Biology

Question 1

The genetic code is said to be degenerate because there are 64 different codons, but translation produces only 20 unique amino acids. The degeneracy of the genetic code is due to which mechanism?

A) exclusion of protein coding regions from mature mRNA
B) errors during tRNA charging
C) ambiguity of the tRNA for the amino acid
D) nontraditional base pairing of the anticodon with the third base of the codon

Answer 1

D) nontraditional base pairing of the anticodon with the third base of the codon

Question 2

Which statement most accurately describes the role of T tubules in skeletal muscle cells?

A) t tubules bind actylcholine at the neuromuscular junction to generate a depolarizing stimulus
B) depolarizing current reaches the sarcoplasmic reticulum by traveling down t tubules
C) muscle contraction is driven by the sliding of t tubules across one another in the sarcomere
D) t tubules sequester Ca2+ out of the cytosol to prevent prolonged muscle contraction

Answer 2

B) depolarizing current reaches the sarcoplasmic reticulum by traveling down t tubules

For a skeletal muslle cell to contract, Ca2+ must be released into the cytosol from the sarcoplasmic reticulum. Ca2+ release is induced when a depolarizing current (action potential) runs along the sarcolemma and travels down the t tubules. This current causes the nearby sarcoplasmic reticulum to open its Ca2+ channels, allowing Ca2+ ions to flow into the cytosol and induce the sarcomeric actin-mypsin interactions required for muscle contraction

Question 3

If a viral antigen were to bind to the receptors on the surface of a B lymphocyte, which of the following immune responses would not occur?

A) division of the B lymphocyte into cells that differentiate to secrete antibodies
B) interaction of the B lymphocyte with other immune cells that stimulate B lymphocyte proliferation
C) secretion of toxins by the B lymphocyte that destroy nearby virus-infected cells
D) proliferation of cells that can recognize the antigen more rapidly in the event of future infection

Answer 3

C) secretion of toxins by the B lymphocyte that destroy nearby virus-infected cells

Activation of B lymphocytes by helper T cells:
1) pathogen binds to B lymphocyte receptor, is endoxytized, and presented on MHC II
B) helper T cell binds foreign antigen and releases cytokines that activate the B lymphocyte
C) activated B lymphocyte divides into many clones that differentiate into plasma cells or memory cells

Natural killer and cytotoxic t tells, not B lymphocytes, respond to antigens by releasing toxins that induce apoptosis in nearby infected cells

Question 4

E. coli K-12, a normal bacterial strain of the human microbiota, is able to cause disease by inducing enterocyte lesions after it is grown in media with EHEC bacteria that contain the F factor plasmid. What process most likely granted virulence to E. coli K-12?

A) Transformation
B) Transduction
C) conjugation
D) transfection

Answer 4

C) conjugation

conjugation = transfer of genetic information from one bacterial cell to another via direct contact. The donor cell contains the F (fertility) factor plasmid, a circular piece of DNA containing genes that direct the formation of the sex pilus. During conjugation, the sex pilus from the donor cell attaches to a recipient cell (one that does not contain F factor) and facilitates the transfer of a single strand of the F factor plasmid DNA to the recipient cell.

According to this question, E. coli K-12 is a nonvirulent bacterial strain. Nonetheless, co-culturing E. coli K-12 with EHEC bactera that contain F factor resulted in E. coli K-12 acquiring virulence. E. coli K-12 virulence most likely resulted from the direct transfer of LEE genes through conjugation.

Question 5

transformation

Answer 5

the cellular uptake of foreign DNA directly from the environment

Question 6

transduction

Answer 6

involves the DNA transfer from one bacterial cell to another by a bacteriophage ( a virus that infects bacteria). During assembly of bacteriophages inside an infected cell, bacterial DNA can become trapped within the capsid of newly created bacteriophages. Subsequent infection of other cells with these new bacteriophages results in the transfer of bacterial DNA into a new host

Question 7

transfection

Answer 7

the process by which genetic material, usually in the form of a plasmid, is introduced into eukaryotic cells

Question 8

EHEC is classified as a bacillus bacterium. Given this information, this type of classification is most likely based on the bacterium’s

Answer 8

morphology

bacilli = rod 
cocci = sphere 
spirilli = spiral

Question 9

What is the size of the pSKII plasmid after it is digested by EcoRI and XhoI and the clndk gene is inserted?

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Answer 9

3.841 base pairs

Figure 1 shows that prior to the insertion of cldnk, the pSKII plasmid contained 2,958 base pairs. Figure 2 shows that the distance between the EcoRI and XhoI restriction sites is 17 + 15 = 32 base pairs; these bases will be removed by the digest.

The passage states that the cldnk gene produces a processed transcript that is 915 base pairs long, so insertion of cldnk cDNA increases the plasmid size by 915 base pairs.

Therefore, the final size can be calculated as 2,958 - 32 + 915 = 3,841

Question 10

What observation could have led researchers to conclude that cldnk is most likely expressed in oligodendrocytes?

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A) a probe made of cldnk mRNA hybridized to RNA in cells that express Sox10
B) An RNA probe that was complementary to cldnk mRNA hybridized in cells that express plp1a
C) a cldnk cDNA probe obtained from oligodendrocytes bound to a microarray with higher affinity than the same probe obtained from other cells
D) a cDNA probe corresponding to cldnk hybrdiized to the nuclei of oligodendrocytes but not the cytosol

Answer 10

B) An RNA probe that was complementary to cldnk mRNA hybridized in cells that express plp1a

If cldnk is found in the same cells in plp1a, then cldnk must be expressed in oligodendrocytes

Question 11

Which experiment could confirm that the cldnk gene product is required for myelination?

A) compare myelin formation in wild-type zebrafish to formation in zebrafish that overexpress cloned cldnk
B) overexpress cldnk in zebrafish cells other than oligodendrocytes and observe whether those cells produce myelin
C) compare myeline formation in wild-type zebrafish to that in zebrafish with cldnk knocked out
D) overexpress a form of cldnk that inhibits the wild-type gene in cells other than oligodendrocytes and observe whether myelin forms

Answer 11

C) compare myeline formation in wild-type zebrafish to that in zebrafish with cldnk knocked out

Knockout organisms have one or more genes that have been made inactive. The biological function of a gene can be inferred by knocking it out and observing the differences between wild-type organisms and knockout organisms.

If cldnk is required for wild-type myelination of oligodendrocytes, then knocking the gene out should result in impaired or abnormal myelination. Therefore, comparing myelination of oligodendrocytes in wild-type zebrafish to myelination in cldnk knockout zebrafish would show whether cldnk is required for proper myelin formation

Question 12

How would expression of cloned cldnk differ from expression of the endogenous cldnk gene?

A) expression of the cloned gene does not involve splicing whereas expression of the endogenous gene does
B) the cloned gene is expressed at lower levels than the endogenous gene
C) the cloned gene produces a protein with a different sequence than the endogenous gene
D) expression of the cloned gene produces a mature mRNA molecule with a different coding region thatn expression of the endogenous gene

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Answer 12

A) expression of the cloned gene does not involve splicing whereas expression of the endogenous gene does

Genes within a eukaryotic chromosome typically consist of exons and introns, both of which are transcribed by RNA polymerase to form pre-mRNA. After transcription, pre-mRNA matures through several processing, including 5’ capping by 7-mehtylguanosine, 3’ polyadenylation, and splicing, to remove introns.

Complementary DNA (cDNA) is derived from mature mRNA through reverse transcription. Because the mature mRNA has already been spliced (introns were removed), the cDNA will not contain any introns. The cDNA can be cloned into expression vectors such as plasmids, which can be introduced into embyronic cells and integrated into the genome. These cloned genes can then be expressed in the organism. However, because the cloned gene does not contain any introns, its expression will not include splicing.

Question 13

Which of the following conclusions is best supported by the results presented in Figure 1?

A) MAGE mRNAs are all composed of the same number of nucleotides
B) MAGE protein classes targeted by immune cells may differ depending on the cancer cell type
C) MAGE-A6 exhibits the highest level of expression in cancer cells
D) MAGE proteins are more active in breast cancer cells than in melanoma cells

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Answer 13

B) MAGE protein classes targeted by immune cells may differ depending on the cancer cell type

(some are expressed and some aren’t)

Question 14

Which steps of the viral life cycle would most likely occur in lytic phages?

I. injection of viral genetic material into the bacterial host
II. integration of viral DNA into the bacterial genome
III. Degradation of the host genome

Answer 14

I and III only

Question 15

Lytic life cycle steps

Answer 15

1. Attachment: the bacteriophage contacts the bacterial cell wall and attaches to the host bacterium using its tail fibers
2. Viral genome entry: the phage uses its tail sheath to inject its genome into the cytoplasm of the bacterial host
3. Host genome degredation: viral enzymes degrade the host genome into its nucleotide components to provide the building blocks for replication of the viral genome
4. Synthesis : Loss of the bacterial (ie host) chromosome ends the synthesis of host molecules, now under the control of the viral genome. As a result, the host machinery (eg ribosomes) begins to synthesize the components needed for new viral progeny, which then assemble inside the host cell
5. Release: many newly assembled viral progeny (virions) are released as the bacterium disintegrates (lysis) due to the action of lysozymes on the host cell wall

Question 16

The scientists hypothesize that genetic leakage has occurred in some of the studied yeast lineages over time. Which of the following best supports this hypothesis?

A) hybrid indivodials express certain phenotypes that differ from those expressed in the parental species
B) hybrid individuals exhibit reproductive isolation with some other Saccharomyces species
C) hybrid individuals can produce some viable offspring when mated with their parental species
D) hybrid individuals frequently pass genetic information to clones of themselves via asexual reproduction

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Answer 16

C) hybrid individuals can produce some viable offspring when mated with their parental species

Genetic leakage: initially, inbreeding of two different parental species occurs, hybrid offspring contains genes from both species 1 and 2.
Over time, inbreeding of hybrid and parental species occurs. This leads to viable offspring which contains mostly parental species 1 genes and some parental species 2 genes
Gene flow will occur from hybrid species and parental species because hybrid genes continue to be transferred to the parental species

Question 17

Compared to crpssing Sp1 individuals with other Sp1 individuals, crossing Sp1 individuals with Sp4 individuals (Table 1) resulted in:

(how many more/fewer spores)

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Answer 17

Approximately 1/2 fewer viable spores

62% vs 29%

Question 18

The researchers hypothesize that the ability of marine teleost kidneys to filter salts is independent of the presence of a glomuerulus. Bassed on the passage, this hypothesis would best be supported if the researchers discovered:

A) higher urinary concentrations of Mg2+ and SO4 2- in the eel than in the goosefish
B) higher urinary concentrations of Cl- and Na+ in the eel than in the goosefish
C) equally high urinary concentrations of Mg2+ and So4 2- in the eel and the goosefish
D) equally high urinary concentrations of Cl- and Na+ in the eel and the goosefish

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Answer 18

C) equally high urinary concentrations of Mg2+ and So4 2- in the eel and the goosefish

In the passage, the passage states that scientists concluded that osmoregulation in marine teleosts involves:

1. seawater ingestion
2. retrieval / absorption of salt and water from intestines
3. excretion of divalent ions primarily through the urine and monovalent ions through the gills

To confirm that the glomerulus does not affect filtration of salts, scientists would have to find similar concentrations of the relevant salts in the urine of both glomeruluar and aglomerular fish. Because divalent ions (mg2+ and sor 2-) are primarily excreted in the urine and monovalent ions (Na+ and Cl-) are primarily excreted through the gills of marine teleosts, researchers would have to find equally high urinary concentrations of the divalent ions in the glomerular eel and the aglomerular goosefish .

Question 19

Which of the following would most likely cause an embryo to implant in a location other than the uterine lining?

A) reduced number of fallopian cilia
B) surge in lutenizing hormone prior to ovulation
C) incomplete gastrulation
D) rupture of follcile from the ovary

Answer 19

A) reduced number of fallopian cilia

Typically, fallopian cilia help propel the fertilized oocyte toward the uterus for implantation. However, with a reduced number for fallopian cilia, improper implantation of the fertilized egg can occur outside the uterus

Question 20

A mutant mouse is found to exhibit normal musculoskeletal function following the administration of anti-nAchR antibodies. Compared to a wild-type mouse, this mutant mouse is most likely to show which of the following changes at the neuromuscular junction?

A) decreased ACh degradation within the synapse
B) reduced expression of nAchR by skeletal muscle fibers
C) decreased reuptake of Ach degredation byproducts into the presynaptic neuron
D) Reduced release of Ach from the presynaptic vesicles

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Answer 20

A) decreased ACh degradation within the synapse

Based on the passage, anti-nAchR antibodies bind nAchR reversibly to block the interaction between Ach and nAchR, resulting in muscle weakness. Accordingly, administration of these antibodies in a wild-type mouse would temporarily prevent binding of Ach to its receptor. However, if a mutant mouse were to exhibit normal muscle contraction despite being treated with anti-nAChR antibodies, the most likely reason in this scenario is that ACh degredation within the mutant ‘s synapse would be decreased compared to the wild-type.

If ACH degredation is decreased, more ACh molecules will remain in the synapse for a longer time period and be able to bind nAChRs after these are no longer bound to anti-nAChR antibodies. This would result in muscle contraction that is delayed but still normal

ACh degradation within the neuromuscular junction is one process by which muscle fiber contraction is terminated. However, decreased Ach degredation would allow more Ach to remain in the synapse, and prolong stimulaiton of nAchR and subsequent muscle contraction

Question 21

To interpret the results of Experiement 2, researchers must assume that:

I. muscle samples were of uniform weight and size
II. no intermolecular interactions occurred between Ach and anti-AChR antibodies in INfusion C
III. co-infusion of ACH and anti-AChR antibodies was performed following ACH infusion

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Answer 21

I and II only

These are both examples of possible confounding variables

Question 22

To further study anti-nAChr antibody function, the researchers wanted to replicate the experiments described in the passage in other mammalian animal models, but their research proposal was rejected by other scientists. Why was the proposal rejected?

A) because resting muscle tension cannot be used as a control for post-infusion tension
B) because the electrolyte-rich baths were infused with inconsistent amounts of ACh
C) because skeletal muscles in the body would not be exposed to autoimmune antibodies
D) because animal models cannot be used to study human diseases

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Answer 22

B) because the electrolyte-rich baths were infused with inconsistent amounts of ACh

*multiply mL of ACH by 2.5 mg/Ml concentration

Question 23

A subset of aggressive cancers has a relatively high growth rate, leading to the formation of large tumours. An effective drug against fast-growing tumours would most likely not target which stage of the cell cycle?

Answer 23

G0

Question 24

A student adds 5 bacterial cells to a test tube containing fresh medium and incubates it for 3 hours at 37C. The bacterial population shows an initial lag phase of 20 minutes followed thereafter by a growth phase with a doubling time of 40 minutes. Given this, what is the approximate number of bacterial cells present at the end of the incubation period?

Answer 24

80 cells

1. calculate the amount of time in which the bacterial population was in log phase: 180 min incubation time - 0 min lag phase = 160 min log phase
2. Find the number of generations in the log phase period:
   160 min log phase / 40 min generation time = 4 generations
3. Multiple the original number of bacterial cells by 2^n to get the final population size:
   5 bacterial cells x 2^4 = 80 cells

Question 25

Which of the following cell-cell junctions would help skin cells withstand mechanical stress by directly anchoring the intermediate filaments of adjacent cells? I. Desmosomes II. Gap junctions III. Tight junctions

Answer 25

I. Desmosomes only

Question 26

Hormones secreted by the adrenal medulla most likely influence energy metabolism by: A) inhibiting proteins that mediate glycogen synthesis B) activating their cytosolic receptors in target cells C) reducing the activity of enzymes that catalyze fat hydrolysis D) increasing the activity of proteins that inhibit gluconeogenesis 401707

Answer 26

A) inhibiting proteins that mediate glycogen synthesis Hormones secreted by the adrenal gland regulate energy metabolism as follows: Glucocorticoids (eg cortisol) released by the adrenal cortex act on the liver to stimualte the synthesis of glucose from other molecules (ie gluconeogenesis) and the breakdown of fats into fatty acids (ie lipolysis). Both glucose and free fatty acids are forms of cellular energy that can be readily utilized. Norepinephrine and epinephrine are released by the adrenal medulla to modulate the metabolism of glycogen, the storage form of glucose. In addition, both hormones promote the breakdown of glycogen into glucose monomers in a process known as glycogenolysis. By inhibiting the enzymes that mediate glycogen synthesis, norepinephrine and epinephrine also inhibit glycogenesis

Question 27

A woman with low serum due to partial adrenal dysfunction is treated with high doses of a synthetic cortisol analogue. According to figure 2, treatment with a cortisol analogue will most likely cause this patient's serum level of endogenous cortisol to: (increase or decrease) 401710

Answer 27

decrease only Figure 2 shows the negative feedback pathway that influences the synthesis and secretion of cortisol (and that cortisol decreases) synthetic cortisol analogues have a molecular structure that resembles that of endogenous cortisol. therefore, the analogue will be able to mimic the effect of cortisol by binding hypothalamic and pituitary cortisol receptors. This interaction will increase negative feedback and reduce the synthesis and secretion of natural cortisol by the adrenal glands. Therefore, after treatment with a synthetic cortisol analogue, the patient's serum level of endogenous cortisol would decrease due to the increase in negative feedback on the hypothalamus and pituitary gland by the cortisol analogue.

Question 28

Cirrhosis of the liver is characterized by hepatic cell death, tissue scarring, and altered liver function. All of the following would be disrupted by cirrhosis of the liver except: A) production of ketone bodies B) metabolism of most drugs C) oncotic pressure within capillaries D) secretion of digestive proteolytic enzymes

Answer 28

D) secretion of digestive proteolytic enzymes

Question 29

Function of liver cells

Answer 29

- Regulation of blood glucose via glycogenesis, glycogenolysis, and gluconeogenesis - Storage of glycogen, minerals (iron) and vitamins - Synthesis of macromolecules such as plasma proteins (clotting factors & albumin), fats, ketone bodies, and cholesterol - Production and secretion of bile - Breakdown / detoxification of numerous drugs and metabolic waste products (bilirubin, ammonia)

Question 30

T. brucei cells and higher eukaryotes are similar in that they both undergo: I. Interphase II. Prophase III. Cytokinesis 402404

Answer 30

I and III only Passage states that T. brucei cells were arrested in S phase. Because S phase is a part of interphase, T. brucei cells undergo interphase. In addition, Figure 1 shows that in M phase, T. brucei cells undergo cytokinesis longitudinally, along the longest axis During prophase, the first phase of mitosis, chromatin condensesinto distinct chromosomes and the nuclear envelop disintegrates. However, according to the passge the chromatin in T. brucei does not condense, and the nuclear envelope remains intact throughout mitosis

Question 31

The number of each type of sex chromosome in a group of 5 male and 5 female rabbits is shown. In the combined population of 15X and 5 Y chromosomes, which of the following types of allele, if any, would be most susceptible to compelte loss from the population due to random chance alone? A) Y-linked allele B) X-linked allele C) either an X or a Y linked D) only an autosomal allele would be subject to complete loss by random chance alone 401951

Answer 31

A) Y-linked allele Y chromosome alleles present at lowest numbers; more susceptible to complete loss by random chance (genetic drift)

Question 32

Based on the passage, the cells that directly stimulate OPCs to differentiate are cells that promote: A) calcium transfer from bone tissue into the bloodstream B) calcium transfer from the bloodstream into bone tissue C) acid production in bone tissue D) proteolytic enzyme production in bone tissue 401899

Answer 32

B) calcium transfer from the bloodstream into bone tissue The passage states that RANKL is a ligand expressed by osteoblasts that binds RANK, a transmembrane receptor expressed by OPCs. The binding of RANKL on osteoblasts to RANK on OPCs induces downstream signaling pathways that lead to the differentiation of OPCs into mature osteoclasts. Accordingly, osteoblasts, which promote the absorption of calcium from blood to bone tissue, directly stimulate OPCs to differentiate

Question 33

A student isolates the cellular components of OPCs in wild-type mice and mutant mice exhibiting a single base substitution that cases reduced osteoclast activity. Western blot analysis is then performed to test for NF-kb and c-FOS expression in the cytoplasm and nucleus of OPCs isolated from these mice. The results of this analysis are shown: Based on the passage, which statement best explains these results? A) Mutant mice have a mutation that disables the ability of Nf-kB to localize the nucleus B) mutant mice have a mutation that prevents c-FOS from interacting with Nf-kB C) Mutant mice have a mutation in the DNA binding region of NF-kB D) Mutant mice have a mutation that prevents the transcription of c-Fos in osteoclast progenitor cells 401901

Answer 33

A) Mutant mice have a mutation that disables the ability of Nf-kB to localize the nucleus Western blot shows that compared to WT mice, MT mice lack nuclear expression of NF-kB and exhibit reduced c-Fos expression. Because Nf-kB requires an intact nuclear localization sequence to enter the nucleus and increase c-Fos expression, a mutation in the NF-kB nuclear localization sequence of MT mice best explains these results

Question 34

Sustained excess nutrient intake can result in resistance of target cells to insulin. Compared to normal patterns after a meal, which of the following best describes an effect of insulin resistance? A) Liver cells form glycogen at a higher rate B) liver cells secrete glucagon at a lower rate C) liver cells produce glucose at a higher rate D) kidney cells secrete glucose at a lower rate

Answer 34

C) liver cells produce glucose at a higher rate In patients with insulin resistance, the liver will increase production of glucose and decrease production of glycogen after a meal. The liver normally decreases production of glucose and increases production of glycogen after a meal

Question 35

When present in exocrine gland cells, specific chloride channels reabsorb chloride ions from fluids in the exocrine gland lumen. Individuals with cystic fibrosis, an inherited disorder involving mucus buildup in certain tissues, exhibit decreased function of these chloride channels. An individual with cystic fibrosis in a hot environment is most likely to demonstrate: A) high plasma chloride ion levels B) low osmolarity of fluid in sweat gland lumens C) high chloride ion concentrations in secreted sweat D) low numbers of chloride channels in exocrine glands

Answer 35

C) high chloride ion concentrations in secreted sweat Few chloride ions will be reabsorbed by the dysfunctional chloride channels in sweat gland cells

Question 36

A certain eukarytoic species is triploid and has 24 total chromosomes. A feature of this species is: A) an inability of some chromosomes to form homologous pairs during meiosis B) high fitness relative to diploid eukaryotic species C) production of viable gametes with low genetic diversity D) somatic cells with 12 chromosome pairs

Answer 36

A) an inability of some chromosomes to form homologous pairs during meiosis 24 chromomes , 8 groups containing 3 chromosomes each As such, 16 of the 24 chromosomes can form homologous fairs during meiosis, 8 pairs, but the remaining 8 chromosomes exist in unpaired form

Question 37

Which of the following molecules is most likely found within the cell membranes of intestinal epithelial cells? A) Peptidoglycan B) Cholesterol C) cytoskeletal filaments D) cellulose

Answer 37

B) Cholesterol cholesterol mitigates the fluid mosaic model

Question 38

A certain eukaryotic organism has multiple, distinct genes with highly similar sequences that are expressed at different times in the organism's life cycle. These genes could have arisen by: A) alternative splicing B) conjugation C) gene duplication D) transformation

Answer 38

C) gene duplication mutations, similar but not identical sequences. These differences allow the gene to carry out distinct roles at various times in an organism's life cycle Not alternative splicing, as this produces multiple protein products by producing distinct mRNA molecules from the same gene / same segment of DNA. In contrast, the question asks about distinct genes / distinct segments of DNA with similar sequences

Question 39

Bacteria are removed from lymph and blood in the

Answer 39

lymph nodes and spleen, respectively

Question 40

A patient found to have abnormally low number of helper T cells in her blood contracts a bacterial infection. During the infection, which of the following is most likely not affected by the patient's low count of helper T cells? A) Expression of MHC proteins that present bacterial antigens on the membrane of infected host cells B) proliferation of B lymphocytes in response to the bacterial infection C) levels of circulating antibodies against the invading bacterium D) Activation of apoptosis-inducing cytotoxic T cells

Answer 40

A) Expression of MHC proteins that present bacterial antigens on the membrane of infected host cells

Question 41

Helper T cell function for B lymphocyte

Answer 41

Promotes B lymphocyte activation and proliferation | -induces differentiation of B lymphocytes in antibody secreting plasma cells and memory cells

Question 42

helper t cell function --> cytotoxic t cell

Answer 42

promotes cytotoxic t cell activation and proliferation

Question 43

helper t cell function --> macrophage

Answer 43

enhances activity of macrophage

Question 44

helper t cell function --> other immune cells

Answer 44

attracts other immune cells to site of infection

Question 45

Cells of the innate immune system and cells of the adaptive immune system differ in that only the latter: A) mount an immediate response against pathogens B) are produced by bone marrow C) bind only one type of antigen D) express antigen-presenting proteins on their cell surface

Answer 45

C) bind only one type of antigen specific immunity

Question 46

Compared to the axons investigated by Biologist 2, the axons studied by Biologist 1 most likely had: A) faster decay of the propagating electrical signal B) axonal membranes that were more exposed to the extracellular space C) slower rates of action potential conduction D) voltage-gated ion channels clustered only at specific locations

Answer 46

D) voltage-gated ion channels clustered only at specific locations According to the passage, Biologist 1 studied brain tissue with neurons containing primarily myelinated axons, whereas Biologist 2 studied unmyelinted axons. Therefore, the myelinated axons investigated by Biologist 1 would contain voltage-gated ion channels clustered at the nodes of Ranvier. However, the unmyelinated axons investigated by Biologist 2 would contain voltage-gated ion channels that are more evenly distributed across the entire axon.

Question 47

Which of the following synaptic events would likely contribute most directly to the generation of an action potential in a postsynaptic neuron? A) the presynaptic neuron releases large quantities of GABA B) postsynaptic neurotransmitter-gated Na+ and Ca2+ channels open C) Neurotransmitter reuptake into the presynaptic cell and surrounding glia is increased D) Neurotransmitter binding activates G protein coupled receptors on the postsynaptic membrane

Answer 47

B) postsynaptic neurotransmitter-gated Na+ and Ca2+ channels open

Question 48

Which of the following cellular processes would most likely lead to increased genetic variation in a population of organisms? A) DNA replication during S phase of the cell cycle B) synapsis involving two homologous chromosomes C) alternative splicing of gene transcripts D) cell division via mitosis

Answer 48

B) synapsis involving two homologous chromosomes This is crossing over

Question 49

One function of endothelial cells is to: A) contract to release blood pressure B) line the inside of the GI tract C) bind sites of vascular damage to form clots D) line the inside of blood vessels

Answer 49

D) line the inside of blood vessels Endothelium is the one-cell-thick layer of specialized epithelial cells that line the interior of the cardiovascular system, which includes the heart, arteries, capillaries, and veins

Question 50

how to differentiate thymine and cytosine

Answer 50

thymine has a methyl group attached to the carbon in the 5th position

Question 51

how to differentiate guanine and adenine

Answer 51

guanine has a carbonyl group at the 6th position

Question 52

Human cardiac muscle differs from skeletal muscle because it: A) apears striated B) requires acetylcholine to contract C) contains intercalated discs D) contains cells with multiple nuclei

Answer 52

C) contains intercalated discs

Question 53

Skeletal muscle appearance

Answer 53

bunched fibers - striated (has sarcomeres) - multinucleated

Question 54

skeletal muscle location and function

Answer 54

- attached to bone via tendons & leverages bones to generate movement - aids circulation by squeezing vessels - heat production (shivering)

Question 55

skeletal muscle gap junctions

Answer 55

not present, cells are electrically isolated

Question 56

skeletal muscle control

Answer 56

voluntary, somatic nervous system (ACh)

Question 57

cardiac muscle appearance

Answer 57

- branched fibers - striated (has sarcomeres) - 1-2 nuclei

Question 58

cardiac muscle location and function

Answer 58

makes up heart walls and pumps blood through circulatory system

Question 59

cardiac muscle gap junctions

Answer 59

present

Question 60

cardiac muscle control

Answer 60

- involuntary - myogenic - regulated by autonomic nervous system

Question 61

smooth muscle appearance

Answer 61

- bunched cells - not striated (no sarcomeres) - uninucleated

Question 62

smooth muscle location & function

Answer 62

- lines follow visceral organs (eg, stomach) and promotes substance (eg, food) movement - lines blood vessels and affects blood pressure

Question 63

smooth muscle gap junctions

Answer 63

present

Question 64

smooth muscle control

Answer 64

- involuntary | - regulated by autonomic nervous system

Question 65

During muscle contraction, the power stroke is the pivoting motion of the myosin head that pulls the actin filament inward to shorten the sarcomere. Given this, which of the following events most immediately precedes this power stroke motion? A) Binding of ATP to myosin heads B) hydrolysis of myosin-bound ATP C) release of Pi from the myosin head D) binding of Ca2+ to tropomyosin

Answer 65

C) release of Pi from the myosin head When ATP bound to a myosin head hydrolyzes, the myosin head shifts back to the high-energy position. Binding of calcium to troponin then allows the mypsin head to attach to the actin filament. Dissociation of Pi leads to a power stroke, which shortens the sarcomere. After the release of ADP, a new ATP molecule binds to the myosin head, causing it to dissociate from the actin filament. The action of many myosin heads cycling through this process leads to muscle contraction.

Question 66

DNA sequencing

Answer 66

- Determines the nucleotide sequence of DNA | - Can be used to determined relative gene count

Question 67

Southern blot

Answer 67

- Detects a particular sequence of DNA | - Can be used to determine relative gene count

Question 68

Northern blot

Answer 68

- Detects a particular sequence of RNA | - Can NOT be used to determine a relative gene count

Question 69

VEGF signaling also leads to the dilation of existing blood vessels. One function of vasodilation is to: A) increase body temperature in cold environments B) increase blood flow to the intestines following a meal C) decrease blood flow to the kidneys during dehydration D) maintain blood pressure following an episode of fluid loss

Answer 69

B) increase blood flow to the intestines following a meal

Question 70

Vasoconstriction

Answer 70

decreased blood vessel diameter, decreased blood flow, increased blood pressure

Question 71

vasodilation

Answer 71

increased blood vessel diameter, increased blood flow, decreased blood pressure

Question 72

All the following findings in sperm would cause lower fertilization rates in IVF except: A) defects in microtubule structure B) mitochondrial deficiency C) cytoplasmic reduction D) misfolded acrosomal enzymes 401087

Answer 72

C) cytoplasmic reduction Sperm cell is divided into 3 segments: 1) head: contains an acrosome and the nucleus. The acrosome is a flattened structure than encaspulates the tip of the nucleus and is rich is specialized lysosome-like enzymes for piercing the outer shell of an oocyte during fertilization 2) Midpiece: packed with mitochondria, essential organelles that produce the ATP required for flagellum-driven sperm motility. This section also contains a pair of central microtubules that are anchored to the cytoskeleton and extend down the length of the flagellum 3) tail / flagellum : singular elongated structure specialized for wavelike movements to propel sperm through a fluid environment. Flagellem-driven motility is derived from the action of ATP-dependent motor proteins that act on the central microtubulues

Question 73

Pathway of sperm development

Answer 73

Spermatogonium, spermatocyte, spermatid, spermatozoon

Question 74

Assume L410 is an acquired mutation that occurs only to hepatocytes of adults who were exposed to a certain mutagen. A women with the L410 mutation has a child with an unaffected man. What is the probability this child has the L410 mutation?

Answer 74

0.00 This acquired mutation occurs only in hepatocytes / liver cells of adults due to exposure to a mutagen. This means the L410 mutation is somatic because it occured sporadically in a minority of cells and not in the cells of the entire organism.

Question 75

Excess post-exercise oxygen consumption is the extra amount of oxygen consumed after the completion of exercise and before respiration returns to resting levels. This extra oxygen is used to restore muscle fibers to their pre-exercise state by doing all of the following except: A) replenishing the ATP stores within the muscle fiber B) facilitating calcium rlease from the sarcoplasmic reticulum in the muscle fiber C) restocking the myoglobin stores withinthe muscle fiber D) supplying the ATP used to replenish glycogen stores in the muscle fiber

Answer 75

B) facilitating calcium release from the sarcoplasmic reticulum in the muscle fiber When oxygen delivery from the blood is insufficient (during rapid, high-intensity exercise) and the demand for contraction exceeds the speed at which aerobic respiration can generate ATP, muscle fibers compenstate by : 1. utilizing ATP that is stored in the muscle fiber 2. breaking down creatine phosphate, a high-energy molecule stored by muscle cells during periods of rest, to generate ATP 3. generating ATP via anaerobic glycolysis, which yields 2 pyruvate molecules and 2 net ATP molecules per glucose molecule. In the absence of oxygen, the 2 pyruvate molecules are converted to lactic acid 4. continuing aerobic respiration using oxygen that is bound to the myoglobin

Question 76

The population of cheetahs that survived the environmental catastrophe that occurred 12,000 years ago were at greater risk of extinction because of: A) changes in allele frequencies due to natural selection B) changes in allele frequencies due to random chance C) changes in allele frequencies due to inbound migration from unaffected populations D) changes in allele frequencies due to random mating 401464

Answer 76

B) changes in allele frequencies due to random chance Genetic drift: random disappearance of genotypes in the population (eg, due to death, lack of reproduction) --> surviving populaiton has decreased genetic diversity --> surviving population reproduces and increases in number but still has less genetic diversity Genetic drift is a mechanism of evolution, however, unlike natural selection, the variations in allele frequency occur randomly by chance. Because smaller populations have a smaller gene pool, the random variations due to genetic drift cannot be buffered

Question 77

Which of the following statements best explains the trend shown in Figure 1? Note: captive cheetahs are generally housed individually A) captive cheetahs are bred with close relatives B) captive cheetahs are selectively outbred C) captive cheetah populations are in HW equilibrium D) captive cheetahs have increased fecundity 401465

Answer 77

B) captive cheetahs are selectively outbred Outbreeding, unrelated individuals, 0% of offspring display reduced fitness

Question 78

Researchers concluded that captive cheetahs have reduced fitness because of their increased susceptibility to diseases. What can best explain the decreased fitness of captive cheetahs describes in the passage? A) captive cheetahs have increased MHC diversity B) captive cheetahs have decreases MHC diversity C) captive cheetahs experience increased environmental stress D) captive cheetahs experience decreased environmental stress 401467

Answer 78

C) captive cheetahs experience increased environmental stress Higher cortisol levels in captive cheetahs lower their ability to fight disease. Therefore captive cheetahs have increased incidence of disease.

Question 79

Compared to the blood flowing into the left ventricle, the blood flowing into the right ventricle: A) has a higher oxygen concentration B) has a higher pH C) is entering a cardiac chamber with thinner walls D) is entering a cardiac chamber that contracts to produce higher pressure

Answer 79

C) is entering a cardiac chamber with thinner walls The right heart pumps blood to the lungs, whereas the left heart pumps blood to the rest of the body. Because the left ventricle must pump blood to a larger structure (all body tissues except the lungs) across longer distances, it must pump blood at a higher pressure. To generate this pressure, the walls of the left ventricle must be extremely muscular and powerful (thicker) In contrast, the right ventricle has thinner walls than the left ventricle because it needs to pump blood at a lower pressure to propel blood to the lungs only

Question 80

Based on the passage, which of the following viruses does NOT have an outer phospholipid bilayer? A) HAV B) HBV C) HCV D) HDV

Question 80

Based on the passage, which of the following viruses does NOT have an outer phospholipid bilayer? A) HAV B) HBV C) HCV D) HDV 401470

Answer 80

A) HAV Bound antibodies were only detected in wells with HAV. or HEV. Because the designed antibodies did not bind in wells containing HBV, HDV< or HDV, the capsid proteins of enveloped viruses are fully enclosed within a phosphplipid bilayer, which would prevent proteins such as antibodies reaching the capsid enclosed inside. Therefore, HBV, HCV, and HDV are likely enclosed viruses whose capsid proteins are enclosed within a phospholipid bilayer In contrast, because wells containing HAV and HEV did bind the flourescent antibodies, their capsid proteins were likely exposed to the environment and were able to interact with the antibodies. Therefore, HAV and HEV must be non-eveloped (naked) viruses that do not have a phospholipid bilayer

Question 81

Researchers perform an experiment in which they infect wild-type mice with either HBV or HEV and subsequently treat them with an antiviral drug that inhibits mouse RNA polymerase activity. Based on the passage, administering this drug to mice infected with either HBV or HEV would most likely result in : I. Decreased viral load in HBV-infected mice II. Decreased viral load in HEV-infected mice III. Reduced number of viral fragments displayed by antigen-presenting cells in HBV-infected mice 401471

Answer 81

I and III only Viral load is measured by the total quantity of viral genetic material inside an organism. DNA viruses such as HBV have double stranded genomes that are similar to the host cell's genome; as a result, this type of virus can directly utilize host enzymes for transcription and replication. After infecting a cell, the dsDNA genome is trafficked to the nuclear, where host DNA-dependent RNA polermase transcribes the viral genes to mRNA that codes for viral proteins, which are subsequently used for viral replication but can also be detected by the immune system In this scenario, the administered drug inhibits mouse Pol II. Inhibition of Pol II in mice infected with HBV would prevent transcription of the virla genome, stopping the production of viral mRNAs and proteins. Consequently, a decrease in viral load is expected as the lack of viral proteins prevents the formation of complete viral particles and viral replicaiton. Similarly, the decrease in viral particles being replicated and released causes a decrease in HBV fragments displayed by antigen-presenting cells in HBV-infected mice.

Question 82

Physician researchering a global hepatitis outbreak discover a new strain of hepatitis (HXV) by analyzing chromosomal DNA of patient's liver cells. They found that HXC has a ++ssRNA genome and replicates through DNA intermediates. What characterisic best describes the new hepatitis virus? A) it integrates its retroviral +ssRNA into the host genome B) it encodes proteins with reverse transcriptase function C) It synthesizes a complementary RNA strand D) It translates its own proteins 401473

Answer 82

B) it encodes proteins with reverse transcriptase function HXV is a retrovirus because it integrated with the host genome. Retroviruses are unique in that they are enveloped and carry two identical +ssRNA molecules. Successful viral replication of HXV depends on enzymes that allow its RNA viral genome to enter the nucleus and integrate with the host's DNA genome. Retroviruses enter host cells through endocytosis, which allows the capsid and envelope to uncoat. Uncoating releaes the viral genetic material inside the host cell. RNA viruses such as HXC must then convert their genomes from +ssRNA into double-stranded DNA (dsDNA) before integration into the host genome can occur. The generation of DNA using an RNA template is accomploshed using a reverse transcripase (ie RNA-dependent DNA polymerase) which is encoded in the viral genome

Question 83

The diagram shows the positions of three different sections of a relaxed sarcomere, I band, A band, H band, During sarcomere contraction, which would shorten in length?

Answer 83

I and H band only

Question 83

The diagram shows the positions of three different sections of a relaxed sarcomere, I band, A band, H band, During sarcomere contraction, which would shorten in length?

Answer 83

I and H band only A band does not change in length during muscle contraction

Question 84

Tissue regeneration has an important role in whcih of the following circumstances? A) generating a multcellular embyro from a zebrafish zygote B) producing neurons from chicken embryonic stem cells C) producing new cartilage cells in a punctured rabbit ear D) creating scar tissue in human skin that has been injured

Answer 84

C) producing new cartilage cells in a punctured rabbit ear Regeneration = regrowth of the same functional tissues or organs previously present in an organism after loss due to injury

Question 85

Vitamin B12 is absorbed by the region of the digestive system that immediately precedes the cecum. Given this, vitamin B12 absorption most likely takes place in the

Answer 85

ileum (final segment of the small intestine)

Question 86

Osteoarthritis commonly affects large, weight-bearing joing of the lower extremitiies such as the knee and hip. Based on the passage, which would most likely occur in a patient with OA involving the knees? A) deterioration of the yellow bone marrow in affected bones B) deposition of excess bone on the epiphyses of affected long bones C) loss of hematopeoetic function in the red marrow of affected bones D) degeneration of compact bone at the diaphysis of affected long bones 401059

Answer 86

B) deposition of excess bone on the epiphyses of affected long bones Structure of long bones: - Epiphyses: rounded ends that have joint surfaces covered by articular cartilage - Diaphysis: hollow shaft (medullary cavity) filled with bone marrow (primarily of the yellow type in adulthood) - Metaphyses: regions where the diaphysis and epiphyses meet - Periosteum : thin layer of connective tissue that covers and protects long bones but does not cover joint surfacs Knee joint is site where the femur and tibia meet.

Question 87

Microscopic examination of tissue samples obtained from the type of bone that becomes exposed in advanced OA is most likely to reveal all of the following except: A) Volkmann canals connecting haversian canals B) mature chondrocytes C) canaliculi D) osteocytes situated within lacunae

Answer 87

B) mature chondrocytes chondrocytes make up the the cellular component of cartilage, not bone

Question 88

Which of the following statements is not true about hyaline cartilage? A) it lacks innervation B) it gives rise to bone through endochondral ossification C) it serves as an attachment that holds bones to muscles D) it is avascular and receives nutrients from surrounding fluids

Answer 88

C) it serves as an attachment that holds bones to muscles

Question 89

A study indicates that endurance exercise performance is related to type 1 skeletal muscle fiber characteristics that allow for efficient performance of aerobic respiration to generate ATP for extended periods of time. Given this, which of the following characteristics is (are) most likely to be associated with type 1 skeletal muscle fibers? I. High mitochondrial density II. High numbers of capillaries around the muscle fiber III. High myoglobin concentration in the muscle fiber

Answer 89

I, II, and III

Question 90

The DNA of telomeres can be most accurately described as: A) noncoding and highly repetitive B) single stranded and self complementing C) able to encode multiple unique proteins D) easily cleaved by restriction enzymes

Answer 90

A) noncoding and highly repetitive

Question 91

In neurons, action potential propagation occurs along the axon. The analogous structure in skeletal muscle fibers that enables action potential propagation is the: A) actin filament B) transverse tubule C) motor endplate D) myelin sheath

Answer 91

B) transverse tubule Electrical depolarization caused by action potential propagation leads to the concentration of skeletal muscle fibers. As extensions of the plasma membrane that penetrate deep into muscle fibers, transverse T tubules are specialized muscle fiber structures that facilitate action potential propagation throughout individual muscle fibers

Question 92

Researchers are studying a single-celled pathogen. Which experimental method would be LEAST effective for determining whether the pathogen is prokaryotic or eukaryotic? A) analyzing the cells for the number of origins of replication B) exposing the pathogen to a flourescent antimitochondrial antibody C) testinf the pathogen for ribosome size D) assessing the pathogen for the presence of pyruvate in the cytoplasm

Answer 92

D) assessing the pathogen for the presence of pyruvate in the cytoplasm Glycolysis is a process that occurs in the cytoplasm of both prokaryotic and eukaryotic cells and results in the synthesis of pyruvate

Question 93

Prokaryotic cells: cell types

Answer 93

bacteria, archea

Question 94

eukaryotic cells: cell types

Answer 94

animals, plants, fungi

Question 95

eukaryotic cells: size

Answer 95

10-100 um

Question 96

eukaryotic cells: DNA

Answer 96

- multiple linear chromosomes - double stranded - multiple origins of replicaiton - diploid or greater - telomeres - located in nucleus

Question 97

eukaryotic cells: cell structure

Answer 97

- membrane-bound organelles - cellular respiration in cytoplasm (anaerobic) & mitochondria (aerobic) - cell wall (eg contains cellulose in plants and chitin in fungi)

Question 98

eukaryotic cells: ribosomes

Answer 98

80s ribosomes (composed of 60s & 40S subunits)

Question 99

eukaryotic cells: cell division

Answer 99

mitosis (cell replication) and meiosis (gamete production)

Question 100

prokaryotic cells: size

Answer 100

1-5 um

Question 101

prokaryotic cells: DNA

Answer 101

- single stranded chromosomes - double-stranded - single origin of replication - haploid - no telomeres - located in cytoplasm

Question 102

prokaryotic cells: cell structure and function

Answer 102

- no membrane-bound organelles - cellular respiration in cytoplasm (aerobic & anaerobic) - cell wall (contains peptidoglycan in bacteria)

Question 103

prokaryotic cells: ribosomes

Answer 103

70S ribosomes (50S & 30S subunits)

Question 104

prokaryotic cells: cell division

Answer 104

binary fission

Question 105

Variable region of the antibody

Answer 105

binds the antigen A single antibody can bind only one type of antigen, and this specifically is determined by the unique amino acid structure of the antibody's variable region

Question 106

A self-directed antibody known as rheumatoid factor (RF) has been found to be involved in RA-mediated inflammation. RF is an antibody that targets a patient's own IgG antibodies, which results in the formation of immune complexes that induce an inflammatory response. High concentrations of self-antibodies such as RF are not typically found in the human body because:

Answer 106

B cells and T cells against self antigens are destroyed

Question 107

A student analyzing dividing cells of a certain type under the microscope finds that each daughter cell produced contains half the number of chromosomes as its parent cell. Given this, cells of this type most likely contain all of the following except: A) peroxisomes B) nuclear pore proteins C) single-stranded genomes D) microtubules

Answer 107

C) single-stranded genomes *not a character of eukaryotic cells

Question 108

Plasmodium falciparum is a protozoan parasite known to cause malaria, an illness characterized by fever, chills, and flu-like symptoms. Which of the following is the most likely reason why P. falciparum would be classified as a eukaryotic organism? A) It is able to inherit genetic information from another individual B) It has a spliceosome-mediated mechanism for intron splicing C) It has a cell wall D) It has a cell membrane

Answer 108

B) It has a spliceosome-mediated mechanism for intron splicing Eukaryotic cells have a unique mechanism for processing mRNA. Eukaryotic mRNA is intiailly transcribed as pre-mRNA, containing in both introns and exons. Following transcription, the spliceosome removes introns from the pre-mRNA molecule to form mature mRNA in a process known as RNA splicing. This mature mRNA is transported out of the nucleus and into the cytoplasm where it is translated into protein not D because all cells are enclosed by a plasma membrane; however eukaryotic cells are unique in that they have membrane-bound organelles as well as a mechanism for intron splicing by the spliceosome

Question 109

Prior to receiving a skin graft, a patient with full-thickness burns on approximately 50% of their total BSA is at risk for all the following except: A) extensive fluid loss B) bacterial infection C) increased sensitivity to touch D) decreased abrasion protection 400932

Answer 109

C) increased sensitivity to touch According to the passage, the damage associated with a full-thickness burn extends through the epidermis and dermis and into the subcutaneous layer (hypodermis). Therefore, this burn involves injury to the epidermal and dermal sensory receptors, resulting in impaired touch sensation, thus this patient would be at risk for decreased sensitivity to touch, not increased

Question 110

Given that scar tissue forms when collagen-producing cells replace normal tissue after an injury, all the following processes would be impaired in the area of scarred skin that forms over a partial-thickness burn EXCEPT: A) resistance to UV radiation B) insulation C) hair growth D) salt excretion

Answer 110

B) insulation

Question 111

Which of the following skin functions will most likely be impaired in tissue with superficial burns? I. Keratinocyte maturation II. Immune surveillance III. Shock absorption 400936

Answer 111

I and II only As stated in the passage, a superficial burn damages only the epidermis. Because the stratum basale, which is responsible for producing keratinocytes that undergo maturation, is part of the epidermis, kerintocyte maturation will be impaired in tissue with superficial burns. Immune serveillance --> epidermal langerhan cells Shock absorption --> hypodermis , not affected but superficial burns

Question 112

Transferrin circulating in the blood consists of 679 amino acids, but it is composed of 698 amino acids when it is post-translationally extracted from liver cells, the primary site of transferrin synthesis. Given this information, researchers would most likely be able to extract and isolate the shorter transferrin peptide from: A) lymph B) the liquid portion of blood C) all cellular elements of blood D) the nucleus of a liver cell 401699

Answer 112

B) the liquid portion of blood

Question 113

Iron buildup in the liver can lead to scarring and loss of liver function, If a change in holoTf excretion rate were to cause this form of liver damage in a patient receiving blood transfusions, the rate at which holoTf is filtered from the glomerular capillaries in this patient would most likely be: A) decreased to increased Na+ reabsorption in the nephron B) decreased due to decreased hydrostatic pressure within the glomerulus C) increased due to increased urinary output from the kidney D) increased due to increased hydrostatic pressure within the glomerulus

Question 113

Iron buildup in the liver can lead to scarring and loss of liver function, If a change in holoTf excretion rate were to cause this form of liver damage in a patient receiving blood transfusions, the rate at which holoTf is filtered from the glomerular capillaries in this patient would most likely be: A) decreased to increased Na+ reabsorption in the nephron B) decreased due to decreased hydrostatic pressure within the glomerulus C) increased due to increased urinary output from the kidney D) increased due to increased hydrostatic pressure within the glomerulus

Question 113

Iron buildup in the liver can lead to scarring and loss of liver function, If a change in holoTf excretion rate were to cause this form of liver damage in a patient receiving blood transfusions, the rate at which holoTf is filtered from the glomerular capillaries in this patient would most likely be: A) decreased to increased Na+ reabsorption in the nephron B) decreased due to decreased hydrostatic pressure within the glomerulus C) increased due to increased urinary output from the kidney D) increased due to increased hydrostatic pressure within the glomerulus 401700

Answer 113

B) decreased due to decreased hydrostatic pressure within the glomerulus In this scenario, a patient receiving blood transfusions exhibits liver damage caused by iron buildup. The passage states that iron-bound transferrin holoTf could be filtered through the kidney. Therefore, if GFR decreased in the kidney, the likely result would be increased retention of iron-bound transferrin in the blood, which would contribute to iron buildup in the liver. Decreased hydrostatic pressure within the glomerulus can lead to decreases GFR (and consequently decreased holoTf filtration rates) because the flow rate of filtered fluid through the kidney will decrease

Question 114

Individuals with beta-thalassemia are at risk for developing an enlarged spleen. The resulting impairment of spleen function in these individuals would not affect: A) immune responses against viral or bacterial infection B) release of hormones regulating blood glucose concentration C) removal of aged red blood cells from the circulation D) storage of red blood cells 401703

Answer 114

B) release of hormones regulating blood glucose concentration

Question 115

spleen function

Answer 115

- filters aged.damaged RBCs - resevoir for blood - immune response (B cell activation site, housing macrophages)

Question 116

Of the genes tested in the passage, which are autosomal? I. Ena II. Abl III. PTP 401617

Answer 116

I, II, and III Genetic traits may be autosomal or sex-linked (X chromosome)

Question 117

Which evidence in Table 1 suggests that PTP and Abl are located close together on the chromosome? A) the two recombinant genotypes occur with similar frequency B) only a small percentage of the progeny are recombinant C) recombination did not occur in male gametes D) The f1 generation has a high recombination frequency 401618

Answer 117

B) only a small percentage of the progeny are recombinant If two genes are located close together on a chromosome, they are relatively unlikely to be separated by a recombination event. Therefore, substantially fewer progeny will have recombinant combinations than parental combinations. As a result, the recombinant progeny can be identified by the fact that there are fewer of them. Table 1 shows that only 43 of 500 flies (~9%) were recombinant. Given this small percentage of recombinant flies, it is likely that the PTP and Abl genes are close together on the chromosome.

Question 118

Suppose a female female heterozygous for both PTP and Ena (EePp) was unintentionally mated with a single F1 male of an unknown genotype from Experiment 1. If 25% of the resulting F2 progeny died in the larval stage, the genotype of the F1 male was most likely: (401619)

Answer 118

Eepp

Question 119

Which of the following conclusions about c-Myc and the miR-17~92 gene cluster is (are) most consistent with the information in the passage? I. c-Myc induces miR-17~92 expression II. c-Myc knockout promotes miR-17~92 function III. miR-17~92 suppresses apoptosis 3020276

Answer 119

I and III only

Question 120

Which statement describes an anatomical characteristic of the uterus? A) the uterus opens into the vagina at the uterine tube B) the uterus includes a layer called myometrium C) the uterus contains a thick layer of striated muscle D) the uterus is generally where fertilization occurs

Answer 120

B) the uterus includes a layer called myometrium As part of the female reproductive system, the uterus nourishes the embryo and fetus. It is lined with endometrium, includes a layer of smooth muscle called myometrium, and connects to the vagina at the cervix

Question 121

Porencephaly type I, a disease caused by a mutation in the autosomal COL4Ai gene, is characterized by the development of noncancerous, fluid-filled growth in brain tissue. A woman expressing this disease is heterozygous for the COL4A1 mutation. If she has a child with a man who is also heterozygous mutation, what is the probability that the child will express the disease?

Answer 121

75% P (child expresses disease) = P (inherits single mutant C allele) + P ( inherits two mutant alleles)

Question 122

In an experiment, a researcher wants to isolate mature mRNA from a sample of total RNA using specialized particles containing long single-stranded nucleic acid molecules (either DNA or RNA) of a synthetic nature. Which of the following single-stranded molecules would most likely be used as a probe to isolate the mature mRNA? A) A DNA molecule consisting of uracil repeats B) A DNA molecule consisting of thymine repeats C) An mRNA molecule consisting of adenine repeats D) An mRNA. molecule consisting of cytosine repeats

Answer 122

B) A DNA molecule consisting of thymine repeats Because a 3' poly-A is added to all mature mRNA molecules, a single-stranded DNA molecule consisting of thymine repeats could be used as a probe because these thymine nucleotides can form complementary hydrogen bonds with the adenine nucleotides of the mature mRNA poly-A-tail

Question 123

Frameshift mutations

Answer 123

result from the insertion or deletion of a number of nucleotide base pairs not divisible by three. This causes the reading frame of the mRNA transcipt to change,leading to multiple amino acids changes downstream of the mutation, which can potentially interfere with protein functions

Question 124

Nonsense mutations

Answer 124

occur when a base pair substitution causes a premature stop codon to be inserted into a gene. This permature stop codon leads to early termination of translation and the production of a truncated (ie shortened) protein

Question 125

Missense mutations

Answer 125

occur when a base pair substituion results in a codon that encodes a different amino acid. This results in the substition of the wild-type amino acid in the protein for this different, mutant amino acid, potentially interfering with protein function. However, there is a possibility that protein function may not be affected if an amino acid with a similar structure (ie, properties) is substitued for the original amino acids

Question 126

Silent mutations

Answer 126

Which occur when a mutation alters the sequence of the gene and mRNA transcript but not the sequence of amino acids in the encoded protein. As a result, the protein retains its wild-type function is not affected by silent mutations

Question 127

ACTC is a gene coding for cardiac actin. Patients exhibiting a specific mutation in this gene, called the E99K mutation, demonstrate a form of cardiac hypertrophy (cell enlargement) thought to be caused by increased senstivity of the contractile apparatus to calcium. Given this, patients with the E99K mutation most likely demonstrate increased binding of: A) myosin to actin B) actin to troponin C) calcium to tropomyosin D) ATP to calcium

Answer 127

A) myosin to actin In this question, ACTC is a gene encoding a protein comprising cardiac actin filaments. Patients with a specific ACTC mutation (E99K) demonstrate a form of cardiac hypertrophy (cell enlargement) thought to be caused by increased sensitivity of the contractile apparatus to calcium. Because of this, a lower intracellular concentration of calcium will be able to bind troponin and cause the conformational change required for the binding of myosin to actin and the subsequent contraction cycle. Therefore, patients with this mutation likely demonstrate binding of myosin to actin .

Question 128

A scientist examining action potentials in mature muscle cells identifies certain muscle cells that, upon receiving an action, can directly transfer ions to adjacent cells. These cells would most likely include: I. smooth muscle cells II. cardiac muscle cells III. skeletal muscle cells

Answer 128

I and II only

Question 129

What statement best explains how osteocytes regulate compact bone remodeling? A) signaling molecules released by osteocytes travel to other bone cells via canaliculi B) Osteocytes travel between lacunae to communicate with other bone cells C) osteocytes differentiate into cells capable of producing signaling molecules that regulate bone remodeling D) haversian canals provide a pathway for osteocyte travel in the bloodstream

Answer 129

A) signaling molecules released by osteocytes travel to other bone cells via canaliculi

Question 130

Huntington's disease is a genetic neurogenerative disorder that causes disorder that causes brain neurons to experience more action potentials than normal. It is known that activity of the neurotransmitter glutamate increases in these patients and activity of the neurotransmitter GABA decreases. Given this, these two neurotransmitters most likely have which of the following effects in the brain? A) Glutamate is an inhibitory neurotransmitter; GABA is an excitary neurotransmitter B) Glutamate is an excitatory neurotransmitter; GABA is an inhibitory neurotransmitter C) both glutamate and GABA are inhibitory D) both glutamate and GABA are excitatory

Answer 130

B) Glutamate is an excitatory neurotransmitter; GABA is an inhibitory neurotransmitter Because excitatory signals increase in these patients, glutamate increases, and the activity of the neurotransmitter GABA decreases. because excitatory signals increase in these patients, glutamate is likely an excitatory neurotransmitter. Conversely, because inhibitory signals decrease in these patients, GABA is likely an inhibitory neurotransmitter

Question 131

A cell from an organism with 78 chromosomes in its somatic cells contains a total of how many sister chromatids and centromeres during metaphase of mitosis?

Answer 131

156 sister chromatids and 78 centromeres Sister chromatids = Somatic cell chromosome number x2 =78 x 2 = 156 Total centromeres = Somatic cell chromosome number = 78

Question 132

Connective tissue examples

Answer 132

Bone, blood, fat (adipose), tendons/ligaments, cartilage

Question 133

epithelial tissue: germ layer precursor

Answer 133

endoderm, mesoderm, ectoderm

Question 134

epithelial tissue: functions

Answer 134

- lines surfaces - protection - absorption , filtration, and secretion

Question 135

epithelial tissue: examples

Answer 135

- skin | - lining of hollow organs (eg digestive tract & trachea)

Question 136

Muscle tissue: germ layer precursor

Answer 136

mesoderm

Question 137

Muscle tissue: functions

Answer 137

- voluntary movement | - involuntary movement

Question 138

connective tissue: germ layer precursor

Answer 138

mesoderm

Question 139

connective tissue: functions

Answer 139

- support and protection - attachment of tissues - insulation & thermoregulation - transport

Question 140

nervous tissue: germ layer precursor

Answer 140

ectoderm

Question 141

nervous tissue: functions

Answer 141

- transmission and receipt of electrical signals | - response to internal and external stimuli (eg, sensory receptors)

Question 142

nervous tissue: examples

Answer 142

brain, spinal cord, nerves

Question 143

The bones of humans and other vertebrates are alike in that they A) consist soley of living tissue B) develop from the embryonic ectoderm C) form a rigid outer covering that protects the soft tissues inside the organism D) form an internal support structure that serves as a scaffold for soft tissues

Answer 143

D) form an internal support structure that serves as a scaffold for soft tissues

Question 144

Endoskeleton

Answer 144

- Consists of both living components (eg, osteocystes, osteoblasts, osteoclasts) & nonliving components (eg, calcium phosphate) - forms an internal scaffold that protects soft tissues (eg organs) & serves as an attachment point for skeletal muscle - grows as the organism grows

Question 145

exoskeleton

Answer 145

- consists only of nonliving components (eg, chitin) - forms a hard outer covering that surrounds & protects soft tissue underneath - shed and regenerated as the organism grows