Exam Master PA-CAT Study Guide

What is the likely diagnosis for a child who reports increased back pain, an uneven gait, and favoritism to the left side?

A. Kyphosis

B. Lordosis

C. Scoliosis

D. Spina bifida

C. Scoliosis

Scoliosis

lateral bending of the vertebral column, often in the thoracic area

Kyphosis

curvature of the spine that produces a "humpback"

Lordosis

having a "hollow back," or being "bent backward." Some describe it as a sway back. It may be caused by poor posture

Spina bifida

defect of the vertebral column where L5 or S1 fails to develop normally, leaving a hole or dimple

A 42-year-old man with a history of anxiety and peptic ulcer disease develops severe back pain. An analysis reveals the presence of amylase, lipase, and peptidase in his stomach. This patient's condition is caused by erosion of a peptic ulcer in what structure?

A. Left gastric artery

B. Pancreas

C. Peritoneal cavity

D. Pleural cavity

E. Splenic artery

B. Pancreas

This patient's sudden onset severe back pain suggests perforation of a peptic ulcer. The location of his pain along with the presence of pancreatic enzymes in the stomach suggest that the ulcer eroded into the pancreas. Since the pancreas is located directly posterior to the stomach, the ulcer would be located on the posterior wall of the stomach.

If the ulcer had eroded into the peritoneal cavity, the leakage of stomach contents into the peritoneal cavity would result in a generalized peritonitis, causing generalized severe abdominal pain and rebound tenderness, not focal back pain. If the ulcer had somehow eroded into the pleural cavity, we would expect to see pleuritic chest pain. If the ulcer had eroded into a blood vessel, we might see blood in the stomach or peritoneum instead of pancreatic enzymes, resulting in a much more serious clinical presentation. Two common blood vessels that a peptic ulcer can

erode into are the splenic artery, which carries blood to the spleen, or the left gastric artery, which supplies blood to the lesser curvature of the stomach.

A patient sustains a myocardial infarction (heart attack) that damages the ventricular septum of the heart. What effect on the heart is most likely to be seen immediately after the heart attack?

A. Blood flowing from the left side of the heart to the right

B. Changes in the electrical conduction of the heart

C. Damage to the valves of the heart

D. Reduction in blood pressure

B. Changes in the electrical conduction of the heart

The septum contains fibers that coordinate the beating of the ventricles. These are sensitive and are damaged almost instantly in myocardial infarction. This damage can be seen via changes in the conduction of electricity through the heart.

Damage to valves, left to right blood flow, and dramatic hypotension are common late complications of myocardial infarction, generally occurring days to weeks following the initial event. This is in contrast to electrical changes, which present near instantly.

If a patient ingests a poison that inhibits protein synthesis, what cellular organelle(s) is/are being targeted?

A. Lysosomes

B. Golgi apparatus

C. Plasma membrane

D. Mitochondria

E. Ribosomes

E. Ribosomes

Ribosomes

responsible for the production of cellular proteins through the process of translation

Lysosomes

contain digestive enzymes and are involved in the breakdown of molecules and cellular maintenance

Golgi apparatus

responsible for sorting, packaging, and sending proteins throughout the cell

plasma membrane

a semi permeable boundary surrounding the cell

Mitochondria

the energy-producing organelles within the cell. Mitochondria produce energy in the form of ATP via aerobic respiration in humans

Case:

Trace evidence retrieved in a crime scene revealed 2 distinct monosaccharides upon infrared spectroscopic analysis. The victim was a scientist who was working with an enzyme that cleaves

disaccharides. The spectroscopic analysis from the crime lab exactly matched the last entry on the victim's electronic notebook. During their investigation, the forensics team analyzed carbohydrate samples from 4 suspects who had entered the crime scene in the past 24 hours. Each of the suspects had a unique carbohydrate sample that they needed to analyze.

Stem:

Based on this evidence, which of the following carbohydrates did the most likely suspect possess?

A. Fructose

B. Maltose

C. Starch

D. Sucrose

D. Sucrose

The victim was working with an enzyme that cleaves disaccharides into its constituent monosaccharides. The spectroscopic analysis revealed the presence of 2 different monosaccharides. The most likely suspect must have been the person who possessed a

disaccharide made of 2 different monomers.

The suspect who possessed sucrose (glucose + fructose), is most likely to have seen the victim

the last.

Maltose is a disaccharide made of 2 glucose monomers and could not have been the relevant

disaccharide.

Fructose (a monosaccharide) and starch (a polysaccharide) cannot be cleaved by the enzyme in

question.

Case:

A graduate student received organellar fractions from 4 different sources. The student was asked to extract DNA from each and was surprised to find 1 sample devoid of any genetic material.

Stem:

Given that the student was careful and there were no mistakes in the extraction procedure, what source would be devoid of genetic material?

A. Beetle leaf extract

B. Onion peel

C. Red blood cells

D. Semen sample

C. Red blood cells

Red blood cells (RBCs) extracted from blood would be devoid of any genetic material. Typically, a blood sample would contain all cellular components (lymphocytes, RBCs, and platelets) from which DNA can be extracted, but RBCs specifically do not contain nuclei or any organelles (to make room for packaging hemoglobin).

Beetle leaf extract and onion peel, both plant sources, and semen sample (human source) would all contain DNA, as they contain nuclei and organelles (mitochondria and chloroplasts), which house various amounts of DNA.

What process occurs during the increase in water solubility of a compound containing functional groups with carboxylic acids as they undergo ionization?

A. Elongation of hydrocarbons

B. Formation of salts

C. Grouping of aromatic rings

D. Standardization of charge

B. Formation of salts

One way in which carboxylic acids increase the water solubility of a compound is through reaction with a strong base to form salts (carboxylate anion salts). Salt formation can increase the water solubility of a compound due to the attraction of the area of the partial positive charge within the compound to the partial negative of water (area surrounding oxygen).

Elongation of hydrocarbons, grouping of aromatic rings, and standardization of charge (reduction of polarity) serve to reduce water solubility.

The acetylation of 0.205 mol of p-aminophenol by acetic anhydride produced acetaminophen and acetic acid. A side reaction resulting from moisture in the reaction vessel hydrolyzed an unknown amount of acetic anhydride before the reaction was complete. After purification, only 29.47 g of acetaminophen was isolated. If 13.35 g of acetic acid was recovered from the products, how much acetic anhydride was consumed in the acetylation reaction and how much underwent hydrolysis?

A. Acetylation - 11.71 g and Hydrolysis - 1.39 g

B. Acetylation - 19.91 g and Hydrolysis - 1.39 g

C. Acetylation - 19.91 g and Hydrolysis - 1.64 g

D. Acetylation - 20.93 g and Hydrolysis - unknown

E. Acetylation - 19.91 g and Hydrolysis - unknown

B. Acetylation - 19.91 g and Hydrolysis - 1.39 g

Since acetaminophen is the product of the acetylation reaction, the amount of acetaminophen produced from the reaction is used to determine how much of the acetic anhydride was consumed by the acetylation reaction. Using the molecular mass, we can determine that 0.195 mol of acetaminophen was produced through acetylation. This would require 19.91 g of acetic

anhydride.

Calculate the amount of acetic acid produced as a side product of the acetylation reaction: 0.195 mol*60.05 g = 11.71 g. Now, subtract the mass of acetic acid produced as a result of acetylation from the mass recovered after reaction to get the mass of acetic acid produced through hydrolysis: 13.35 g - 11.71 g = 1.64 g acetic acid. The hydrolysis reaction proceeds as follows:

(CH3CO)2O + H2O → 2CH3COH

Therefore, 1.64 g of acetic acid is produced by 1.39 g of acetic anhydride.

What generally results from a point mutation, deletion, or insertion in the promoter region of a

proto-oncogene?

A. Increased transcription of protein product

B. Stimulation of cell mitosis

C. Transduction of continuous cell growth signals

D. Uncontrolled stimulation of kinase signaling pathway

A. Increased transcription of protein product

A point mutation, deletion, or insertion in the promoter region of a proto-oncogene is an activation mechanism that generally causes increased transcription.

Stimulation of cell mitosis is an example of a general process involving proteins encoded by proto-oncogenes.

Transduction of continuous cell growth signals relates to a mutation within an oncogene rather than a mutation to the promoter region; specifically, this refers to the mutated ras oncogene, which causes a protein to remain in an active state and transduces continuous cell growth signals.

Uncontrolled stimulation of kinase signaling pathway similarly relates to a mutation within an oncogene rather than a mutation to the promoter region; specifically, this refers to the mutated braf oncogene, which encodes for a protein with a modified kinase domain.

Which hormone produced by the hypothalamus and secreted by the pituitary gland would you expect to be effective in improving social interactions in children with autism?

A. Human growth hormone

B. Oxytocin

C. Thyroid-stimulating hormone

D. Vasopressin

B. Oxytocin

Oxytocin is involved in bonding and trust; low levels of this hormone are correlated with autism in children. Human growth hormone is involved with physical development of children. Thyroid-stimulating hormone activates production of thyroid hormones. Vasopressin helps regulate water and electrolyte homeostasis.

An 8-month-old boy who has never been vaccinated presents with a 3-day history of fever and watery non-bloody diarrhea. On physical examination, he appears dehydrated. What is the genome of the most likely infecting organism?

A. Double-stranded DNA

B. Non-segmented single-stranded positive-sense RNA

C. Segmented double-stranded RNA

D. Single-stranded DNA

C. Segmented double-stranded RNA

Rotaviruses cause most of the watery diarrheal illness in infants and children worldwide. They are non-enveloped RNA viruses containing 11 segments of double-stranded RNA genome within a double-shelled capsid.

Rotaviruses, like the influenza viruses that have a segmented genome, can undergo genetic reassortment. Other viruses that cause gastroenteritis are Adenovirus, a double-stranded DNA virus, and Norovirus, a non-segmented single-stranded positive-sense RNA. Single-stranded DNA viruses do not cause gastroenteritis.

Case:

A 13-year-old boy presents in the ED with fever, tender joints, and rapid heartbeat. His mother says he has been sick with a sore throat, which she thought was a cold; now she is concerned it may actually be the flu. Rapid strep test and flu test in the ED are both negative. The physician assistant finds a rash with pink rings and a clear center, orders an Antistreptolysin O antibody test and EKG, and gives the patient penicillin and a round of steroids. The lab test shows a high level of antibodies against Streptococcus.

Stem:

What organism is the most likely pathogen?

A. Escherichia coli

B. Staphylococcus aureus

C. Streptococcus pyogenes

D. Streptococcus viridans

C. Streptococcus pyogenes

Streptococcus pyogenes typically causes "strep throat" and would cause a positive rapid strep test. A rapid strep test looks for Group A strep, which causes beta-hemolysis or full hemolysis on blood agar after culture. Untreated strep throat can lead to rheumatic fever in children and adolescents, caused by the immune system's response from an earlier strep throat or scarlet fever infection; it is thought to be caused by a generalized inflammatory response.

Escherichia coli is a Gram-negative rod that does not typically cause pharyngitis. E. coli can grow on blood agar as a smooth round gray-white colony. Some strains are beta-hemolytic, or they show no hemolysis at all.

Staphylococcus aureus is a Gram-positive coccus seen in clusters that does not typically cause pharyngitis. On a culture, this organism grows as a medium-sized round creamy yellow colony and shows beta hemolysis.

Streptococcus viridans is an alpha-hemolytic (green incomplete hemolysis) Gram-positive coccus seen in chains. Streptococcus viridans is part of the normal flora of a human mouth. On blood agar, it grows in small grayish alpha-hemolytic colonies.

Case:

A young boy being observed at his daycare. The observer noted that he was quite relaxed and not very interested in his surroundings.

Stem:

According to the EAS Temperament Model, how would the child have scored?

A. Low on emotionality

B. Low on sociability

C. Low on activity

D. Low on affection

A. Low on emotionality

The EAS Temperament Model uses 3 dimensions of temperament: emotionality, activity, and sociability. Affection is not one of the 3 dimensions. The boy's relaxed and non-interested demeanor is part of the emotionality dimension, which measures the intensity of emotional reactions. Activity measures a person's energy level and sociability measures a person's ability to affiliate and interact with others.

Which Gestalt principle describes the following example? Refer to the image.

A. Similarity

B. Camouflage

C. Continuity

D. Closure

D. Closure

Closure is when we tend to see complete figures even when part of the information is missing. In this case, the square image in white is not really there, but rather, formed by the missing pieces of the surrounding circles.

Similarity is when things that share visual characteristics—such as shape, size, color, texture, or value—are seen as belonging together.

Continuity is when the edge of one shape continues into space and meets with other shapes or the edges of the picture plane; the viewer will follow the established pattern.

Camouflage is when the figure blends into the background, making the image visually disruptive.

Which statement applies to the elongation stage of transcription?

A. RNA polymerase binds to a sequence of DNA known as a promoter.

B. RNA polymerase builds an RNA molecule, making a chain.

C. RNA polymerase separates the DNA strands, creating a single-strand template.

D. Sequences send signals that release the transcript from the RNA polymerase.

B. RNA polymerase builds an RNA molecule, making a chain.

During the elongation stage of transcription, the RNA polymerase decodes the template strand, building an RNA molecule, creating a chain. In the initiation stage of transcription RNA polymerase is bound to a sequence of DNA referred to as the promoter. Once bound, the polymerase separates the DNA strands, creating the template strand. In the termination stage,

sequences called terminators signal completion of the RNA transcript. This causes the transcript to be released from the RNA.

A couple expecting a child have different blood types; the mother is AO and the father is BO. Which blood type in the infant would be an example of codominance?

A. AB

B. AO

C. BO

D. O

A. AB

The blood type AB is an example of codominance. The A and B allele are both fully dominant, and the phenotype has both traits expressed equally. AO and BO blood types are examples of simple dominance, where either the A or B allele is the dominant trait and the O allele is a recessive trait. With type O blood, the offspring has received 2 copies of the recessive trait, one from each parent.

What is the standard deviation calculated for the following 10 fasting blood sugar levels of patients with diabetes?

84, 87, 93, 99, 103, 116, 126, 129, 130, 133

A. 0

B. 2.86

C. 18.02

D. 18.99

C. 18.02

A standard deviation is the square root of a variance, so the variance must be calculated first.

Variance is the squared difference from the mean (average).

First, calculate the mean of the values.

84 + 87 + 93 + 99 + 103 + 116 + 126 + 129 + 130 + 133 = 1100

1100 ÷ 10 = 110

For each value, subtract the mean and square the result (the squared difference).

For example,

84 - 110 = -26

-262 = 676

After completing all the calculations, take the average of the squared difference to find the

variance.

3246/10 = 324.6 Therefore, the variance is 324.6.

The standard deviation (the square root of the variance) is √ 324.6=18.02324.6=18.02

2.86 would result from squaring the negative numbers and getting negative numbers instead of positive numbers. 0 would result from forgetting to square the differences. 18.99 would result from dividing the squared difference sum by 9 instead of by 10.

Researchers often wish to reduce their sample size to save money in conducting studies. What factor might make a study's sample size smaller?

A. Measure central tendency and dispersion.

B. Reduce statistical power and use chi-squared models for analysis.

C. Reduce the nonresponse rate and stratify the population.

D. Use inferential statistics rather than descriptive statistics.

C. Reduce the nonresponse rate and stratify the population.

A study that has a nonresponse bias of 50% will need a large sample size in comparison to one with a nonresponse rate of 1%. Putting resources into follow-up can reduce the nonresponse rate and reduce sample size. Stratifying the population reduces variation within groups, allowing a smaller sample size to adequately represent a population.

Measuring central tendency and dispersion would not affect the sample size. This is synonymous with analyzing mean and standard deviation, so these calculations would not affect the experimental setup. Similar to the case of central tendency and dispersion, our method of analysis would not affect our experimental design.

Reducing statistical power would allow us to reduce our sample size, but using chi-squared models for analysis would not.

Using inferential statistics rather than descriptive statistics is another mode of analysis, and would not achieve the desired reduction in sample size.