AAMC 3 - Bio/Biochem

Question 1

Based on the information in the passage, what is the most likely mechanism of inheritance for HPRCC?

A. Autosomal dominant

B. Autosomal recessive

C. X-linked recessive

D. Y linked

Answer 1

A. Autosomal dominant

AAMC: The answer to this question is A because the allele must be inherited in a autosomal dominant pattern in order for individuals II-1 and II-2 to produce both affected and unaffected offspring. The inheritance pattern for the offspring of these individuals is consistent with autosomal dominant if the parents are heterozygous. The rest of the pedigree is consistent with this mode of inheritance as well. In addition, the offspring of individuals II-1 and II-2 rule out autosomal recessive, X-linked recessive, and Y-linked. It is a Scientific Reasoning and Problem Solving question because in addition to interpreting the pedigree shown in the figure, it requires forming a hypothesis regarding the genetic transmission pattern of HPRCC.

Jack Westin Advanced Solution:

This aligns with our pedigree analysis and mechanism of inheritance flowchart. Males are not truly disproportionately affected (see above) so autosomal is more likely than sex-linked. Also, two affected parents had unaffected offspring which means that it is a dominant mechanism of inheritance. If it were recessive, II-1 and II-2 would only be able to pass down recessive alleles and all of their children would be homozygous recessive and affected. Because III-1 and III-3 are not affected, the parents must both be heterozygous and only have one copy of the dominant allele.

Question 2

DNA sequence analysis was performed on normal kidney tissue from an unaffected individual and on tumor tissue from an affected individual. The region of the DNA containing the MET mutation is shown.

Based on these results, the MET mutation is what type of mutation?

A. Purine to purine

B. Purine to pyrimidine

C. Pyrimidine to purine

D. Pyrimidine to pyrimidine

Answer 2

D. Pyrimidine to pyrimidine

AAMC: The answer to this question is D because by comparing the band patterns of the two analyses, the only difference between the two sequences is the ninth base; the normal sample contains a T, and the tumor sample contains a C. Both T and C are pyrimidines, which makes the mutation pyrimidine to pyrimidine.

Question 3

Which amino acid residue is involved in the bond that joins the α and β subunits of HGF through a disulfide linkage?

A. A

B. C

C. S

D. Y

Answer 3

B. C

Disulfide bonds form between cysteine residues.

Question 4

Which of the following animal pairs best illustrates the outcome of convergent evolution?

A. The dolphin and the shark

B. The domestic sheep and the mountain goat

C. The polar bear and the panda bear

D. The light-colored and the dark-colored forms of the peppered moth

Answer 4

A.The dolphin and the shark

AAMC: The answer to this question is A because convergent evolution is defined as a process whereby distantly related organisms independently evolve similar traits to adapt to similar needs.

Question 5

Kallman Syndrome is a disease in which gonadotropin-releasing hormone producing neurons fail to migrate from the olfactory area to the hypothalamus during embryonic development. Which endocrine axis is disrupted in individuals with Kallman Syndrome?

A. The stress axis

B. The growth hormone axis

C. The thyroid axis

D. The reproductive axis

Answer 5

D. The reproductive axis

AAMC: The answer to this question is D because gonadotropin-releasing hormone regulates pituitary gonadotropin (luteinizing hormone and follicle-stimulating hormone) secretion.

Thus, if gonadotropin-releasing hormone is not able to regulate luteinizing hormone and follicle-stimulating hormone secretion from the pituitary, the reproductive axis will remain quiescent.

Question 6

Which statement correctly identifies an enzyme involved in DNA replication and describes its primary function?

A. Ligase catalyzes the binding of RNA primers to DNA via phosphodiester bonds.

B. Primase catalyzes the replacement of RNA primer nucleotides with DNA nucleotides.

C. Helicase catalyzes the separation of the parent DNA strands at the origin of replication.

D. Topoisomerase catalyzes the joining of adjacent Okazaki fragments into a continuous strand of DNA.

Answer 6

C. Helicase catalyzes the separation of the parent DNA strands at the origin of replication.

AAMC: The answer to this question is C because during DNA replication, helicase catalyzes the unwinding and separation of the parental DNA strands, so that each can be replicated.

Jack Westin Advanced Solution:

Helicase “unzips” the strands of DNA at the beginning of replication. This occurs at the origin of replication. This is the best answer.

Question 7

Proteases fall under what major class of enzymes?

Answer 7

Proteases are Hydrolases

They break peptide bonds by the addition of water

Question 8

Which type of membrane transport is directly affected by cardiac glycoside Na+/K+ ATPases?

A. Simple diffusion

B. Facilitated diffusion

C. Primary active transport

D. Secondary active transport

Answer 8

C. Primary active transport

Question 9

___________ of histone proteins decreases the positive charge on lysine residues and weakens the interaction of the histone with DNA, resulting in a(n) ______ chromatin conformation that allows for easier access of the transcriptional machinery to the DNA.

Answer 9

Acetylation of histone proteins decreases the positive charge on lysine residues and weakens the interaction of the histone with DNA, resulting in an open chromatin conformation that allows for easier access of the transcriptional machinery to the DNA.

Histone acetylases (acetyltransferases) loosen chromatin and promote transcription

Question 10

Histone _________ are proteins that function to remove acetyl groups from histones, which results in a(n) ______ chromatin conformation and overall ________ in gene expression levels in the cell.

Answer 10

Histone deacetylases are proteins that function to remove acetyl groups from histones, which results in a closed chromatin conformation and overall decrease in gene expression levels in the cell.

Histone deacetylases condense chromatin and decrease transcription

Question 11

Histone deacetylases (HDACs) change chromatin by:

A. decreasing its coiling and promoting DNA replication.

B. increasing its condensation and inhibiting transcription.

C. decreasing charge repulsion between acetyl groups, which increases transcription.

D. loosening the attachment of DNA to nucleosome core particles.

Answer 11

B. increasing its condensation and inhibiting transcription.

Histone deacetylases condense chromatin and decrease transcription

Question 12

The adrenal medulla is part of which branch(es) of the peripheral nervous system?

A. The somatic nervous system only

B. The sympathetic branch of the autonomic nervous system only

C. The parasympathetic branch of the autonomic nervous system only

D. Both the sympathetic and the parasympathetic branches of the autonomic nervous system

Answer 12

B. The sympathetic branch of the autonomic nervous system only

AAMC: The answer to this question is B because the adrenal medulla secretes epinephrine and norepinephrine in response to short-term stress. Reactions to short-term stress are mediated by the sympathetic branch of the autonomic nervous system.

Question 13

Which of the following best describes the chemical energy that is derived from the Krebs cycle? Energy is produced in the forms of:

A. ATP, which directly supplies energy for many cellular processes, and NAD⁺, which supplies energy for the electron transport chain.

B. NAD⁺, which directly supplies energy for many cellular processes, and ATP, which supplies energy for the electron transport chain.

C. ATP, which directly supplies energy for many cellular processes, and NADH, which supplies energy for the electron transport chain.

D. NADH, which directly supplies energy for many cellular processes, and ATP, which supplies energy for the electron transport chain.

Answer 13

C.ATP, which directly supplies energy for many cellular processes, and NADH, which supplies energy for the electron transport chain.

AAMC: The answer to this question is C because the Krebs cycle produces both ATP and NADH but not NAD⁺. ATP directly supplies energy for many cellular processes, such as muscle contraction, and NADH, which is used in the electron transport chain.

Question 14

The average osmotic pressure of ocean water is 28 atm corresponding to a concentration of 0.50 M solutes (approximated as NaCl). What is the approximate concentration of solutes (also approximated as NaCl) present in blood with an osmotic pressure of 7 atm?

A. 0.12 M

B. 0.25 M

C. 2.0 M

D. 3.5 M

Answer 14

A. 0.12 M

π = MRTi

AAMC: The answer to this question is A because osmotic pressure is directly proportional to solute concentration. Since the osmotic pressure of blood is one-fourth that of ocean water, the solute concentration is also one-fourth that of ocean water, or 0.25 × 0.50 M = 0.12 M.

Question 15

Describe an imprinted gene

Answer 15

An imprinted gene is one in which only one copy is expressed based on parent of origin.

Individuals have 2 copies of the gene as usual, one from mom and one from dad. However, the gene will only be expressed from the maternal or paternal chromosome. i.e. in Angelman’s syndrome, only the maternally-inherited copy of the gene is active. The silencing of the other (paternal) copy is achieved via epigenetic marking.

Question 16

Certain viruses contain RNA as their genetic material. One of the ways these RNA viruses replicate themselves is to:

A. code for or carry a transcriptase that copies viral RNA.

B. infect microorganisms possessing RNA as their genetic material.

C. alter the host cell’s polymerase in order to synthesize progeny viral RNA from the viral RNA template.

D. stimulate the transcription of specific sequences of the host’s DNA, which, in turn, direct the assembly of viral particles.

Answer 16

A. code for or carry a transcriptase that copies viral RNA.

This is correct. RNA viruses contain transcriptases, namely reverse transcriptase, that can use the viral RNA transcript to make cDNA or complementary DNA that can be recognized by the host for replication.

Another name for reverse transcriptase that you might see on test day is RNA dependent DNA polymerase because it depends on RNA and will produce DNA.

Question 17

Large amounts of protein are found in the urine of a patient. Based on this information, which portion of the nephron is most likely malfunctioning?

A. Collecting duct

B. Distal tubule

C. Glomerulus

D. Loop of Henle

Answer 17

C. Glomerulus

AAMC: The answer to this question is C because in healthy individuals, the structure of the glomerular capillaries prevents the entry of large molecules, such as proteins, into the filtrate.

Question 18

The rate of a typical enzymatic reaction is increased by which of the following changes?

A. Decrease in a substrate concentration

B. Increase in pH from 6.8 to 7.4

C. Increase in the energy of activation

D. Increase in temperature from 20°C to 37°C

Answer 18

D. Increase in temperature from 20°C to 37°C

AAMC: The answer to this question is D because while the optimum pH value varies greatly from one enzyme to another, the optimum temperature for enzymes is normally 37 °C.

Question 19

Mucous secretions in the respiratory tract inhibit microbial infections. These secretions are produced by which of the following tissue types found in the lungs?

A. Smooth muscle

B. Epithelial

C. Nervous

D. Connective

Answer 19

B. Epithelial

Epithelial cells can be found in the luminal aspect of internal organs including the lungs. There are many types of epithelial cells: squamous, cuboidal, columnar, transitional and Goblet cells. Goblet cells in particular secrete mucus that can trap irritants and smaller foreign bodies. These Goblet epithelial cells can be found in the lungs to trap and inhibit microbial infections making this the correct answer.

It should also be noted that the cilia in the bronchi will move the debris and mucus up and out of the lungs in a process called mucociliary clearance.

Question 20

Dewlaps that reflect UV light would evolve by natural selection only if:

A. individuals with UV-reflective dewlaps mated more frequently than did individuals without them.

B. individuals with UV-reflective dewlaps were less subject to predation than individuals without them.

C. individuals with UV-reflective dewlaps were better able to communicate than individuals without them.

D. individuals with UV-reflective dewlaps produced more offspring than did individuals without them.

Answer 20

A. individuals with UV-reflective dewlaps produced more offspring than did individuals without them.

AAMC: Although many different types of adaptations may help an individual organism survive, they will not be passed on to the next generation unless the organism produces offspring, passing on the genes that cause the advantageous phenotype. To evolve by natural selection and become a general characteristic of the species, the genes that cause dewlaps to reflect UV light must become a significant portion of the gene pool, which will most likely occur if individuals with UV-reflective dewlaps produce more offspring than do individuals without them. Thus, A is the best answer.

Jack Westin: This answer choice is consistent with the breakdown I did of the question. It ties into that definition of natural selection perfectly.

Question 21

Aldosterone stimulates Na⁺ reabsorption by the kidneys. What changes in blood volume and pressure would be expected as a result of aldosterone deficiency?

A. Decreased volume and decreased pressure

B. Decreased volume and increased pressure

C. Increased volume and decreased pressure

D. Increased volume and increased pressure

Answer 21

A. Decreased volume and decreased pressure

AAMC: The question asks the examinee to determine the result of an aldosterone deficiency on blood volume and blood pressure, given that aldosterone stimulates Na⁺ reabsorption in the kidney. An aldosterone deficiency would result in lower Na⁺ reabsorption into the bloodstream. Because H₂O passively follows Na⁺ during reabsorption in the kidney, less Na⁺ reabsorption would result in less H₂O reabsorption into the bloodstream. This would result in decreased blood volume. Blood volume would also be affected by lower blood Na⁺ levels because there would be less of this ion to osmotically hold water in the extracellular fluid. Decreased blood volume would result in decreased blood pressure as well.

Question 22

In almost all vertebrates, when the optic cup fails to develop in the embryo, the lens also fails to form. This constitutes evidence that:

A. the process of neurulation follows gastrulation.

B. the eye develops early in vertebrate morphogenesis.

C. cells may induce neighboring cells to differentiate.

D. cell differentiation is an “all or none” phenomenon.

Answer 22

C. cells may induce neighboring cells to differentiate.

AAMC: The optic cup develops from a bulge on the side of the developing brain, which influences the overlying ectoderm to produce the lens. It is therefore an example of cells inducing neighboring cells to differentiate, so option C is the correct answer. The other response choices are irrelevant to the question, so they are not good answers. The absence of the optic cup and lens has no influence on timing of neurulation relative to gastrulation (choice A). The presence or absence of the optic cup and lens has no effect on the timing of eye development (choice B). Cell differentiation is not an “all or none” phenomenon (choice D). In most cases, cells progressively differentiate to achieve their final specification.

Question 23

In mammals, which of the following events occurs during mitosis but does NOT occur during meiosis I?

A. Synapsis

B. The splitting of centromeres

C. The pairing of homologous chromosomes

D. The breaking down of the nuclear membrane

Answer 23

B. The splitting of centromeres

AAMC: One of the key differences between mitosis and meiosis occurs during their respective anaphases. During anaphase of mitosis, sister chromatids are pulled apart at the centromeres, each becoming an independent chromosome in the two diploid daughter cells. During anaphase I of meiosis I, homologous pairs of chromosomes are separated into the two daughter cells. However, each chromosome still consists of two sister chromatids joined to each other at the centromere. It is not until anaphase II of meiosis II that the centromere is split and the sister chromatids separate. Thus, B is the best answer.

Jack Westin: This answer choice ties into answer choice A. In meiosis I, we have the splitting of those homologous chromosomes we talked about in answer choice A. In mitosis, during anaphase, we have actual separation of centromeres toward opposite poles of the cell. Answer choice B is the best answer choice so far.