb F^n ~~,~~ \vec u \mapsto \vec c =(P_{\vec u}(a_1),P_{\vec u}(a_2),\ldots,P_{\vec u}(a_n))$$ for $a_i \in \mathbb F$ all distinct and $P$ a polynomial of degree $k-1$ with coefficient vector $\vec u\in\mathbb F^k$. Now, we construct a $(n,k',d'_{min})$ code $\mathcal C'$ similarly to the above one by assigning $a_1\leftarrow a_2$ while leaving $n,P$ and $a_2,\ldots,a_n$ unchanged. As before, the code is generated by evaluating $P_{\vec u}(a_2,a_2,a_3,\dots,a_n)$ over all possible coefficients vectors $\vec u \in \mathbb F^k$. This is by definition not an RS code, however it is still a well-defined linear block code. True or false: We know for certain that $d'_{min}=d_{min}-1$.
Choices:
False
True
Answer:
|
Let’s delve into the intricacies of the original problem involving Reed-Solomon codes, carefully unraveling the changes made to the code and assessing their impact on the minimum distance. We will take a structured approach to analyze each aspect step-by-step, ensuring clarity and coherence as we arrive at the conclusion.
### Step 1: Understanding the Original Reed-Solomon Code
The original Reed-Solomon code is defined as follows:
\[ \mathbb{F}^k \rightarrow \mathbb{F}^n, \quad \vec{u} \mapsto \vec{c} = (P_{\vec{u}}(a_1), P_{\vec{u}}(a_2), \ldots, P_{\vec{u}}(a_n)) \]
where:
- \( \vec{u} \) is a coefficient vector in \( \mathbb{F}^k \)
- \( P_{\vec{u}} \) is a polynomial of degree \( k-1 \)
- \( a_i \) are distinct points in the field \( \mathbb{F} \)
**Importance:** The minimum distance \( d_{min} \) of this code is a critical measure of its error-correcting capability. It is tied to the number of distinct evaluations of the polynomial at distinct points, which ensures that codewords can be recognized and errors can be corrected.
### Step 2: Modifying the Code
Now, we modify this code by setting \( a_1 \) to be equal to \( a_2 \):
\[ P_{\vec{u}}(a_2, a_2, a_3, \ldots, a_n) \]
This means we have:
- Two evaluations at the same point \( a_2 \), which means our code now evaluates the polynomial at \( a_2 \) twice.
**Impact of Change:** By forcing two evaluations to be the same, we inherently reduce the number of distinct outputs generated by the polynomial, which leads to potential overlapping in the codewords produced.
### Step 3: Analyzing Minimum Distance
The minimum distance \( d'_{min} \) of the new code \( \mathcal{C}' \) is affected by this overlap. The minimum distance is defined as the smallest Hamming distance between distinct codewords.
- Since \( P_{\vec{u}}(a_2) \) is evaluated twice, two distinct input vectors \( \vec{u_1} \) and \( \vec{u_2} \) that would previously yield different outputs for distinct \( a_1 \) and \( a_2 \) may now yield the same output.
- This overlap can lead to fewer unique codewords, decreasing the minimum distance.
### Step 4: Concluding the Minimum Distance Relationship
Given that \( d'_{min} \) is likely less than \( d_{min} \) due to the loss of distinct evaluations, we must evaluate the claim that \( d'_{min} = d_{min} - 1 \).
- The statement does not hold universally. While it is plausible that the minimum distance has decreased, it does not follow that it decreases by exactly 1. The extent of the decrease depends on the specific structure of the polynomial and the coefficients used.
- There could be scenarios where the minimum distance drops by more than 1 or even remains the same, depending on how the polynomial behaves at the overlapping point.
### Final Conclusion
In light of this reasoning, we can confidently assert that the original statement is **False**. The change from \( a_1 \) to \( a_2 \) does not guarantee that the minimum distance \( d'_{min} \) will equal \( d_{min} - 1 \). Instead, it simply indicates the potential for a decrease, the magnitude of which cannot be definitively quantified without further information about the specific polynomials and fields in use.
### Why This Matters
Understanding the implications of such modifications in coding theory is crucial for practical applications, such as digital communications and data storage. It underscores the need for careful design in error-correcting codes, where even small changes can significantly affect performance and reliability.
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would be present after one hour? The question implies a = 1, b = 2 and τ = 10 min.
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With work schedules, employees will often use a tradeoff of "9/80" where an 80-hour work period is compressed into a narrow group of 9 nearly-9 hour working days over the traditional 10 8-hour working days, allowing the employee to take every second Friday off.
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There is a complex project that is composed of several sub-tasks. We would like to assign tasks to workers, such that the project finishes in the shortest possible time. As an example, suppose the "project" is to feed the goats. There are three goats to feed, one child can only feed one goat at a time, and there are two children that can feed them: Shmuel feeds each goat in 10 minutes and Shifra feeds each goat in 12 minutes. Several schedules are possible: If we let Shmuel feed all goats, then the makespan
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A can do a work in 9 days. B can do the same work in 18 days. If both A & B are working together in how many days they will finish the work?
A)3
B)5
C)4
D)2
E)6
Answer:
|
E)6
Justification: A rate = 1/9
B rate = 1/18
(A+B) rate = (1/9)+(1/18) = 1/6
A & B finish the work in 6 days
correct option is E
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As for all parallelograms, the area K of a rhombus is the product of its base and its height (h). The base is simply any side length a: K = a ⋅ h . {\displaystyle K=a\cdot h.} The area can also be expressed as the base squared times the sine of any angle: K = a 2 ⋅ sin α = a 2 ⋅ sin β , {\displaystyle K=a^{2}\cdot \sin \alpha =a^{2}\cdot \sin \beta ,} or in terms of the height and a vertex angle: K = h 2 sin α , {\displaystyle K={\frac {h^{2}}{\sin \alpha }},} or as half the product of the diagonals
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The diagonals cut the quadrilateral into four triangles of which one opposite pair have equal areas.
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A rhombus has an axis of symmetry through each pair of opposite vertex angles, while a rectangle has an axis of symmetry through each pair of opposite sides. The diagonals of a rhombus intersect at equal angles, while the diagonals of a rectangle are equal in length. The figure formed by joining the midpoints of the sides of a rhombus is a rectangle, and vice versa.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
The diagonals of a rhombus are 15 cm and 20 cm. Find its area?
A)399
B)266
C)150
D)277
E)281
Answer:
|
C)150
Justification: 1/2 * 15 * 20 = 150
Answer:C
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In mathematics and statistics, the arithmetic mean ( arr-ith-MET-ik), arithmetic average, or just the mean or average (when the context is clear) is the sum of a collection of numbers divided by the count of numbers in the collection. The collection is often a set of results from an experiment, an observational study, or a survey. The term "arithmetic mean" is preferred in some mathematics and statistics contexts because it helps distinguish it from other types of means, such as geometric and harmonic. In
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In ordinary language, an average is a single number taken as representative of a list of numbers, usually the sum of the numbers divided by how many numbers are in the list (the arithmetic mean). For example, the average of the numbers 2, 3, 4, 7, and 9 (summing to 25) is 5. Depending on the context, an average might be another statistic such as the median, or mode. For example, the average personal income is often given as the median—the number below which are 50% of personal incomes and above which are
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differences from the overall average is 9640. The difference of 6780 between these is also the weighted sum of the squares of the differences between the subject averages and the overall average: 5 ( 36 − 52 ) 2 + 4 ( 33 − 52 ) 2 + 6 ( 78 − 52 ) 2 = 6780.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
If the average (arithmetic mean) of x and y is 40, and z – x = 60, what is the average of y and z?
A)20
B)50
C)65
D)70
E)140
Answer:
|
D)70
Justification: x+y/2 = 40
=> x+y = 80
x = z - 60...sub this value
z-60+y = 80
=>z+y = 140
=>z+y/2 = 70.
ANSWER:D
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brain tumour, with no evidence of toxicity or side effects, and some long-term survivors.
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can metastasis widely). It is not just rapid growth that characterizes a cancer, but their ability to secrete enzymes, angiogeneic factors, invasion factors, growth factors and many other factors that allow it to spread.
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identified as having a potential link to various additional cancers such as oral cancer, lung cancer, gastric cancer, and gliomas.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Which of the following carcinoma most frequently metastasizes to brain?
A. Small cell carcinoma lung
B. Prostate cancer
C. Rectal carcinoma
D. Endometrial cancer
Answer:
|
A. Small cell carcinoma lung
Justification: . Small cell carcinoma lung
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eigenspinor. First we multiply: c + = ∗ χ = 1 5 {\displaystyle c_{+}={\begin{bmatrix}1\ 0\\\end{bmatrix}}*\chi ={1 \over {\sqrt {5}}}} . Now, we simply square this value to obtain the probability of the particle being found in a spin up state: P + = 1 5 {\displaystyle P_{+}={1 \over 5}}
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.Hence | P | = 9 {\displaystyle \left|{\mathcal {P}}\right|=9} and | Z | = 6 {\displaystyle \left|{\mathcal {Z}}\right|=6} .
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B is that D1 + D2 ≤ 5, and the event A is D1 = 2. We have P ( A ∣ B ) = P ( A ∩ B ) P ( B ) = 3 / 36 10 / 36 = 3 10 , {\displaystyle P(A\mid B)={\tfrac {P(A\cap B)}{P(B)}}={\tfrac {3/36}{10/36}}={\tfrac {3}{10}},} as seen in the table.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
The spinner shown is divided into 6 sections of equal
size. What is the probability of landing on a section that
contains the letter P using this spinner?
A)3/6
B)4/6
C)5/6
D)2/6
E)1/6
Answer:
|
D)2/6
Justification: The spinner has 6 sections in total and 2 of these sections contain the letter P.
Sections are equal to one another in size and thus they are each equally likely to be landed on.
Therefore, the probability of landing on a section that contains the letter P is 2/6 .
correct answer D
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A triglyceride is broken down into two fatty acids and a monoglyceride, which are absorbed by the villi on the intestine walls. After being transferred across the intestinal membrane, the fatty acids reform into triglycerides (re-esterified), before being absorbed into the lymphatic system through lacteals. Without bile salts, most of the lipids in food would be excreted in feces, undigested.Since bile increases the absorption of fats, it is an important part of the absorption of the fat-soluble
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cells lining the intestines, bile acids from the gallbladder, and by products of digestion.
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bile acids, predominantly deoxycholic acid (DCA) and lithocholic acid (LCA).
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Medium chain fatty acids
A. Require pancreatic lipase for digestion
B. Absorbed through lymphatics
C. Also digested in stomach
D. Require bile salts for absorption and digestion
Answer:
|
C. Also digested in stomach
Justification: Ref: Textbook of Medical Biochemistry 8th Edition Dr (Brig) MN Chatterjea, Rana Shinde page no: 392-396
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is often claimed, erroneously, that he said, "The definition of insanity is doing the same thing over and over and expecting different results."
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Neuroethics also encompasses the ethical issues raised by neuroscience as it affects our understanding of the world and of ourselves in the world. For example, if everything we do is physically caused by our brains, which are in turn a product of our genes and our life experiences, how can we be held responsible for our actions? A crime in the United States requires a "guilty act" and a "guilty mind". As neuropsychiatry evaluations have become more commonly used in the criminal justice system and
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and the early argument systems of Donald Nute and of Guillermo Simari and Ronald Loui. Defeasible reasoning accounts of precedent (stare decisis and case-based reasoning) also make use of specificity (e.g., Joseph Raz and the work of Kevin D. Ashley and Edwina Rissland). Meanwhile, the argument systems of Henry Prakken and Giovanni Sartor, of Bart Verheij and Jaap Hage, and the system of Phan Minh Dung do not adopt such a rule.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Legal responsibility of insane is/are all except:
A. Rule of Haase
B. Mc Naughten's rule
C. Curren's rule
D. Durham rule
Answer:
|
A. Rule of Haase
Justification: A i.e. Rule of Haase
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Severe limb injuries in which the efforts to save the limb fail or the limb cannot be saved. Traumatic amputation (an unexpected amputation that occurs at the scene of an accident, where the limb is partially or entirely severed as a direct result of the accident, for example, a finger that is severed from the blade of a table saw) Amputation in utero (Amniotic band)
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Whereas an avulsion is characterized by a "flap" of skin being removed, an amputation is characterized by a complete loss of a limb. This can occur at any point on the extremity, and is usually followed by significant arterial bleeding. However, as serious as this injury is, an amputated limb that is cooled and transported to the hospital can sometimes be surgically reattached.Types of wounds
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The word amputation is borrowed from Latin amputātus, past participle of amputāre "to prune back (a plant), prune away, remove by cutting (unwanted parts or features), cut off (a branch, limb, body part)," from am-, assimilated variant of amb- "about, around" + putāre "to prune, make clean or tidy, scour (wool)". The English word "Poes" was first applied to surgery in the 17th century, possibly first in Peter Lowe's A discourse of the Whole Art of Chirurgerie (published in either 1597 or 1612); his work was
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Tarso Metatarsal amputation is also known as?
A. Chopa's amputation
B. Lisfranc amputation
C. Pirogoff amputation
D. Symes amputation
Answer:
|
B. Lisfranc amputation
Justification: Tarsometatarsal joint is known as Lisfranc joint and amputation through this joint is known as Lisfranc amputation. Amputation of foot Mid foot amputation: Type Level of amputation Lisfranc Tarsometatarsal joint Chopa Midtarsal joint Pirogoff Calcaneus is rotated forward to be fused to tibia after veical section through the middle Hind Foot Amputation: Type Level of amputation Syme Distal tibia fibula 0.6 cm proximal to the periphery of ankle joint passing through the dome of ankle Sarmiento Distal tibia and fibula approximately 1.3 cm proximal to the ankle joint and excision of medial and lateral malleoli Boyd talectomy, forward shift of the calcaneus and calcaneotibial ahrodesis Ref: Campbell's Operative Ohopedics 10/e, Page 557-70; Current Diagnosis & Treatment in Ohopedics 3/e, Page 654-57.
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dimensions of 50 women between the ages of 18 and 50, with a mean age of 35.6:
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, XN, and add up their values. By Wald's equation, the resulting value on average is E E = 1 + 2 + 3 + 4 + 5 + 6 6 ⋅ 1 + 2 + 3 + 4 + 5 + 6 6 = 441 36 = 49 4 = 12.25 . {\displaystyle \operatorname {E} \operatorname {E} ={\frac {1+2+3+4+5+6}{6}}\cdot {\frac {1+2+3+4+5+6}{6}}={\frac {441}{36}}={\frac {49}{4}}=12.25\,.}
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(20 years or less) and patients between 41 and 60 years of age, fared better than those who were in the 21 to 40 years age group.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Five years ago, the average age of A, B, C and D was 45 years. With E Joining them now, the average of all the five is 48 years. The age of E is?
A)45
B)47
C)48
D)49
E)40
Answer:
|
E)40
Justification: Solution
5 years ago average age of A,B,C,D = 45 years
=> 5 years ago total age of A, B, C, D = 45 x 4 = 180 years
=> Total present age of A, B, C, D = 180 + 5x4 = 200 years
If E's present age is x years = 200+x/5= 48
x=40 years. Answer E
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be maximised over the manufacturer's production set. This is interpreted as profit maximisation.
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that (i) price should equal marginal cost and (ii) output should be maximised subject to (i).
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Each seller would vary her output based on the output of the other and the market price would be determined by the total quantity supplied. The profit for each firm would be determined by multiplying their output by the per unit market price. Differentiating the profit function with respect to quantity supplied for each firm left a system of linear equations, the simultaneous solution of which gave the equilibrium quantity, price and profits.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A factory produces x widgets per day. The factory's fixed costs are $8000 per day. The price per widget is $70 and the variable costs are $30 per widget. How many widgets need to be produced for profits of $5440 a day?
A)42.33
B)90.33
C)168
D)224
E)311
Answer:
|
E)311
Justification: profits=5440=40x-8000 --> x=311
Answer E.
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at that time, the accepted standard of care for HIV-positive mothers was known as the 076 regimen and involved five daily doses of AZT from the second trimester onwards, as well as AZT intravenously administered during labour. As this treatment was lengthy and expensive, it was deemed unfeasible in the Global South, where mother-to-child transmission was a significant problem. A number of studies were initiated in the late 1990s that sought to test the efficacy of a shorter, simpler regimen for use in
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Consistent and proactive precautionary measures, such as the rigorous use of antiretroviral medications, cesarean section, face masks, heavy-duty rubber gloves, clinically segregated disposable diapers, and avoidance of mouth contact will further reduce child-attendant transmission of HIV to as little as 1–2%.During 1994 to 1999, AZT was the primary form of prevention of mother-to-child HIV transmission. AZT prophylaxis prevented more than 1000 parental and infant deaths from AIDS in the United States. In
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of mother-to-child transmission of HIV during pregnancy, labor, and delivery and has been proven to be integral to uninfected siblings' perinatal and neonatal development.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Dose of Nevirapine to Newborn for Prevention of Mother to child transmission of HIV is
A. 2 mg/ kg
B. 5 mg/kg
C. 10 mg/kg
D. 20 mg/kg
Answer:
|
A. 2 mg/ kg
Justification: Bih Weight NVP daily dose(in mg) NVP daily dose(in ml) Duration Infants with bih weight <2000gm 2 mg/kg once daily 0.2 ml/kg once daily Upto 6 weeks irrespective of exclusively breast fed or exclusive replacement fed Bih weight 2000 -2500gm 10 mg once daily 1 ml once daily Bih weight more than 2500gm 15 mg once daily 1.5 ml once daily
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The presence of hematochezia is six-times greater in a LGIB than with a UGIB.Occasionally, a person with a LGIB will not present with any signs of internal bleeding, especially if there is a chronic bleed with ongoing low levels of blood loss. In these cases, a diagnostic assessment or pre-assessment should watch for other signs and symptoms that the patient may present with. These include, but are not limited to, hypotension, tachycardia, angina, syncope, weakness, confusion, stroke, myocardial
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and hemoglobin readings, with hematocrit levels >49% in men and >43% in women, not because of an absolute increase in them but because of the leak of plasma); very low blood pressure (profound arterial hypotension, with systolic blood pressure levels <90 mm Hg); albumin deficiency (hypoalbuminemia measuring <3.0 g/dL); partial or generalized edema, and cold extremities; a paraprotein in the blood (an MGUS in approximately 80% of cases).
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The presence of hemochromatosis may be discovered incidentally on blood testing, or a diagnosis suspected based on symptoms may be supported or ruled out by blood testing. Elevated serum ferritin, an indicator of blood iron levels, and transferrin saturation, which is involved with absorption of iron from the gut, are very common. Transferrin saturation may approach or reach 100%, where a normal value would lie between 16% and 45%. If transferrin saturation is normal, juvenile hemochromatosis can be ruled
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A 60 yr old female with history 8 blood transfusions in 2yrs.Her Hb-60g/I , TLC-5800, platelet-3.4 Iakhs, MCV-60, RBC-2.1 lakhs/mm3. She is having hypochromic mierocytic anemia. Which investigation is not needed?-
A. Evaluation for pulmonary hemosiderosis
B. Urinary hemosiderin
C. Bone marrow examination
D. G1 endoscopy
Answer:
|
A. Evaluation for pulmonary hemosiderosis
Justification: Idiopathic pulmonary hemosiderosisis a rare disease characterized by repeated episodes of bleeding into the lungs, which can cause anemia and lung disease. The body is able to remove most of the blood from thelungs, but a large amount of iron is left behind. Ref Robbins 9/e pg 419
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2004 ranks growth deficiency as follows: Severe: Height and weight at or below the 3rd percentile.
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is below this amount: so, the remaining 5% of the time, the usage is above that amount. Physicians will often use infant and children's weight and height to assess their growth in comparison to national averages and percentiles which are found in growth charts. The 85th percentile speed of traffic on a road is often used as a guideline in setting speed limits and assessing whether such a limit is too high or low.In finance, value at risk is a standard measure to assess (in a model-dependent way) the
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below the 10th percentile of standardized growth charts appropriate to the population. Prenatal or postnatal presentation of growth deficits can occur, but are most often postnatal.Criteria for FASD are least specific in the Institute of Medicine (IOM) diagnostic system ("low birth weight..., decelerating weight not due to nutrition..., disproportional low weight to height" p.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
All are true regarding WHO growth cha, EXCEPT:
A. It is an impoant tool for educating
B. The position of dot is more impoant than direction
C. Between top 2 lines it shows road-to-health
D. Lowermost line represents children below third percentile
Answer:
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B. The position of dot is more impoant than direction
Justification: Growth cha is a visible display of the child's physical growth and development. The WHO prototype growth cha has 2 reference curves. The upper reference curve represents the median (50th percentile) for boys (slightly higher than that of girls), and the lower reference curve the 3rd percentile for girls (slightly lower than that for boys). The cha can be used for both sexes. The space between the two growth curves has been called 'road-to-health'. It is the direction of growth that is more impoant than the position of dots on the line. It used as an educational tool for mothers. Ref: Park's Textbook of Social and Preventive Medicine, 19th edition, Page 435-437.
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In most competitions the sponsor or "coach" is allowed to select 4 students per division to participate in a "team" test (formally called "Team Bowl".) Each team member sits with the rest of their team and is allowed to communicate and collaborate during the team round. A few competitions do not allow the team members to sit together; rather every member of the division takes the team test alone and without conversing, then the 4 highest scores are averaged together; these 4 people are on team. Some
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When there is enough space, schools may take advantage of this multiple team rule and have up to four teams in one division's team round, though only the first two teams are considered for Sweepstakes. The grading scale is different for the team round. Questions are given one by one, whereas in the individual round students are given the test in its entirety.
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together; these 4 people are on team. Some competitions allow each school to have a second team for each division, "Team II".
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
A certain team consists of 5 professors and 4 teaching assistants. How many different teams of 3 can be formed in which at least one member of the group is a professor? (Two groups are considered different if at least one group member is different.)
A)74
B)80
C)86
D)92
E)98
Answer:
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B)80
Justification: The total number of ways to form a group of 3 from 9 people is 9C3 = 84
The number of ways to form a group without a professor is 4C3 = 4
The total number of ways to form a valid group is 84 - 4 = 80.
The answer is B.
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Pupillary reflexes, particularly the pupillary light reflex, are a powerful diagnostic tool often employed in clinical and emergency medical practice. A lack of equal consensual pupillary constriction to a light stimulus, especially a Marcus Gunn pupil, can be indicative of optic nerve damage, brainstem death, or optic tract damage in between.
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In some neurological disorders, the pupil does not react to light, but it does react to accommodation. This is called “light-near dissociation”. In Adie syndrome, damage involving the ciliary ganglion manifests light-near dissociation and a tonically dilated pupil (usually on the same side).
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The pupillary light reflex is an autonomic reflex that controls pupil diameter to accommodate for increases in illumination as perceived by the retina. Higher light intensity causes pupil constriction, and the increase of light stimulation of one eye will cause pupillary constriction of both eyes. The neural circuitry of the pupillary light reflex includes the optic tract which joins the optic nerve to the brachium of the superior colliculus.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Light reflex absent but accommodation reflex present is seen in -
A. Hutchison's pupil
B. Argyl Robeson pupil
C. Adies pupil
D. Horner's syndrome
Answer:
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B. Argyl Robeson pupil
Justification: Ref khurana 6/e
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enlarged liver and/or spleen, clinical evidence of hepatitis, and/or central nervous system disturbances such as irritability, decreased levels of consciousness, seizures, meningitis (i.e. neck stiffness, photophobia, and headache), impaired cranial nerve function, hemiplegia, ataxia (i.e. poor coordination of complex muscle movements), and reduced muscle tone. Laboratory studies show abnormal liver function tests, reduced levels of blood fibrinogen, impaired blood clotting, and high levels of blood
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liver disease have normal transaminases. Elevated transaminases that persist less than six months are termed "acute" in nature, and those values that persist for six months or more are termed "chronic" in nature.
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In terms of the diagnosis of hepatitis E, only a laboratory blood test that confirms the presence of HEV RNA or IgM antibodies to HEV can be trusted. In the United States no serologic tests for diagnosis of HEV infection have ever been authorized by the Food and Drug Administration. The World Health Organization has developed an international standard strain for detection and quantification of HEV RNA. In acute infection the viremic window for detection of HEV RNA closes 3 weeks after symptoms begin.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Acute viral hepatitis — A is diagnosed by
A. Hepatic injury related with transaminase level
B. Variable increase in transaminase level
C. IgG anti HAV used for diagnosis
D. No rise of transaminase at all
Answer:
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B. Variable increase in transaminase level
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After one hour, or six ten-minute intervals, there would be sixty-four bacteria. Many pairs (b, τ) of a dimensionless non-negative number b and an amount of time τ (a physical quantity which can be expressed as the product of a number of units and a unit of time) represent the same growth rate, with τ proportional to log b. For any fixed b not equal to 1 (e.g. e or 2), the growth rate is given by the non-zero time τ. For any non-zero time τ the growth rate is given by the dimensionless positive number b.
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A quantity x depends exponentially on time t if where the constant a is the initial value of x, the constant b is a positive growth factor, and τ is the time constant—the time required for x to increase by one factor of b: If τ > 0 and b > 1, then x has exponential growth. If τ < 0 and b > 1, or τ > 0 and 0 < b < 1, then x has exponential decay. Example: If a species of bacteria doubles every ten minutes, starting out with only one bacterium, how many bacteria would be present after one hour? The question
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colonies can be counted after 24 to 48 hours depending on the type of bacteria. Counts are reported as colony forming units per 100 mL (cfu/100 mL).
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
It takes 30 days to fill a laboratory dish with bacteria. If the size of the bacteria doubles each day, how long did it take for the bacteria to fill 1/32 of the dish?
A)25
B)26
C)27
D)28
E)29
Answer:
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A)25
Justification: The bacteria doubles each day, so after 29 days, the dish was half full.
After 28 days, the dish was one quarter full.
After 27 days, the dish was one eighth full.
After 26 days, the dish was one sixteenth full.
After 25 days, the dish was 1/32 full.
The answer is A.
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the teeth (dentinogenesis imperfecta). Potentially life-threatening complications, all of which become more common in more severe OI, include: tearing (dissection) of the major arteries, such as the aorta;: 333 pulmonary valve insufficiency secondary to distortion of the ribcage;: 335–341 and basilar invagination.
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As CLP can make oral hygiene more difficult, there is an increased rate of cavities. In addition, abnormal positioning of individual teeth may affect occlusion, which can create an open bite or cross bite. This in turn can then affect the patient's speech.
Dysbaric osteonecrosis, also known as aseptic bone necrosis, is generally a longer term effect on the bones and joints of divers caused by decompression bubbles and may occur even if no clinical decompression sickness has been diagnosed.
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speaking, damage to clothing and books, and mouth infections.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
If the necessary interocclusal space is not provided, long term sequelae will be:
A. Clicking of dentures
B. Ridge resorption
C. Soreness of tissues
D. Problems with speech and mastication
Answer:
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B. Ridge resorption
Justification: Effects of increase in vertical dimension
Discomfort: Increasing the vertical dimension alters the environment in which unconscious movements of muscles during chewing take place and until the original condition is restored, discomfort will result. The jarring effect of teeth coming into contact sooner than expected also causes discomfort.
Trauma: The sudden and frequent contacting of teeth causes trauma to the denture-bearing area, especially under the lower denture, where the area to resist pressure is less.
TMJ problem: The constant tooth contact will also affect the TMJ causing soreness and pain.
Bone resorption: The increased vertical height does not allow the muscles that close the mouth to complete their contraction. They will continue to exert force to overcome this obstruction and this will lead to resorption of supporting tissues.
Muscular fatigue: Due to the constant effort of the muscles to close the mouth, muscular fatigue will also occur.
Clicking of teeth: The premature contact of teeth sooner than what the individual is used to, will cause clicking of teeth.
Facial distortion: There will be an inability to close the lips, which will produce a strained expression and elongation of face.
Difficulty in swallowing and speech: The inability to close the lips will also cause difficulty in swallowing and speech.
Effects of decrease in vertical dimension
Inefficiency: The biting force exerted by the teeth in occlusion decreases which causes inefficient mastication.
Cheek biting: The loss of muscle tone and reduced vertical height causes the flabby cheeks to become trapped during mastication.
TMJ problem: The patient has to often protrude the mandible to occlude the teeth and this causes pain and clicking in the TMJ.
Facial distortion: The following effects are seen:
Nose appears closer to the chin
Loss of lip fullness
Loss of tonicity of muscles of facial expression
Face appears flabby
Patient appears older
Angular cheilitis: The corners of the mouth form deep folds, which are bathed in saliva. This becomes infected and sore.
Key Concept:
A lack of interocclusal space results in excess pressure on ridges continuously. As a result of which, ridge resorption takes place.
Ref: Textbook of prosthodontics V Rangarajan Ed. 2nd
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before physics, and in fact many familiar processes like fires and chemical explosions are chemical chain reactions.
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The general characteristics of chemical reactions are: Evolution of a gas Formation of a precipitate Change in temperature Change in state
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reactions at the timescales on which they occur, opening up the field of femtochemistry.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Which is evidence of a chemical reaction?
A. the light produced by magnesium when burned
B. the evaporation of water from a solution
C. the fizzing of a soft drink
D. the heat from a light bulb
Answer:
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A. the light produced by magnesium when burned
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Dual serotonin-norepinephrine reuptake inhibitors in particular duloxetine, as well as tricyclic antidepressants in particular amitriptyline, and nortriptyline are considered first-line medications for this condition.
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reuptake inhibitors (SSRI's) are a class of drugs demonstrated to be an effective treatment in major depressive disorder and are the most prescribed class of antidepressants.
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Molloy and fellow Eli Lilly chemist Klaus Schmiegel synthesized a series of dozens of its derivatives. Hoping to find a derivative inhibiting only serotonin reuptake, another Eli Lilly scientist, David T. Wong, proposed to retest the series for the in vitro reuptake of serotonin, norepinephrine and dopamine, using a technique developed by neuroscientist Solomon Snyder. This test showed the compound later named fluoxetine to be the most potent and selective inhibitor of serotonin reuptake of the series.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
Which of the following drug is serotonin-norepinephrine reuptake inhibitor?
A. Venlafaxine
B. Amphetamine
C. Doxepin
D. Miazapine
Answer:
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A. Venlafaxine
Justification: Venlafaxine, Milnacipran are considered as SNRIs Amphetamine is a psycho stimulant drug. Mitrazapine is an atypical anti depressent. Doxepin is a TCA tricyclic anti deprerssent. Ref: HL Sharma 3rd ed. Classification Pg: 469
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Shock index (SI) has been defined as heart rate/systolic blood pressure ; SI≥0.6 is a clinical shock. Such ratio value is clinically employed to determine the scope or emergence of shock. The SI correlates with the extent of hypovolemia and thus may facilitate the early identification of severely injured patients threatened by complications due to blood loss and therefore need urgent treatment, i.e. blood transfusion.
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within 3–5 minutes of sudden VF cardiac arrest dramatically improves survival. In cities such as Seattle where CPR training is widespread and defibrillation by EMS personnel follows quickly, the survival rate is about 20 percent for all causes and as high as 57 percent if a witnessed "shockable" arrest.
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for critically ill patients, especially those in cardiac arrest.
The following are multiple choice questions (with answers) about knowledge and skills in advanced master-level STEM courses.
What is the strength of shock given to below victim of cardiac arrest with shockable rhythm:
A. 200 J asynchronous Shock
B. 300 AC shock
C. 200 Syn. Shock
D. 300 J Syn. Shock
Answer:
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A. 200 J asynchronous Shock
Justification: It should be between 120-200 J asynchronized shock.
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