FL4 B/B

Question 1

Most cytochrome P450 enzymes (important for metabolizing) alter the activity of drugs by:

A. phosphorylating them.

B. dephosphorylating them.

C. oxidizing them.

D. reducing them.

Answer 1

C. oxidizing them.

Question 2

JAK2 phosphorylates the tail of the GH receptor on several tyrosine residues, thereby forming a docking site for molecules that have phosphotyrosine-binding domains, such as STAT5b. JAK2 then phosphorylates receptor-associated STAT5b molecules.

During phosphorylation of STAT5b proteins, phosphate groups are exchanged for what atoms on tyrosine residues?

A.Hydrogen atoms of hydroxyl groups

B.Hydrogen atoms of methyl groups

C.Oxygen atoms of hydroxyl groups

D.Carbon atoms of methyl groups

Answer 2

A.Hydrogen atoms of hydroxyl groups

Question 3

A defining characteristic of proteins that act as transcription factors (such as STAT5b) is that they:

A.can dimerize.

B.can phosphorylate other proteins.

C.contain a DNA binding domain.

D.are present within the nucleus of the cell.

Answer 3

C.contain a DNA binding domain. that allows it to bind to regulatory nucleic acid sequences in a gene to alter transcription

Question 4

Fatty acid oxidation takes place in the:

A. the mitochondria.

B. the cell membrane.

C. the cytosol.

D. the lysosome.

Answer 4

A. the mitochondria.

Question 5

The acetyl-CoA carboxylases ACC1 and ACC2 regulate fatty acid synthesis and fatty acid oxidation, respectively, through the conversion of acetyl-CoA to malonyl-CoA

During the ACC1- or ACC2-catalyzed formation of malonyl-CoA, what is the structure of the functional group added to acetyl-CoA?

Answer 5

C

Carboxylation

Question 6

Bacteriophages were labeled with radioactive phosphorous (32P) and sulfur (35S). The labeled bacteriophages were then allowed to infect their host cells. At the end of the experiment, the 32P label was found only inside the host cells and the 35S label was found only outside the host cells. This experiment shows that:

A. bacteriophages consist only of DNA and protein.

B. only DNA, not protein, can enter the host cell.

C. only protein, not DNA, can enter the host cell.

D. both DNA and protein can enter the host cell.

Answer 6

B. only DNA, not protein, can enter the host cell.

Question 7

Which statement describes ATP consumption and production during the preparatory and payoff phases of glycolysis, respectively?

A. Two ATP molecules consumed, four ATP molecules produced

B. Four ATP molecules consumed, eight ATP molecules produced

C. Four ATP molecules consumed, two ATP molecules produced

D. Eight ATP molecules consumed, four ATP molecules produced

Answer 7

A. Two ATP molecules consumed, four ATP molecules produced

Question 8

As blood passes through actively contracting skeletal muscle tissue, the affinity of hemoglobin for oxygen in the muscle tissue:

A. increases as a result of an increase in plasma temperature.

B. increases as a result of an increase in plasma PO2.

C. decreases as a result of a decrease in plasma pH.

D. decreases as a result of a decrease in plasma PCO2.

Answer 8

C. decreases as a result of a decrease in plasma pH.

Question 9

A researcher measures the concentration of CO2(aq) in a solution at equilibrium. Which additional quantity must be measured in order to calculate the Henry’s Law constant kH for the gas?

A. Atmospheric pressure

B. Volume of the solvent

C. Partial pressure of the gas

D. Vapor pressure of water

Answer 9

C. Partial pressure of the gas

The Henry’s Law constant kH relates the solubility of a gas S to the pressure of that gas Pg above the solution and is written as S = kH•Pg.

Question 10

Ghrelin is a 28-amino-acid peptide hormone that is cleaved from a longer protein produced by the stomach and small intestine.

Based on the passage, what can be determined about the composition of the mRNA that encodes the protein from which ghrelin is cleaved?

It is composed of:

A. more than 28 amino acids.

B. exactly 84 nucleotides.

C. exactly 87 nucleotides.

D. more than 87 nucleotides.

Answer 10

D. more than 87 nucleotides.

28 x 3 = 84

it comes from a longer peptide so D

Question 11

Hormone replacement therapy is often given to children who have Prader–Willi syndrome (growth-hormone deficiency) once they have reached the hyperphagic stage. Based on the passage, the most likely reason for this therapy would be to prevent:

A.giantism.

B.weight loss.

C.short stature.

D.muscle rigidity.

Answer 11

C.short stature.

Question 12

Eating suppresses ghrelin secretion. However, this suppression does not appear to be caused by the presence of food in the stomach or duodenum, nor by stomach distension. Given this, which of the following physiological conditions is likely to suppress ghrelin secretion following a meal?

A.Increased levels of circulating glucagon

B.Decreased osmolarity in the ileum

C.Increased levels of circulating insulin

D.Decreased levels of circulating insulin

Answer 12

C.Increased levels of circulating insulin

Question 13

Which subunit is LEAST suitable for generation of vaccine against B. pertussis?

Answer 13

the toxin subunit

Question 14

Treatment of pertussis includes erythromycin, an antibiotic that inhibits bacterial protein synthesis by binding to the 23S rRNA component of the large subunit of the bacterial ribosome.

Erythromycin interferes with protein synthesis by binding to which ribosomal subunit?

A. 50S

B. 60S

C. 70S

D. 80S

Answer 14

A. 50S

70 - intact prokaryote ribosome

60 - large subunit for eukaryotes

Question 15

Treatment of pertussis includes erythromycin, an antibiotic that inhibits bacterial protein synthesis by binding to the 23S rRNA component of the large subunit of the bacterial ribosome. The binding interferes with aminoacyl transferase function, preventing tRNA transfer between the tRNA sites of the ribosome.

Binding of erythromycin to the large subunit of the bacterial ribosome most likely blocks the transfer of the tRNA bound at:

A. the E site to the P site of the ribosome.

B. the P site to the A site of the ribosome.

C. the E site to the A site of the ribosome.

D. the A site to the P site of the ribosome.

Answer 15

D. the A site to the P site of the ribosome.

During protein translation, aminoacyl transferase functions to transfer the tRNA originally bound at the A (amino acid) site to the P (peptide) site and later to the E (exit) site of the ribosome.

Question 16

Which inhibitor is this?

Answer 16

mixed

Question 17

Which inhibitor is this?

Answer 17

competitive

Question 18

Which inhibitor is this?

Answer 18

uncompetitive

Question 19

Which inhibitor is this?

Answer 19

Noncompetitive

Question 20

Which experiment can be used to investigate the transcriptional regulation of the Cdkn1a protein?

A. Assessing Cdkn1a mRNA levels by RT-PCR

B. Assessing Cdkn1a mRNA levels by Southern blotting

C. Assessing Cdkn1a protein levels by quantitative PCR

D. Assessing Cdkn1a protein levels by Western blotting

Answer 20

A. Assessing Cdkn1a mRNA levels by RT-PCR

As the question is focused on the transcriptional regulation, it is logical to assess the mRNA levels as opposed to protein levels. RT-PCR is a molecular technique that measures mRNA levels of specific protein.

Quantitative PCR is a technique that measures the levels of DNA, not mRNA, and cannot be used to measure the transcriptional regulation of a gene

Western blot is a technique that measures the translational levels of a protein, not the transcriptional regulation of a gene.

Question 21

The nuclear localization sequence of MYOD1 contains how many potential phosphorylation sites? NDAFEITKRC

A. None

B. One

C. Two

D. Three

Answer 21

B. One

There are three amino acids that are phosphorylated in eukaryotes: serine (S), threonine (T), and tyrosine (Y).

Question 22

Which technique separates proteins independently of their charge?

A. Native PAGE

B. Gel filtration chromatography

C. Ion exchange chromatography

D. Isoelectric focusing

Answer 22

B. Gel filtration chromatography - separates based on size: uses beads!

NATIVE separates molecules based on their electrophoretic mobility, relying on length, conformation, and charge

Question 23

A peptide consisting of nine amino acids was partially hydrolysed. Three different tripeptides were isolated. None of the tripeptides share a common amino acid. Based on the data, what is the total number of possible structures possible for the full-length peptide?

A. 3

B. 4

C. 6

D. 27

Answer 23

C. 6

If the sequences were ABC, DEF, and GHI, they can only be joined in 6 different possible ways to make a nonapeptide. Each of the tripeptides can appear in the first position and can combine in two possible ways with the other two tripeptides: 3 × 2 = 6.

the product would be 9x3 = 27 but this does not represent the number of possibilities of the full-length sequence

Question 24

A woman with cystic fibrosis considers having children with an asymptomatic man who is heterozygous for the cystic fibrosis allele. What is probability of having a child with cystic fibrosis?

A. 25%

B. 50%

C. 75%

D. 100%

Answer 24

A heterozygous genotype that results in an asymptomatic phenotype implies that the disorder is due to a recessive allele. Therefore, a homozygous individual and a heterozygous individual have 50% probability of having a child with the disorder.

h h

H. Hh Hh

h. hh hh

Question 25

Which of the following statements best applies to the inactive X chromosome in mammalian females?

A. It does not replicate.

B. Its chromatin structure is less condensed than that of an active X chromosome.

C. It is one of the last chromosomes to replicate during mitosis.

D. It is highly transcribed.

Answer 25

C. It is one of the last chromosomes to replicate during mitosis.

Question 26

What is the equilibrium constant expression for Reaction 1?

A. [Acetylphosphate][CO2][H2O2]/[Pyruvate][Pi][O2]

B. [Pyruvate][Pi][O2]/[Acetylphosphate][CO2][H2O2]

C. [Acetylphosphate][CO2][H2O2]/[Pyruvate][Pi][O2][H2O]

D. [Pyruvate][Pi][O2][H2O]/[Acetylphosphate][CO2][H2O2]

Answer 26

A. [Acetylphosphate][CO2][H2O2]/[Pyruvate][Pi][O2]

Question 27

The negative control used in Experiment 1 was most likely:

A. unconditioned media without AECs.

B. unconditioned media with AECs.

C. conditioned media without AECs.

D. conditioned media with AECs.

Answer 27

B. unconditioned media with AECs.

Question 28

Cholecystokinin (CCK) is a peptide hormone that was first isolated from the small intestine, where its release is stimulated by the presence of food. Once secreted by the small intestine, CCK facilitates digestion by releasing into the duodenum bile from the gallbladder and pancreatic juices rich in enzymes (such as amylase and trypsin).

Which of the following events does the release of CCK likely trigger?

A. Relaxation of muscle in wall of gallbladder and relaxation of hepatopancreatic sphincter

B. Contraction of muscle in wall of gallbladder and relaxation of hepatopancreatic sphincter

C. Relaxation of muscle in wall of gallbladder and contraction of hepatopancreatic sphincter

D. Contraction of muscle in wall of gallbladder and contraction of hepatopancreatic sphincter

Answer 28

B. Contraction of muscle in wall of gallbladder and relaxation of hepatopancreatic sphincter

Question 29

Extra water could be excreted through the intestine if the subjects were having

Answer 29

diarrhea

Question 30

The osmolarity of urine in these subjects was 958 milliosmoles/L, compared with an average blood osmolarity of 284 milliosmoles/L. The higher osmolarity of urine suggests that:

A. the kidneys are secreting very little Na+.

B. the kidneys are acting to conserve water.

C. the subjects are on a low-protein diet.

D. the subjects are dehydrated.

Answer 30

B. the kidneys are acting to conserve water.

Question 31

In a different experiment, healthy human volunteers received an intravenous infusion of 100 mL of hyperosmotic NaCl solution, which induced thirst and a desire to drink. In this experiment, thirst probably was stimulated via:

A. blood volume receptors, as a result of increased blood volume after the infusion.

B. osmoreceptors, as a result of increased blood osmolarity after the infusion.

C. a sensation of dryness in the mouth by oral receptors.

D. the NaCl itself, because salt makes a person thirsty regardless of osmolarity.

Answer 31

B. osmoreceptors, as a result of increased blood osmolarity after the infusion.

Osmoreceptors are activated by the increase in osmolarity of the blood. This process triggers the sensation of thirst.

Question 32

Which of the following responses could maintain cardiac output under dehydration conditions that reduce blood volume?

A. A decrease in stroke volume

B. A decrease in heart rate

C. An increase in heart rate

D. An increase in blood viscosity

Answer 32

C. An increase in heart rate (heart contractions per minute)

In dehydration conditions, a decrease in SV will decrease, not maintain, the cardiac output. SV is the amount of blood thats pumped by the left ventricle per beat

Question 33

During exercise, the osmolarity of venous blood from active muscles will increase as a result of an increase in:

A. lactate concentration in plasma.

B. O2 concentration in plasma.

C. oxyhemoglobin concentration in red blood cells.

D. blood pressure in systemic arteries.

Answer 33

A. lactate concentration in plasma.

Question 34

Like other retroviruses, HIV contains reverse transcriptase, an enzyme that converts the viral genome from:

A. RNA to DNA.

B. DNA to RNA.

C. RNA to protein.

D. protein to RNA.

Answer 34

A. RNA to DNA.

Question 35

Given that the solid line in each graph represents an uncatalyzed chemical reaction, in which of the following graphs does the dashed line best represent the same chemical reaction when catalyzed by an enzyme?

Answer 35

B

The role of an enzyme is to reduce the activation energy of a reaction, energy of the reactants and products are not modified by the enzyme. This graph shows exactly what happens when an enzyme catalyzes a reaction.

Question 36

Answer 36

D

Please note that the values on the y-axis are on Log10. Based on the graph, cell division just starts to slow down after the 10,000-cell stage, which corresponds to the value of 4 in the y-axis of the graph. For this reason, gastrulation should start between 10,000 and 100,000 cells.

Log10 10,000 = 4

Question 37

pptB encodes pilin phosphotransferase B (PptB), which adds phosphoglycerols onto pilin subunits primarily at residue 93.

Which change in a property of pilin will be observed after modification by PptB?

A. Increase in number of disulfide bonds

B. Decrease in molecular weight

C. Increase in enzymatic activity

D. Decrease in isoelectric point

Answer 37

D. Decrease in isoelectric point

PptB is a phosphotransferase, which adds a phosphoglycerol to pilin subunits. The addition of this group (phosphoglycerol) adds a negative charge to pilin and thus decreases its isoelectric point.

Not B because - The addition of a phosphoglycerol will increase the molecular weight of pilin.

Question 38

During the proliferation phase, researchers observed increased transcription of the gene pptB after the bacteria adhered to nasopharyngeal epithelial cells. pptB encodes pilin phosphotransferase B (PptB), which adds phosphoglycerols onto pilin subunits primarily at residue 93.

Which amino acid residue in the pilin subunit is most likely modified by PptB?

A. Cysteine

B. Phenylalanine

C. Serine

D. Tryptophan

Answer 38

C. Serine

PptB is a phosphotransferase. Of the amino acid options, only serine can be modified by a phosphotransferase.

Question 39

Although colonization of the nasopharynx is usually asymptomatic, in some cases the bacteria enter the blood, migrate to the brain, and cross the blood–brain barrier. The subsequent immune response causes meningitis, or inflammation of the meninges, which is a life-threatening condition.

Based on the information in the passage, which immune cells would mount the initial immune response to N. meningitidis that results in meningitis?

A. B cells

B. Cytotoxic T cells

C. Helper T cells

D. Microglia

Answer 39

D. Microglia

The other options are not present in the brain

Question 40

The presence of which type of intercellular connections between endothelial cells of brain capillaries results in the blood–brain barrier?

A. Desmosomes

B. Gap junctions

C. Intercalated discs

D. Tight junctions

Answer 40

D. Tight junctions

Tight junctions are intercellular junctions that prevent the movement of solutes within the space between adjacent cells. In blood capillaries, neighboring endothelial cells form tight junctions with one another to restrict the diffusion of harmful substances and large molecules into the interstitial fluid surrounding the brain.

Question 41

These are intracell junctions that function as anchors to form strong sheets of cells

A. Desmosomes

B. Gap junctions

C. Intercalated discs

D. Tight junctions

Answer 41

A. Desmosomes

Question 42

These are intercell junctions that provide cytoplasmic channels between adjacent cells

A. Desmosomes

B. Gap junctions

C. Intercalated discs

D. Tight junctions

Answer 42

B. Gap junctions

Question 43

These are specialized junctions between cardiac muscle cells that provide direct electrical coupling among cells

A. Desmosomes

B. Gap junctions

C. Intercalated discs

D. Tight junctions

Answer 43

C. Intercalated discs

Question 44

If the third codon in the coding region of the Rdl GABA receptor cDNA is replaced with an amber codon (for example, TAG) and the modified RdlGABA receptor cDNA is expressed in frog oocytes, functional full-length receptors will:

A. accumulate in the Golgi complex.

B. accumulate in the rough endoplasmic reticulum.

C. not be synthesized.

D. lose resistance to dieldrin.

Answer 44

C. not be synthesized.

An amber codon is a stop codon

Question 45

Which pathway depicts the sequence of cellular compartments traversed by newly synthesized GABA-receptor subunits as they move to the cell surface?

A. Rough endoplasmic reticulum, endosome, and Golgi complex

B. Rough endoplasmic reticulum, Golgi complex, and secretory transport vesicle

C. Golgi complex, endosome, and lysosome

D. Golgi complex, rough endoplasmic reticulum, and secretory transport vesicle

Answer 45

B. Rough endoplasmic reticulum, Golgi complex, and secretory transport vesicle

As GABA receptors are located in the cell membrane, they will be synthesized in the RER, modified in the Golgi apparatus, and transported to the membrane by transport vesicles.

Endosomes are formed to transport a substance from the outside of the cell, not inside the cell.

Question 46

Assume that a mutant allele is the result of the deletion of two entire codons. The base pair length of the mutant allele, compared to that of the wild-type allele, would have:

A. two fewer nucleotides.

B. three fewer nucleotides.

C. four fewer nucleotides.

D. six fewer nucleotides.

Answer 46

D. six fewer nucleotides.

Question 47

Which process is most likely responsible for the failure of a gamete to receive a copy of a particular chromosome?

A. Recombination

B. Linkage

C. X inactivation

D. Non-disjunction

Answer 47

D. Non-disjunction when sister chromatids fail to separate during cell division

Question 48

When genetic material is broken and brought back together

A. Recombination

B. Linkage

C. X inactivation

D. Non-disjunction

Answer 48

A. Recombination

Question 49

Occurs when genes tend to be inherited together

A. Recombination

B. Linkage

C. X inactivation

D. Non-disjunction

Answer 49

B. Linkage