Biochem Finals (Batch 2027) Flashcards

1
Q

Which transporter of lipid is lowest in density but largest in size, containing the highest percentage of lipid and the smallest amount of protein?
A. Chylomicron
B. VLDL
C. LDL
D. HDL

A

a. Chylomicron
Rationale: Chylomicrons are lipoproteins that transport dietary lipids from the intestines to other locations in the body. They are the largest and least dense of the lipoprotein particles, consisting mostly of triglycerides and having a very small protein content.

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2
Q

A plasmid consisting of its own DNA with foreign DNA inserted into it is called:
A. Junk DNA
B. Recombinant DNA
C. Non-coding DNA
D. None of the above

A

B. recombinant DNA
Rationale: Recombinant DNA is formed by combining DNA from two different sources, creating new genetic combinations.

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2
Q

Endonucleases, a group of enzymes, cleave DNA:
A. Externally
B. Internally
C. Both A and B
D. Neither A nor B

A

B. Internally
Rationale: Endonucleases cleave the phosphodiester bond within a polynucleotide chain, acting internally rather than at the ends.

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3
Q

The first human protein produced through recombinant DNA technology is:
A. Somatostatin
B. Erythropoietin
C. Interferon
D. Insulin

A

A. somatostatin

Rationale: While insulin is one of the most well-known and impactful proteins produced using recombinant DNA technology, the first human protein to be produced in this manner was actually somatostatin. Somatostatin, a smaller peptide hormone that regulates the endocrine system, was successfully synthesized using recombinant DNA techniques in the early 1970s before the production of recombinant insulin.

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3
Q

An example of an autonomously replicating mini chromosome is:
A. Virus
B. Plasmid
C. Phage
D. Lichen

A

B. plasmid
Rationale: Plasmids are small, circular, double-stranded DNA molecules found in bacteria that replicate independently of the chromosomal DNA.

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3
Q

What is the metabolic fate of the carbon skeleton of amino acids upon degradation?
A. Used for ATP synthesis
B. Used for fatty acid synthesis
C. Used for glucose synthesis
D. All of the above
E. None of the above – it is converted to urea

A

D. All of the above
Rationale: The carbon skeletons of amino acids can be used for ATP synthesis, fatty acid synthesis, and glucose synthesis, depending on the body’s needs and metabolic state.

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3
Q

Prior to the production of recombinant insulin, insulin obtained from cows and pigs was given to patients. Some problems faced by this treatment were:
A. The insulin was not active
B. In some humans, it induced antibody production
C. It reduced the weight of patients
D. Loss of memory power

A

B. in some humans it induced antibody production
Rationale: Insulin from cows and pigs sometimes caused allergic reactions or antibody production in humans due to slight differences in the protein structure compared to human insulin.

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4
Q

Electrophoresis, a technique used in DNA fingerprinting, helps to separate:
A. Tissues
B. DNA segments
C. Cells from DNA
D. RNA from DNA

A

B. DNA segments
Rationale: Electrophoresis is used to separate DNA fragments based on their size and charge, a crucial step in DNA fingerprinting.

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4
Q

A segment of DNA that reads the same forward and backward is called:
A. Plasmid DNA
B. Palindromic DNA
C. Complementary DNA
D. Copy DNA

A

B. palindromic DNA
Rationale: Palindromic DNA sequences read the same in both directions (5’ to 3’ and 3’ to 5’), and are often recognition sites for restriction enzymes.

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4
Q

cDNA, a term used in recombinant DNA technology, means:
A. Competitive DNA
B. Complementary DNA
C. Chemical DNA
D. Complex DNA

A

B. complementary DNA
Rationale: Complementary DNA (cDNA) is synthesized from an mRNA template using the enzyme reverse transcriptase, often used in cloning eukaryotic genes in prokaryotes.

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4
Q

What is an important function of fiber?
A. It adds flavor to food
B. It aids in the absorption of fat-soluble vitamins
C. It can bind toxigenic substances
D. It is an important energy source in the absence of lipids

A

C. It can bind toxigenic substances
Rationale: Dietary fiber can bind to and help eliminate toxins from the body, aiding in digestive health.

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5
Q

How many nitrogen atoms does urea have?
A. One
B. Two
C. Three
D. Four

A

B. Two
Rationale: Urea has two nitrogen atoms in its structure, which is why it is an efficient way for the body to excrete excess nitrogen.

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6
Q

A defect in the enzyme carbamoyl phosphate synthase I results in:
A. Hyperammonemia
B. Hyperuricemia
C. Hypercitrullinemia
D. Hyperammonuria

A

A. Hyperammonemia
Rationale: Carbamoyl phosphate synthase I is involved in the urea cycle, which helps to remove ammonia from the body. A defect in this enzyme leads to the accumulation of ammonia, resulting in hyperammonemia.

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7
Q

Amino acid oxidases remove nitrogen from amino acids in the form of:
A. Urea
B. Ammonia
C. Uric acid
D. Glutamate

A

B. Ammonia
Rationale: Amino acid oxidases catalyze the oxidative deamination of amino acids, producing ammonia as a byproduct.

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8
Q

In amino acid catabolism, the α amino nitrogen is:
A. Used as a substrate for glycogenesis
B. Converted to a less toxic substance in the liver
C. Excreted in the form of uric acid in humans
D. Used as a substrate for fatty acid synthesis

A

B. converted to a less toxic substance in the liver
Rationale: The α amino nitrogen is converted to urea in the liver, a less toxic substance that can be excreted by the kidneys.

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9
Q

Glutamate dehydrogenase catalyzes what type of reaction?
A. Reductive biosynthesis
B. Transketolation
C. Methyl group transfer
D. Oxidative deamination

A

D. Oxidative deamination
Rationale: Glutamate dehydrogenase catalyzes the oxidative deamination of glutamate, producing α-ketoglutarate and ammonia.

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10
Q

A patient was said to be gaining nitrogen and is in a state of positive nitrogen balance. Which condition fulfills this criteria?
A. An elderly patient in a cachectic state
B. Child undergoing a growth spurt
C. Infant with protein energy malnutrition
D. A post-partum mother

A

B. Child undergoing a growth spurt
Rationale: A positive nitrogen balance occurs when nitrogen intake exceeds nitrogen loss, which is typical during periods of growth, such as in a child undergoing a growth spurt.

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11
Q

Absence of this enzyme causes dietary cellulose to remain undigested.
A. Glucoamylase
B. Lactase B-glycosidase
C. Sucroamylase
D. Trehalase

A

b. lactase B-glycosidase

Rationale: Lactase β-glycosidase is not involved in the digestion of cellulose. In fact, none of the listed enzymes (glucoamylase, lactase β-glycosidase, sucroamylase, or trehalase) are responsible for breaking down cellulose. Humans lack the enzyme cellulase, which is necessary to digest cellulose. Therefore, the absence of cellulase is the reason dietary cellulose remains undigested in humans. However, since cellulase is not listed as an option, the question might be misleading as none of the provided enzymes can digest cellulose.

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12
Q

Procarcinogens need to be activated by an enzyme to become which of the following?
A. True carcinogens
B. Pre-carcinogens
C. Ultimate carcinogens
D. All of the above

A

C. Ultimate carcinogens
Rationale: Procarcinogens are converted to active carcinogens, known as ultimate carcinogens, through metabolic activation by enzymes.

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13
Q

A mechanism of oncogene activation where a piece of one chromosome is split off and joined to another, as seen in Burkitt’s lymphoma.
A. Promoter insertion
B. Enhancer insertion
C. Gene amplification
D. Chromosomal translocation

A

D. Chromosomal translocation
Rationale: Burkitt’s lymphoma is characterized by a chromosomal translocation, where a part of chromosome 8 is transferred to chromosome 14, leading to the activation of the MYC oncogene.

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14
Q

Calcitonin is a tumor biomarker that is associated with:
A. Medullary carcinoma of the thyroid
B. Myeloma
C. Hepatocellular carcinoma
D. Germ cell tumor

A

A. Medullary carcinoma of thyroid
Rationale: Calcitonin is produced by the parafollicular C cells of the thyroid gland and is a specific marker for medullary thyroid carcinoma.

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15
Q

Imatinib, a tyrosine kinase inhibitor, belongs to which of the following classes of drugs?
A. Inhibitors of hormone receptors
B. Monoclonal antibodies
C. Inhibitors of signal transduction
D. Anti-angiogenesis agents

A

C. Inhibitors of signal transduction
Rationale: Imatinib (Gleevec) is a tyrosine kinase inhibitor that specifically inhibits the BCR-ABL tyrosine kinase, which is involved in signal transduction pathways that lead to cancer cell proliferation.

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16
Q

Loss of tumor suppression in a cell usually results from:
A. A deletion of a tumor suppressor gene
B. A translocation of a tumor suppressor gene
C. An inversion involving a tumor suppressor gene
D. Cytokine activation of a tumor suppressor gene

A

A. A deletion of a tumor suppressor gene
Rationale: Loss of function in tumor suppressor genes often occurs due to deletions or mutations, leading to uncontrolled cell growth and cancer development.

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17
Q

Growth of new blood vessels in and around tumors is called:
A. Invasiveness
B. Dedifferentiation
C. Metastasis
D. Angiogenesis

A

D. Angiogenesis
Rationale: Angiogenesis is the process by which new blood vessels form from pre-existing vessels, providing tumors with the necessary oxygen and nutrients to grow.

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18
Q

Which hormone/s will be released upon stimulation of Thyroid Stimulating Hormone (TSH)?
A. Thyroxine
B. Triiodothyronine
C. Thyrotropin-releasing hormone
D. Only A and B

A

D. Only A and B
Rationale: Thyroid-stimulating hormone (TSH) stimulates the thyroid gland to produce and release thyroxine (T4) and triiodothyronine (T3).

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19
Q

The following are true of steroid hormone action, EXCEPT:
A. Hydrophobic steroid hormones bound to plasma protein carriers diffuse into the target cell
B. Steroid hormone receptors are in the cytoplasm or nucleus
C. The receptor-hormone complex binds to DNA and affects the genes
D. Activated genes produce new mRNA that moves back into the cytoplasm

A

a. Hydrophobic steroid hormones bound to plasma protein carriers diffuse into the target cell
Rationale: Hydrophobic steroid hormones do not diffuse into target cells while bound to plasma protein carriers. They must first dissociate from these carriers to enter the cell.

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20
Q

TRUE of primary hypersecretion due to a problem with the adrenal cortex:
A. Low CRH levels
B. Low ACTH levels
C. High cortisol levels
D. All of the above

A

d. All of the above
Rationale: In primary hypersecretion due to an adrenal cortex problem, cortisol levels are high, which leads to feedback inhibition resulting in low levels of CRH (corticotropin-releasing hormone) and ACTH (adrenocorticotropic hormone).

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21
Q

After an overnight fast, a diabetic woman feels nauseous and skips breakfast but takes her insulin shot. This results in:
A. Increased glycogenolysis
B. Hypoglycemia
C. Increased lipolysis
D. Glycosuria

A

b. Hypoglycemia
Rationale: Taking insulin without eating can lead to a significant drop in blood glucose levels, resulting in hypoglycemia.

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22
Q

Hormonal stimulation of the formation of the second messenger IP3 leads to the release of which other intracellular messenger?
A. Cyclic AMP
B. Prostaglandin
C. Calcium
D. Leukotriene

A

c. Calcium
Rationale: Inositol triphosphate (IP3) stimulates the release of calcium ions from the endoplasmic reticulum, acting as a second messenger in various signaling pathways.

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23
Q

The following hormones use cAMP as a second messenger, EXCEPT:
A. FSH
B. LH
C. Glucagon
D. Estrogen

A

d. Estrogen
Rationale: Estrogen primarily acts through intracellular receptors that directly influence gene transcription, rather than using cAMP as a second messenger.

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24
Q

Hormones that act on cells near the secreting cell are classified as:
A. Endocrine
B. Paracrine
C. Autocrine
D. Neurocrine

A

b. Paracrine
Rationale: Paracrine signaling involves the release of hormones that act on nearby cells, unlike endocrine signaling which involves hormones traveling through the bloodstream to distant cells.

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25
Q

Biotin is involved in which of the following types of reactions?
A. Hydroxylations
B. Dehydrations
C. Decarboxylations
D. Carboxylations

A

D. Carboxylations
Rationale: Biotin acts as a coenzyme for carboxylase enzymes, which are involved in carboxylation reactions.

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26
Q

If there is a dysfunction in the oxidase enzyme system (like cytochrome oxidase, ascorbic acid oxidase), which mineral is most likely affected?
A. Zinc
B. Copper
C. Chloride
D. Selenium

A

B. Copper
Rationale: Copper is a key component of many oxidase enzymes, including cytochrome oxidase and ascorbic acid oxidase.

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27
Q

A deficiency of vitamin B12 causes:
A. Cheilosis
B. Pernicious anemia
C. Beriberi
D. Scurvy

A

B. Pernicious anemia
Rationale: Vitamin B12 deficiency can lead to pernicious anemia, a condition in which the body can’t make enough healthy red blood cells.

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28
Q

All the following vitamins give rise to cofactors that are phosphorylated in the active form, except:
A. Pyridoxine
B. Niacin
C. Lipoamide
D. Riboflavin

A

C. lipoamide
Rationale: Lipoamide is not a vitamin that gives rise to a phosphorylated cofactor; it is derived from lipoic acid and acts as a cofactor in its own right.

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29
Q

All the following statements describing vitamin K are true, except that:
A. It is synthesized by intestinal bacteria
B. It prevents thrombosis
C. It is obtained by eating spinach and cabbage
D. It is required for liver synthesis of prothrombin

A

B. it prevents thrombosis
Rationale: Vitamin K is essential for blood clotting and helps in the synthesis of clotting factors; it does not prevent thrombosis but rather supports the clotting process.

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30
Q

All the following statements are true of vitamin A, except:
A. It is synthesized in the skin
B. It promotes maintenance of epithelial tissue
C. It is used to treat severe acne
D. It promotes maintenance of vision

A

A. it is synthesized in skin
Rationale: Vitamin A is not synthesized in the skin; this statement is true for vitamin D. Vitamin A is obtained from the diet and is essential for vision, epithelial maintenance, and treatment of severe acne.

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31
Q

This is a toxic protein aggregate that is the hallmark of diseases such as Parkinson’s and Alzheimer’s Disease:
A. Amyloid
B. Schiff base
C. Amylopectin
D. Amadori intermediate

A

A. Amyloid
Rationale: Amyloid proteins aggregate to form plaques, which are associated with neurodegenerative diseases such as Parkinson’s and Alzheimer’s.

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32
Q

What is the role of telomeres in the aging process?
A. Telomeres cause a progressive slowing down of an individual’s heart rate when they accumulate with age
B. In theory, telomeres generate protein aggregates which when expressed will cause DNA damage
C. Telomeres generate Okazaki fragments, which are not detected by DNA polymerases during cell replication of DNA
D. Telomeres shorten after every cell division until senescence sets in

A

D. Telomeres shorten after every cell division until senescence sets in
Rationale: Telomeres protect chromosome ends but shorten with each cell division, leading to cellular aging and eventual senescence.

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33
Q

Which of the following is an important cofactor for vitamins C, E, and Glutathione Peroxidase in the neutralization of free radicals?
A. Copper
B. NADPH
C. Selenium
D. Bioflavonoids

A

C. Selenium

Rationale: Selenium is an essential cofactor for the enzyme glutathione peroxidase, which plays a critical role in protecting cells from oxidative damage by neutralizing free radicals. While NADPH is important for the regeneration of glutathione, it is not a cofactor for vitamins C, E, and glutathione peroxidase in the same direct sense as selenium.

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34
Q

True of cytochrome P450:
A. It is a selenium-containing enzyme
B. It catalyzes phase I of xenobiotic metabolism
C. It helps synthesize steroid hormones
D. It catalyzes phase II of xenobiotic metabolism

A

D. It helps synthesize steroid hormones
Rationale: Cytochrome P450 enzymes are involved in the synthesis of steroid hormones, as well as in the metabolism of xenobiotics. They do not contain selenium and are primarily involved in phase I metabolism.

35
Q

The main goal of xenobiotic metabolism is to:
A. Avoid liver catabolism to enable a higher bioavailability
B. Render the molecule water-soluble
C. Excrete the xenobiotic via the fecal route
D. Increase the lipophilic character of the xenobiotic for better cellular absorption

A

B. Render the molecule water-soluble
Rationale: The main goal of xenobiotic metabolism is to render the molecule more water-soluble, facilitating its excretion from the body, typically via the kidneys.

36
Q

In general, xenobiotic drugs are manufactured as which of the following to facilitate absorption?
A. Water-soluble
B. Fat-soluble
C. Highly polar
D. Hydrophilic

A

B. Fat-soluble
Rationale: Xenobiotic drugs are often designed to be fat-soluble (lipophilic) to facilitate absorption through cell membranes, which are composed of lipid bilayers.

37
Q

The hypervariable region of an antibody consists of:
A. Amino acids that form the antigen-binding site
B. Phospholipids that form the antigen-binding site
C. Oligosaccharides that form the antigen-binding site
D. A part of the constant region of heavy and light chains

A

A. amino acids that form antigen binding site
Rationale: The hypervariable regions of an antibody, also known as complementarity-determining regions (CDRs), consist of amino acids that form the antigen-binding site.

38
Q

Monomers of certain immunoglobulins are joined by:
A. Variable chain
B. Constant chain
C. J-chain
D. All of the choices

A

C. J-chain
Rationale: The J-chain (joining chain) links monomers of IgA and pentamers of IgM.

39
Q

Which of the following statements is true regarding the Fc region?
A. Fragment crystallization and is the constant region
B. Fragment constant and is the variable region
C. Fragment crystallization and is the variable region
D. Fragment crystallization and has both variable and constant regions

A

A. fragment crystallization & is the constant region
Rationale: The Fc region (fragment crystallizable) is the constant region of an antibody that interacts with cell surface receptors and complement proteins.

40
Q

Which is not a function of IgG?
A. Major antibody in serum
B. First antibody type produced against an antigen during the primary antibody response
C. Activates or fixes complement
D. Involved in opsonization

A

B. First antibody type produced against an antigen during the primary antibody response
Rationale: IgM is the first antibody produced during the primary immune response, not IgG. IgG is the major antibody in serum and is involved in complement activation and opsonization.

41
Q

Which is the most efficient complement-fixing class of antibody?
A. IgE
B. IgA
C. IgM
D. IgG

A

C. IgM
Rationale: IgM is the most efficient at fixing complement due to its pentameric structure, which allows for multiple binding sites for the complement system.

42
Q

Antigen-binding sites of an immunoglobulin are located in:
A. Light chain alone
B. Heavy chain alone
C. Fc region of the antibody
D. Fab regions of the antibody

A

D. Fab regions of the antibody
Rationale: The antigen-binding sites are located in the Fab (fragment antigen-binding) regions of the antibody, which include the variable regions of both the light and heavy chains.

43
Q

The final common pathway in the complement system involves which of the following?
A. The formation of the membrane attack complex
B. Increased vasodilation and permeability of capillary beds
C. Vasodilation and permeability of capillary beds
D. All of the above

A

A. the formation of the membrane attack complex
Rationale: The final common pathway in the complement system results in the formation of the membrane attack complex (MAC), which creates pores in the target cell membrane, leading to cell lysis.

44
Q

Activation of which factor provides an important link between the intrinsic and extrinsic pathways in the coagulation cascade?
A. Factor IV
B. Factor V
C. Factor VII
D. Factor X

A

D. Factor X
Rationale: Factor X is the convergence point of the intrinsic and extrinsic pathways in the coagulation cascade. Activation of Factor X leads to the conversion of prothrombin to thrombin, a key step in the formation of a blood clot.

45
Q

Which of the following molecules from the endothelium inhibits platelet aggregation by increasing levels of cAMP?
A. Nitric oxide
B. Prostacyclin
C. ADPase
D. Thrombomodulin

A

B. Prostacyclin
Rationale: Prostacyclin (PGI2) is a potent vasodilator and inhibitor of platelet aggregation, which works by increasing cyclic AMP (cAMP) levels in platelets.

46
Q

Which is not a feature of von Willebrand factor?
A. Glues platelets to damaged endothelium
B. Binds and protects Factor IX
C. Is secreted by endothelial cells into the plasma
D. All of the above

A

B. Binds and protects Factor IX
Rationale: von Willebrand factor (vWF) glues platelets to damaged endothelium and binds and protects Factor VIII, not Factor IX. It is also secreted by endothelial cells into the plasma.

47
Q

This stage of hemostasis is considered to be the primary hemostatic mechanism:
A. Vascular phase
B. Platelet phase
C. Coagulation phase
D. Bleeding phase

A

B. Platelet phase
Rationale: The platelet phase, involving the adhesion, activation, and aggregation of platelets to form a platelet plug, is considered the primary mechanism of hemostasis.

48
Q

Thrombosis occurs when the endothelium lining the blood vessels is damaged or removed. Which among the following is not characteristic of the resulting red thrombus from this event?
A. Predominantly composed of erythrocytes
B. Consists primarily of platelets and fibrin
C. It forms at the site of an injury or abnormal vessel wall, particularly in areas where blood flow is stagnant
D. Seen mostly in venous areas

A

B. Consists primarily of platelets and fibrin
Rationale: A red thrombus, which typically forms in venous areas, is predominantly composed of erythrocytes trapped in a fibrin mesh. Platelet-rich thrombi (white thrombi) are more common in arterial thrombosis.

49
Q

Which statement is incorrect?
A. Gene regulation is fundamental to cell specialization in multicellular organisms
B. Patterns of gene expression established during the early developmental stages are not permanent, enabling one cell type to differentiate into a different cell type later in the life cycle
C. Some gene family members are expressed at different phases of development
D. Post-transcriptional and post-translational processing events can regulate the synthesis rate of gene products

A

b. Patterns of gene expression established during the early developmental stages are not permanent, enabling one cell type to differentiate into a different cell type later in the life cycle
Rationale: This statement is incorrect because gene expression patterns established during early development are usually stable and contribute to cell specialization, preventing cells from changing type later in life.

50
Q

Which of the following statements is true of RNA interference?
A. Is a normal way for organisms to regulate gene expression
B. Is a mechanism for combating virus infection in plants
C. Occurs only in vertebrates
D. Is already used therapeutically in many disorders

A

a. is a normal way for organisms to regulate gene expression
Rationale: RNA interference (RNAi) is a natural process used by cells to regulate gene expression by degrading specific mRNA molecules or inhibiting their translation.

51
Q

Which of the following is true of the lac operon in E. Coli?
A. The operon is only switched on in the absence of lactose in the growth medium
B. The lac operon messenger RNA is a polycistronic mRNA
C. The enzyme β-galactosidase is only produced in large quantities when the lac repressor is bound to the operator
D. The promoter is the binding site for the lac repressor

A

b. The lac operon messenger RNA is a polycistronic mRNA
Rationale: The lac operon produces a single polycistronic mRNA that encodes multiple proteins involved in lactose metabolism.

52
Q

Which of the following statements regarding the regulation of trp operon expression by attenuation is correct?
A. The leader peptide sequence encodes enzymes related to tryptophan synthesis
B. The leader peptide sequence contains no tryptophan residues
C. Rapid translation of the leader peptide allows completion of the mRNA transcript
D. Rapid translation of the leader peptide prevents completion of the mRNA transcript

A

d. Rapid translation of the leader peptide prevents completion of the mRNA transcript
Rationale: In the trp operon, rapid translation of the leader peptide (indicating high tryptophan levels) causes the formation of a terminator structure in the mRNA, preventing completion of the transcript.

53
Q

In terms of lac operon regulation, what happens when E. coli is grown in medium containing glucose and lactose?
A. Both CAP and the lac repressor are bound to the DNA
B. CAP is bound to the DNA but the lac repressor is not
C. Lac repressor is bound to the DNA but CAP is not
D. Neither CAP nor the lac repressor gene is bound to the DNA

A

d. Neither CAP nor the lac repressor gene are bound to the DNA
Rationale: When both glucose and lactose are present, the catabolite activator protein (CAP) is not bound to the DNA due to low cAMP levels, and the lac repressor is not bound because lactose (or its isomer allolactose) inactivates it.

54
Q

A type of base pair substitution that converts purine-pyrimidine to a pyrimidine-purine:
A. Transition
B. Base analogs
C. Transversions
D. Viruses

A

C. Transversions
Rationale: A transversion mutation is a type of base pair substitution where a purine (A or G) is replaced by a pyrimidine (C or T), or vice versa.

55
Q

This mutation results from deletion or insertion of nucleotides in DNA that generates altered mRNAs:
A. Silent mutations
B. Missense mutations
C. Nonsense mutations
D. Frameshift mutations

A

D. Frameshift mutations
Rationale: Frameshift mutations occur when nucleotides are inserted or deleted from the DNA sequence, causing a shift in the reading frame of the mRNA and resulting in altered proteins.

56
Q

The type of mutation that only affects the individual in which the mutation arises:
A. Somatic mutations
B. Missense mutations
C. Germline mutations
D. Frameshift mutations

A

A. Somatic mutations
Rationale: Somatic mutations occur in non-reproductive cells and are not passed on to offspring, affecting only the individual in which the mutation occurs.

57
Q

This type of mutation may have no detectable effect because of the degeneracy of the genetic code:
A. Silent mutations
B. Missense mutations
C. Nonsense mutations
D. Frameshift mutations

A

A. Silent mutations
Rationale: Silent mutations do not alter the amino acid sequence of a protein due to the redundancy (degeneracy) of the genetic code, where multiple codons can encode the same amino acid.

58
Q

This type of mutation results in the premature termination of amino acid incorporation into a peptide chain and leads to a shorter polypeptide chain:
A. Silent mutations
B. Missense mutations
C. Nonsense mutations
D. Splice site mutations

A

C. Nonsense mutations
Rationale: Nonsense mutations introduce a premature stop codon, leading to the early termination of translation and resulting in a truncated, often nonfunctional protein.

59
Q

What is the role of the A site in translation?
A. Binds to the tRNA holding the growing polypeptide chain of amino acids.
B. Binds to incoming aminoacyl-tRNA which has the anticodon for the corresponding codon in the mRNA.
C. Binds to the amino acid that corresponds to the amino-terminal end of the growing peptide chain.
D. Binds to the tRNA on the acceptor arm of the translation exit site.

A

B. binds to incoming to aminoacyl-tRNA which has the anti-codon for the corresponding codon in the mRNA
Rationale: The A (aminoacyl) site of the ribosome binds to the incoming aminoacyl-tRNA, which carries the anticodon that matches the codon on the mRNA.

60
Q

The wobble hypothesis allows for non-standard base pairing between:
A. DNA template and RNA transcript
B. Codon and anticodon
C. Exons and introns
D. Template and coding strands

A

B. codon and anticodon
Rationale: The wobble hypothesis explains that the third base of the codon and the first base of the anticodon can form non-standard base pairs, allowing for flexibility in the genetic code.

61
Q

A mutation in the Shine-Dalgarno sequence will result in:
A. Termination of translation in bacteria
B. Termination of transcription in eukaryotes
C. Inhibition or activation of translation in bacteria
D. Inhibition of translation in eukaryotes

A

C. inhibition or activation of translation in bacteria
Rationale: The Shine-Dalgarno sequence is important for the initiation of translation in bacteria. A mutation in this sequence can inhibit or alter the efficiency of translation initiation.

62
Q

Which statement is true regarding eukaryotic protein synthesis?
A. Many ribosomes can translate the same mRNA molecule simultaneously.
B. Even before transcription is completed, translation of the mRNA can begin.
C. The 18s rRNA of the 50s ribosomal subunit is responsible for the peptidyltransferase activity that joins the amino acids.
D. Diphtheria toxin inhibits prokaryotic but not eukaryotic protein synthesis.

A

A. Many ribosomes can translate the same mRNA molecule simultaneously.

Rationale: In eukaryotic cells, multiple ribosomes can attach to a single mRNA molecule, forming a polysome, and translate it simultaneously, allowing for efficient protein synthesis.

The other statements are incorrect because:

B: In eukaryotes, transcription and translation are separated processes; translation does not begin before transcription is completed.
C: The 18S rRNA is part of the small ribosomal subunit (40S) in eukaryotes, not the 50S subunit (which is found in prokaryotes). The peptidyltransferase activity is associated with the 28S rRNA in the 60S large ribosomal subunit in eukaryotes.
D: Diphtheria toxin inhibits eukaryotic protein synthesis by inactivating elongation factor-2 (EF-2), not prokaryotic protein synthesis.

63
Q

Which of the following is the first anticodon read during translation?
A. UAB
B. UAC
C. GAG
D. AAG

A

B. UAC
Rationale: The first codon of the mRNA during translation is usually the start codon AUG, which is recognized by the anticodon UAC on the initiator tRNA carrying methionine.

64
Q

What is the role of the poly-A tail during translation initiation?
A. It stimulates recruitment of the 40s ribosomal subunit to mRNA
B. It recognizes the stop codons during the termination phase of translation
C. It binds to transfer RNA and facilitates peptide bond formation
D. It attaches to the methionine bound to the methylguanosine cap structure and initiates the transcription of the upstream nucleotide sequences

A

A. It stimulates recruitment of the 40s ribosomal subunit to mRNA
Rationale: The poly-A tail enhances the stability of the mRNA and aids in the recruitment of the ribosomal subunits during translation initiation. It interacts with the poly-A binding proteins and other factors to facilitate the assembly of the translation initiation complex.

65
Q

What molecule, once incorporated into the growing chain of DNA, halts replication?
A. Novobiocin
B. 2’, 3’ dideoxyadenosine triphosphate
C. 2’ deoxycytidine triphosphate
D. Actinomycin D

A

B. 2’, 3’ dideoxyadenosine triphosphate
Rationale: 2’, 3’ dideoxyadenosine triphosphate (ddATP) lacks the 3’-OH group necessary for forming a phosphodiester bond with the next nucleotide, thus halting DNA chain elongation.

66
Q

Primase is the enzyme responsible for:
A. Introducing nicks into the DNA double strand to prevent the formation of knots
B. Hydrolyzing ATP to facilitate DNA unwinding
C. Making short strands of RNA at the site of replication initiation
D. Forming a replication fork in the DNA double helix

A

C. making short strands of RNA at the site of replication initiation
Rationale: Primase synthesizes short RNA primers that are necessary for DNA polymerase to begin DNA synthesis.

67
Q

Okazaki fragments are joined together by:
A. RNA polymerase
B. DNA ligase
C. DNA polymerase
D. RNA ligase

A

B. DNA ligase
Rationale: DNA ligase is responsible for sealing the nicks between Okazaki fragments on the lagging strand, forming a continuous DNA strand.

68
Q

A replication fork is:
A. Only seen in bacterial cells
B. A Y-shaped structure where both DNA strands are replicated simultaneously
C. A site where one DNA strand serves as a template, but the other strand is not replicated
D. Created by the action of the enzyme RNA polymerase

A

B. a Y-shaped structure where both DNA strands are replicated simultaneously
Rationale: The replication fork is the Y-shaped region where the DNA double helix is unwound and both strands are replicated.

69
Q

Which of the following causes the unwinding of the DNA double helix?
A. DNA polymerase
B. DNA helicase
C. RNA primer
D. Primosome

A

B. DNA helicase
Rationale: DNA helicase unwinds the DNA double helix ahead of the replication fork, allowing the strands to be replicated.

70
Q

Which of the following best describes semiconservative replication?
A. The replication of DNA takes place at a defined period in the cell cycle
B. A DNA molecule consists of one parental strand and one new strand
C. The number of DNA molecules is doubled with every other replication
D. The replication of DNA never takes place with 100% accuracy

A

B. A DNA molecule consists of one parental strand and one new strand
Rationale: In semiconservative replication, each new DNA molecule consists of one old (parental) strand and one newly synthesized strand.

71
Q

At the 5’ end, mRNA molecules have a:
A. Poly (A) tail
B. TATA cap
C. 7-methylguanosine cap
D. GCCT tail

A

C. 7-methylguanosine cap
Rationale: The 5’ end of eukaryotic mRNA molecules is capped with a 7-methylguanosine cap, which is important for mRNA stability, splicing, and translation initiation.

72
Q

The RNA sequence that appears in mature RNAs:
A. Exons
B. Introns
C. Cap
D. Tail

A

A. Exons
Rationale: Exons are the coding sequences that remain in mature mRNA after introns have been spliced out.

73
Q

The TATA box in eukaryotes is located where?
A. 10 bp upstream from the transcription start site (TSS)
B. 35 bp upstream from the TSS
C. 25-30 bp upstream from TSS
D. 5 bp upstream from the TSS

A

C. 25-30 bp upstream from TSS
Rationale: The TATA box is a DNA sequence found about 25-30 base pairs upstream of the transcription start site and is important for the initiation of transcription by RNA polymerase II.

74
Q

The type of RNA polymerase that catalyzes the synthesis of mRNA:
A. Pol I
B. Pol II
C. Pol III
D. Pol IV

A

B. Pol II
Rationale: RNA polymerase II (Pol II) is responsible for transcribing mRNA in eukaryotic cells.

75
Q

This includes the promoter, RNA-coding region, and a terminator:
A. Transcription unit
B. Primary transcript
C. 5’ Flanking sequences
D. 3’ Flanking sequences

A

A. Transcription unit
Rationale: A transcription unit consists of the promoter, the RNA-coding region, and the terminator, encompassing the entire region transcribed into RNA.

76
Q

The most abundant of all types of RNA comprising about 80% of the total:
A. mRNA
B. tRNA
C. rRNA
D. snRNA

A

C. rRNA
Rationale: Ribosomal RNA (rRNA) is the most abundant type of RNA, making up about 80% of the total RNA in cells.

77
Q

This structural form of DNA has a left-handed helix that contains about 12 base pairs per turn:
A. A form
B. B form
C. C form
D. Z form

A

D. Z form
Rationale: Z-DNA is a left-handed helix with about 12 base pairs per turn.

78
Q

Which of these histones is not a part of the nucleosome core?
A. H1
B. H2A
C. H2B
D. H3

A

A. H1
Rationale: Histone H1 is not part of the nucleosome core; it binds to the linker DNA between nucleosomes, stabilizing the chromatin structure.

79
Q

Which statement is not true?
A. The single-stranded DNA has a higher relative absorbance at 260 nm wavelength than does double-stranded DNA
B. The melting temperature (Tm) is affected by the number of hydrogen bonds between bases
C. DNA that contains high concentrations of Adenine and Thymine denatures at a lower temperature than Guanine and Cytosine-rich DNA
D. There are three hydrogen bonds between Adenine and Thymine but only two between Guanine and Cytosine

A

D. There are three hydrogen bonds between Adenine and Thymine but only two between Guanine and Cytosine
Rationale: The statement is incorrect because there are two hydrogen bonds between Adenine (A) and Thymine (T), and three hydrogen bonds between Guanine (G) and Cytosine (C).

80
Q

True of Chargaff’s Rule, except:
A. In any sample of dsDNA molecules, the concentration of deoxyadenosine nucleotides equals that of thymidine nucleotides
B. The concentration of deoxyguanosine nucleotides equals that of deoxycytidine nucleotides
C. The concentration of deoxyadenosine nucleotides equals that of Uracil nucleotides
D. None of the above

A

C. the concentration of deoxyadenosine nucleotides equals that of Uracil nucleotides
Rationale: Chargaff’s Rule states that in any sample of double-stranded DNA, the concentration of adenine (A) equals thymine (T), and the concentration of guanine (G) equals cytosine (C). Uracil is found in RNA, not DNA.

81
Q

Thymine is present in:
A. DNA
B. RNA
C. DNA and RNA
D. Neither DNA nor RNA

A

A. DNA
Rationale: Thymine is one of the four nucleotides in DNA but is not present in RNA, where it is replaced by uracil.

82
Q

This is the first stage in the expression of genetic information:
A. Replication
B. DNA synthesis
C. Translation
D. Transcription

A

D. Transcription
Rationale: Transcription is the first stage in the expression of genetic information, where DNA is transcribed into RNA.

83
Q

What is the second messenger utilized in the regulation of the cytidyl transferase enzyme used in the synthesis of phospholipids?
A. cGMP
B. IP3
C. 1,2 DAG
D. cAMP

A

C. 1,2 DAG
Rationale: Diacylglycerol (1,2-DAG) is a second messenger involved in the regulation of cytidyl transferase, an enzyme in the synthesis of phospholipids.

84
Q

What are components of ceramide?
A. Sphingosine + fatty acid
B. Glycerol + fatty acid + phosphoric acid
C. Glycerol + fatty acid + phosphoric acid + nitrogenous base
D. Sphingosine + fatty acids + phosphoric acid

A

A. sphingosine + fatty acid
Rationale: Ceramides are composed of a sphingosine backbone linked to a fatty acid via an amide bond.

85
Q

What is the significance of L ratio determination of amniotic fluid in a pregnant female?
A. Fetal heart rate
B. Fetal lung maturity
C. Fetal head size
D. Expected date of delivery

A

B. fetal lung maturity
Rationale: The lecithin-to-sphingomyelin (L) ratio in amniotic fluid is used to assess fetal lung maturity, with a higher ratio indicating greater maturity and a reduced risk of respiratory distress syndrome.

86
Q

What is the major component of lung surfactant?
A. Dipalmitoyl lecithin
B. Dipalmitoylphosphatidyl serine
C. Dipalmtoylcephalin
D. Dipalmitoylphosphatidyl inositol

A

A. dipalmitoyl lecithin
Rationale: Dipalmitoyl lecithin, also known as dipalmitoylphosphatidylcholine (DPPC), is the major component of lung surfactant, which reduces surface tension in the alveoli.

87
Q

The enzyme that is deficient in Niemann-Pick’s disease?
A. Glucocerebrosidase
B. Hexosaminidase A
C. Sphingomyelinase
D. β-galactocerebrosidase

A

C. sphingomyelinase
Rationale: Niemann-Pick disease is caused by a deficiency of the enzyme sphingomyelinase, leading to the accumulation of sphingomyelin in cells.

88
Q

Which of the following is an intermediate for the synthesis of phospholipids and triacylglycerols?
A. Diacylglycerol
B. Cholesterol
C. Choline
D. Inositol

A

A. diacylglycerol
Rationale: Diacylglycerol is an intermediate in the synthesis of both phospholipids and triacylglycerols.

89
Q

Which of the following lipids is NOT part of a lipoprotein?
A. Phospholipids
B. Triacylglycerols
C. Esterified cholesterol
D. Unesterified cholesterol

A

d. unesterified cholesterol
Rationale: Unesterified cholesterol (free cholesterol) is indeed a part of lipoproteins. The correct answer should be different, but given the options, none are completely accurate.

90
Q

Which of the following is true about the metabolism of low-density lipoproteins (LDL)?
A. LDL receptors are positively charged glycoprotein molecules clustered in pits on cell membranes.
B. After binding, the LDL undergoes a process called exocytosis.
C. The pH of the contents of the endosomes increases, allowing separation of the LDL from its receptor, with receptors migrating to one side of the endosome and the LDLs staying free within the lumen of the vesicle.
D. The receptors can be recycled, whereas the lipoprotein remnants in the vesicles are degraded by lysosomal enzymes, releasing cholesterol, fatty acids, amino acids, and phospholipids.

A

d. The receptors can be recycled whereas the lipoprotein remnants in the vesicles are degraded by lysosomal enzymes releasing cholesterol, fatty acids, amino acids, and phospholipids.
Rationale: LDL receptors are indeed recycled, and the lipoprotein remnants are degraded by lysosomal enzymes, releasing their components.

91
Q

They are composed predominantly of triacylglycerol and carry this lipid to the peripheral tissues from the liver:
A. Chylomicron
B. VLDL
C. LDL
D. HDL

A

b. VLDL
Rationale: Very low-density lipoproteins (VLDL) are composed predominantly of triacylglycerol and transport lipids from the liver to peripheral tissues.

92
Q

As the lipid-to-protein ratio decreases, particles become smaller and more dense in the following order:
A. Chylomicron > HDL > VLDL > LDL
B. Chylomicron > VLDL > LDL > HDL
C. HDL > LDL > VLDL > Chylomicron
D. HDL > LDL > VLDL > Chylomicron

A

b. chylomicron > VLDL > LDL > HDL
Rationale: Lipoproteins decrease in size and increase in density in the order of chylomicrons, VLDL, LDL, and HDL as the lipid-to-protein ratio decreases.

93
Q

This apoprotein is found in the particles released by the intestinal mucosal cells called “nascent” chylomicrons:
A. ApoCII
B. ApoB48
C. ApoB100
D. ApoC

A

b. apoB48
- Rationale: ApoB48 is the apoprotein found in nascent chylomicrons, which are released by intestinal mucosal cells.