FL 8 - Bio/Biochem

Question 1

If something decreases insulin sensitivity, what would the body do to respond to this insensitivity?

Answer 1

The body has to secrete more insulin to try to compensate for the insensitivity. This leads to hyperinsulinemia, which is characterized by excess insulin in the blood.

Question 2

What two conditions could arise due to an excess consumption of sucrose?

Answer 2

Sucrose breaks down into glucose and fructose.

Therefore, its overconsumption would increase blood glucose and fructose levels, resulting in hyperglycemia and hyperfructosemia.

Question 3

What is hypotension?

Answer 3

Hypotension is characterized by low blood pressure.

Question 4

Describe the role of gastrin

Answer 4

Gastrin is a peptide hormone that stimulates secretion of HCl by the parietal cells of the stomach.

Question 5

Describe the role of somatostatin and what triggers it’s release

Answer 5

Somatostatin is a peptide hormone secreted by delta cells of the pancreas. It inhibits both insulin and glucagon and its release is triggered by high glucose and amino acid levels.

Question 6

What cells secrete somatostatin?

Answer 6

Somatostatin is a peptide hormone secreted by delta cells of the pancreas. It inhibits both insulin and glucagon and its release is triggered by high glucose and amino acid levels.

Question 7

CMIIs are typically inherited in an autosomal recessive fashion in which 75% of the alleles in the population are normal. What proportion of the population would exhibit CMIIs?

A. 6%

B. 14%

C. 25%

D. 56%

Answer 7

A. 6%

This question requires one to know the Hardy-Weinberg formulas:

p + q = 1 (alleles in a population)

and p² + 2pq + q² = 1 (phenotypes in a population)

where p is the frequency of the dominant allele and q is the frequency of the recessive allele. The figure below shows a Punnett square emphasizing the relationship between phenotypes and alleles.

The problem states that p = 0.75. Thus, q = 1 - 0.75 = 0.25. The proportion of the population exhibiting an autosomal recessive phenotype would then be equal to q² = 0.25² = 0.0625 = 6.25%. Choice A is closest.

Question 8

Shown here is the Developmental quotient for neurological deficiencies.

All of the following DQs could reasonably exist EXCEPT:

A. 0.25

B. 1

C. 85

D. 220

Answer 8

A. 0.25

As illustrated in Equation 1, the developmental quotient is a ratio of two positive numbers multiplied by 100. In order to arrive at a DQ value of 0.25, the age ratio would need to be 0.0025. Assuming a human could live to 150 and only progressed to the level of a 1-year-old, their DQ would be 1/150 x 100 = .66. It is unreasonable to expect a human to live so long, and a DQ of 0.25 would be even less reasonable. Therefore, no fractional DQ is likely to be calculated.

B. 1 - This could exist (e.g., a 100-year-old with the mental development of a 1-year-old).

C. 85 -This could exist (e.g., a 20-year-old with the developmental age of 17), and values very close to this exist in Figure 1.

D. 220 -This could exist (e.g., a 10-year-old with the mental development of a 22-year-old).

Question 9

The protein sequence of a particular human gene begins with Met-Gly-Leu-Leu-Val-Pro, and that of its bacterial analogue starts with Met-Gly-Leu-Val-Val-Pro. Which type of mutation is most likely responsible for the discrepancy between the two sequences?

A. Conservative mutation

B. Nonsense mutation

C. Silent mutation

D. Frameshift mutation

Answer 9

A. Conservative mutation

There is only a single amino acid difference between the two sequences, and moreover, the mutation resulted in a similar amino acid (Leu > Val, both hydrophobic with a alkane side chain). Therefore, it is most likely a single base pair that has been changed in the DNA. Such mutations are known as point mutations, and more specifically, mutations that change an amino acid to another similar one are known as conservative. Types of point mutations are shown below:

Question 10

The Markov model formalism is a bioinformatic tool that is used to search for a specific peptide sequence. For which of the following hormones would this test be most applicable?

A. Cortisol

B. Aldosterone

C. Glucagon

D. Estrogen

Answer 10

C. Glucagon

All of the choices are steroid hormones except glucagon, which is a peptide hormone.

Question 11

In a remote hospital in Asia, a patient is suspected to have dengue fever, a life-threatening disease that causes increased vascular permeability. As a result, her physician can expect to find:

I. generalized edema.

II. decreased heart rate.

III. increased renin levels.

IV. decreased blood pressure.

A. III only

B. I and II only

C. I, II, and IV only

D. I, III, and IV only

Answer 11

D. I, III, and IV only

If our patient has increased vascular permeability, her capillaries will leak fluid into extracellular compartments. This fluid excess will show up as edema (swelling) (I). A drop in blood pressure will be sensed by the kidneys, which will activate the renin-angiotensin-aldosterone system. This will cause fluid retention in an attempt to elevate blood pressure (III). Now, we need to think about what’s going to happen to the blood pressure with all that fluid leaving the circulation: blood pressure will drop (IV).

Question 12

Hepatic scarring that develops as a result of cirrhosis can lead to progressive loss of liver function. All of the following processes may be impaired due to cirrhosis EXCEPT:

A. cholesterol production.

B. blood clotting.

C. bilirubin excretion.

D. digestive enzyme synthesis.

Answer 12

D. digestive enzyme synthesis.

With EXCEPT questions, we often need to use process of elimination to arrive at the correct answer. Since the question asks about liver complications, we can deduce that the disease can interfere with, among other things, hepatic cholesterol and clotting factor synthesis as well as bilirubin (the breakdown product of heme) conjugation and excretion. No digestive enzymes are synthesized in the liver.

Bile, which is synthesized in the liver and is involved in lipid digestion, is not a digestive enzyme but an emulsifier.

Question 13

Oxocarbenium is shown on the left. A student wishes to dissolve the oxocarbenium for later analysis. Which solvent would be best for this process?

A. Ph-CH₂CH₃

B. CH₃(CH₂)₄CHCH₂

C. CH₃CH₂CH₂CN

D. CCl₄

Answer 13

C. CH₃CH₂CH₂CN

This question is testing our ability to use standard organic separation techniques. Solubility follows the rule that like dissolves like.

The oxocarbenium, shown in Fig 2, is a charged molecule. Charged and polar molecules will have a preference for, and will dissolve in, other charged/polar solvents. We need to identify a charged or polar molecule. Choice C shows a polar solvent, as shown below:

Question 14

Describe constitutional isomers

Answer 14

Constitutional isomers are compounds that have the same molecular formula and different connectivity, as shown below:

Question 15

What is the difference between a prosthetic group and a cofactor?

Answer 15

A cofactor is a non-protein component of a molecule required for the molecule’s biological activity. When a cofactor is tightly bound it is often referred to as a prosthetic group.

Prosthetic groups are tightly-bound organic molecules, such as vitamins, or inorganic compounds, such as metal ions, that are required for the hormonal or enzymatic function of a molecule. They are not the protein component.

Question 16

Gene analysis indicating increased expression of which enzyme by mice fed an high-fat diet would most support the finding of increased hepatic gluconeogenesis in these mice?

A. PEP carboxykinase

B. Pyruvate kinase

C. Phosphofructokinase-1

D. Glucokinase

Answer 16

A. PEP carboxykinase

Phosphoenolpyruvate carboxykinase (PEPCK) is an enzyme in gluconeogenesis. It converts oxaloacetate into phosphoenolpyruvate and carbon dioxide. Of the choices listed, it is the only enzyme that catalyzes a gluconeogenic reaction; thus, only an indication of its increased expression would support a finding of increased hepatic gluconeogenesis.

Question 17

Blood type in humans is a classic example of codominant inheritance. An AB⁺ male has a child with an O⁺ female. What is the likelihood that their first child will be able to receive a blood transfusion from a B^- donor?

A. 25%

B. 50%

C. 75%

D. 100%

Answer 17

B. 50%

The child has a 50% chance of inheriting his father’s A allele, and a 50% chance of inheriting his father’s B copy. On the mother’s side, the child is guaranteed to inherit an O allele (as type O individuals are homozygous for that recessive allele). In other words, the child has a 50% chance of being type A and a 50% chance of being type B. Only a type B individual can receive blood from a type B donor without complications, so the answer is 50%. The presence of the Rh factor is irrelevant here, as the donor lacks the factor. The Punnett square for this scenario is shown below:

Question 18

How many stereoisomers does the antibiotic doxycycline (shown below) have?

A. 0

B. 32

C. 64

D. 128

Answer 18

D. 128

Stereoisomers are molecules that have the same atoms bonded in the same way, but differ in the arrangement of those atoms in 3D space. This definition includes both enantiomers (compounds that are mirror images of each other) and diastereomers (compounds that are not mirror images of each other). To calculate the number of stereoisomers, we need to know the number of stereogenic centers, which includes chiral centers (R, S configuration) and double bonds (which can be assigned an E or Z configuration). Note that double bonds with two identical substituents bonded to either carbon do not count.

In doxycycline (see diagram below), there are 7 stereogenic centers: 6 chiral centers and 1 double bond. The double bond is in the Z configuration because, according to the priority rules, we assign the “upper” ring path (C(=O)C-OH) a higher priority than the “lower” ring path (C(=O)C-N(CH₃)₂).

The number of stereoisomers can be calculated by 2ⁿ - (# of meso compounds), where n is the number of stereogenic centers. Due to doxycycline’s asymmetry, there are no meso compounds, so the number of stereoisomers is 2⁷ = 128.

Question 19

All of the following are differences between the genetic material in a parasitic bacterium and that of its eukaryotic hosts EXCEPT:

A. supercoiling of DNA.

B. presence of introns.

C. size.

D. number of origins of replication.

Answer 19

A. supercoiling of DNA.

Both eukaryotic and prokaryotic DNA are supercoiled to save space. Eukaryotes have an additional level of supercoiling achieved through histones to squeeze more DNA into their nuclei.

Question 20

Compare eukaryotes and prokaryotes, in regard to the number of origins of replication each has.

Answer 20

Bacterial chromosomes, owing to their shape and size, have only one origin of replication.

Each eukaryotic chromosome has multiple origins of replication to speed up the process of copying all that DNA.

Question 21

Subsequent solvation studies employing other haloalcohols revealed the strongest interaction between Trp19 in monomeric melittin and the hydrophobic component of what solvent molecule?

A. 2,2-difluoroethanol

B. 2-fluoroethanol

C. 2,2-dichloroethanol

D. 2-chloroethanol

Answer 21

D. 2-chloroethanol

The four molecules contained in the answer choices differ in the number and strength of electron-withdrawing halogen atoms present. Specifically, as the number and electronegativity of the halogens present decreases, the polarity of the molecule will decrease as well, and its hydrophobicity will increase. This will strengthen the interaction of the hydrophobic component of the molecule with Trp19.

On this basis, 2-choloroethanol should be most hydrophobic and have the strongest interactions with Trp19. The ordering of these solvents by priority is shown below:

Question 22

It is likely that a substance that interferes with the effects of steroid hormones would do so by interacting with ________ receptors.

Answer 22

It is likely that a substance that interferes with the effects of steroid hormones would do so by interacting with nuclear receptors.

Question 23

It is likely that a substance that interferes with the effects of peptide hormones would do so by interacting with __________ receptors.

Answer 23

It is likely that a substance that interferes with the effects of peptide hormones would do so by interacting with membrane-bound receptors.

Question 24

What is the role of Human chorionic gonadotropin?

Answer 24

Human chorionic gonadotropin is secreted by the placenta after implantation, allowing for the maintenance of the corpus luteum, which secretes progesterone during the first trimester of pregnancy.

It plays no role in the development of female secondary sex characteristics or puberty.

Question 25

What is the role of Human chorionic gonadotropin in puberty?

Answer 25

It plays no role in the development of female secondary sex characteristics or puberty.

Human chorionic gonadotropin is secreted by the placenta after implantation, allowing for the maintenance of the corpus luteum, which secretes progesterone during the first trimester of pregnancy.

Question 26

_________ is involved in the development of female secondary sex characteristics during puberty.

Answer 26

Estrogen is involved in the development of female secondary sex characteristics during puberty.

Question 27

What mechanism of gene transfer is most likely to be responsible for the transfer of genes from bacteria to a eukaryotic host?

A. Conjugation

B. Transformation

C. Nuclear-phage recombination

D. Transposable elements

Answer 27

C. Nuclear-phage recombination

The mechanism underlying nuclear-phage recombination can be deduced from the term “phage,” which indicates that this mechanism involves a virus acquiring bacterial genetic material and transferring it to a eukaryotic host.

Question 28

Which of the following is the LEAST plausible explanation for a patient experiencing hypoglycemia?

A. Accidental self-injection of excess insulin

B. Deficient gluconeogenesis

C. Frequent and prolonged fasting

D. Increased rate of glycogenolysis

Answer 28

D. Increased rate of glycogenolysis

Glycogenolysis, the breakdown of glycogen, yields glucose 6-phosphate, which in turn gets converted to glucose. Therefore, increased rate of glycogenolysis leads to increased blood glucose level.

Question 29

Which of the following proteins found in the blood is likely to have significant free fatty acid binding capability?

A. Hemoglobin

B. Fibrinogen

C. Immunoglobulins

D. Albumin

Answer 29

D. Albumin

Serum albumin is the main carrier of free fatty acids in the blood.

Question 30

In order for immersion of a neuron into a KCl solution to induce depolarization, what must be true of the relative ion permeability of the neuronal membrane?

A. It is low for K⁺ and low for Cl^–.

B. It is low for K⁺ and high for Cl^–.

C. It is high for K⁺ and high for Cl^–.

D. It is high for K⁺ and low for Cl^–.

Answer 30

D. It is high for K⁺ and low for Cl^–.

In order for KCl to depolarize the cell membrane, it must modify the cell’s resting potential in such a way as to make the interior of the neuron more positive relative to the extraneuronal environment. Of the choices given, the only arrangement that would guarantee this change is if the permeability of the neuronal membrane is high for K⁺ and low for Cl^–. If so, inward K⁺ will increase the neuronal interior’s net positive charge, while the increase in extraneuronal Cl^– concentration will make the neuronal interior more positive relative to the increased extraneuronal negative charge concentration.

Question 31

A mutant L-FABP protein (which normally binds the anionic component of LCFA’s) with greatly reduced LCFA binding affinity differs from wild-type LCFA at a single critical amino acid residue. Which of the following substitutions is most likely to account for this change in L-FABP-LCFA binding?

A. Lys → Thr

B. Asp → Ser

C. Glu → Asp

D. Val → Tyr

Answer 31

A. Lys → Thr

The passage states that L-FABP binds the anionic component of LCFAs. An anionic region is best stabilized by a cationic residue. At physiological pH, only basic proteins have cationic side chains. The substitution shown in choice A, Lys → Thr, represents the replacement of a basic amino acid with a polar non-ionizable residue. This would reduce this stabilizing effect and decrease L-FABP’s ability to bind LCFAs.

Question 32

If miRNA is classified as non-coding RNA, then which of the following RNA molecules is LEAST likely to be classified in the same manner?

A. tRNA

B. snRNA

C. snoRNA

D. hnRNA

Answer 32

D. hnRNA

We are looking for the RNA molecule that is not noncoding (or basically, the RNA molecule that is coding), which means that it is translated into protein.

hnRNA is heterogeneous RNA, which is a precursor to mRNA. mRNA is the only coding RNA.

Question 33

What is snRNA?

Answer 33

snRNA, or small nuclear RNA, is a class of RNA molecules found within the nucleus of eukaryotic cells associated with specific proteins, referred to as snRNPs.

Question 34

What is snoRNA?

Answer 34

snoRNA, or small nucleolar RNA, are a subset of snRNA and play an essential role in RNA biogenesis and the modifications of rRNA and tRNAs.

Question 35

Consider a mixture of lysine, histidine, and aspartate in a buffered solution at pH 4.0. You perform an isoelectric focusing procedure on these amino acids. What would be the order of the amino acids in terms of migration distance from greatest to smallest?

A. Lysine > Aspartate > Histidine

B. Aspartate > Histidine > Lysine

C. Aspartate > Lysine > Histidine

D. Lysine > Histidine > Aspartate

Answer 35

D. Lysine > Histidine > Aspartate

You should know that lysine and histidine are basic amino acids, but histidine has a pI around 7 since it is a common residue used to transfer protons in enzymes. Lysine has a pI in the basic range. You should recognize aspartate as an acidic amino acid that would have a pI in the acidic range. At pH = 4.0, we are closest to the pI of aspartate, then histidine, and finally lysine.

This can be reasoned without knowing the actual values. Here are the actual pI values for the amino acids:

pI of lysine: 9.74

pI of histidine: 7.59

pI of aspartate: 2.77

The structures of these compounds, as well as the pKa values that are relevant for the pI, are shown below.

The final migration distance will be: lysine > histidine > aspartate

Remember LAH (like the strong base Lithium Aluminum Hydride) is the acronym for the basic amino acids: Lysine, Arginine, and Histidine.

Question 36

All of the following are glia in the CNS EXCEPT:

A. oligodendrocytes.

B. astrocytes.

C. ependymal cells.

D. Schwann cells.

Answer 36

D. Schwann cells.

Schwann cells make and maintain the myelin sheaths of peripheral nerves. In other words, they exist in the PNS, not the CNS.

Question 37

Describe the role of ependymal cells.

Answer 37

Ependymal cells make cerebrospinal fluid.

Question 38

Describe a role of astrocytes.

Answer 38

Astrocytes are a key component of the integrity of the blood-brain barrier.

Question 39

Describe the role of oligodendrocytes.

Answer 39

Oligodendrocytes form and maintain myelin in the CNS