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Terms in this set (53)
Transposons are mobile genetic elements that can "jump" around the genome, causing chromosomal aberrations. All of the following are true of transposons EXCEPT:
Two transposons in opposite orientations= chromosomal inversion
Two transposons in the same orientation= chromosomal deletions and translocations
If a single transposon inserts into a protein-coding region of the DNA= protein coding region will be disrupted and proteins levels will likely decrease
However, if a single transposon inserts into the promoter or regulatory region of a protein coding gene, protein levels can either increase or decrease (choice D is not true of transposons and is the correct answer choice).
α-ketoglutarate dehydrogenase (α-KGDH) is an enzyme used in cellular respiration that catalyzes the decarboxylation of α-ketoglutarate (a 5-carbon compound) to succinyl-CoA (a 4-carbon compound). Inherited deficiencies of α-KGDH result in permanent lactic acidosis, among other symptoms, and lead to death at an early age. α-KGDH is mostly likely found in the:
the fact that individuals with α-KGDH deficiency suffer from permanent lactic acidosis suggests that the lack of this enzyme shuts down the Krebs cycle and puts the individual into permanent fermentation. Thus, this enzyme is most likely found in the
*, where the Krebs cycle takes place
High levels of ATP would:
inhibit phosphofructokinase and inhibit pyruvate kinase, thus inhibiting glycolysis
Phosphofructokinase (PFK) catalyzes the phosphorylation of fructose-6-P in the third step of glycolysis. If ATP levels in the cell are high, ATP can bind to PFK (at a site other than the active site) to inhibit the reaction. This is most accurately described as:
When a molecule binds to an enzyme at a site other than the active site and regulates the activity of that enzyme, it is best described as
Which of the following differences between RNA and DNA could help explain the differences in secondary structure observed between the two types of nucleic acids?
This question is asking for a difference between RNA and DNA and only one of the answer choices provided is a valid difference between the two nucleic acids.
RNA contains a 2' hydroxyl group
, possesses uracil in place of thymine, and generally only exists in a single-stranded form in the cell
When glucose levels are low, glucagon stimulates the enzyme that breaks down fructose-2,6-bisphosphate. This molecule typically:
stimulates glycolysis and inhibits gluconeogenesis, so it's absence due to increased breakdown inhibits glycolysis and stimulates gluconeogenesis, which is exactly what is needed when glucose levels are low
All of the following are examples of a reduction EXCEPT:
A. the conversion of glucose to (ultimately) carbon dioxide during cellular respiration.
B. the creation of NADH during glycolysis and the Krebs cycle. (reduction)
C. the conversion of oxygen to water along the electron transport chain. (reduction)
D. the creation of ethanol from pyruvate during alcoholic fermentation. (reduction)
A graduate student in a yeast lab that studies double-strand break (DSB) repair has a mutant strain that is unable to complete repair via nonhomologous end joining. Which of the following is true about this strain?
This mutant strain is unable to complete nonhomologous end joining and is thus relying on homologous recombination for DSB. This is a specific repair process that involves formation of a joint molecule and uses DNA polymerase.
Which forms of eukaryotic post-transcriptional modification is most critical for ribosome association with an mRNA transcript?
The 5' cap contains the ribosome binding site in eukaryotes and is critical for ribosome association
β-Oxidation is a means of creating acetyl-CoA from fatty acids. This acetyl-CoA can then enter the Krebs cycle. Each turn of the β-oxidation cycle produces one acetyl-CoA molecule and a fatty acid two carbons shorter than it was at the beginning of the cycle. Additionally, one NADH and one FADH2 are generated per turn. Lauric acid is an 12-carbon saturated fatty acid. Including those produced in the Krebs cycle, how many total NADH and FADH2 molecules would be generated from the complete β-oxidation of lauric acid and subsequent entry of the acetyl-CoA into the Krebs cycle?
Each turn of the β-oxidation cycle produces one acetyl-CoA and a fatty acid two carbons shorter than before. Lauric acid, with 12 carbons, would ultimately produce 6 acetyl-CoA. If lauric acid enters the β-oxidation cycle, then after one turn of the cycle we would have one acetyl-CoA and a 10C fatty acid. After two turns of the cycle we would have two acetyl-CoA and an 8C fatty acid. This would continue in this manner until after the 5th turn of the cycle we would produce our 5th acetyl-CoA molecule, and all that would be left over would be another two-carbon acetyl-CoA molecule (the 6th acetyl-CoA). This last 2-carbon molecule does not need to go through the β-oxidation cycle again, so only 5 turns of cycle are necessary. Thus, 5 NADH and 5 FADH2 would be generated during β-oxidation. When the 6 acetyl-CoA molecules go through the Krebs cycle, and additional 18 NADH would be generate (3 per turn of the Krebs cycle) and and additional 6 FADH2 would be generated (1 per turn), for a total of 23 NADH and 11 FADH2.
A yeast colony is subject to the mutagen EMS (ethyl methanesulfonate) and subsequently fails to divide. Further analysis reveals excessive supercoiling in the S-phase and failure to progress in the cell cycle. Which of the following was the most likely gene target of EMS?
Hemoglobin is an oxygen-carrying protein found in red blood cells. It is made up of four protein subunits that display cooperative binding. Myoglobin is also an oxygen-carrying protein, however it is found in muscle cells and it is made of only a single protein subunit. How would the saturation curves for hemoglobin and myoglobin compare?
Hemoglobin would have a sigmoidal curve while myoglobin would have a simple curve.
An error was made during DNA replication that ultimately resulted in the synthesis of an mRNA with guanine substituted for cytosine in the codon UCA. This would represent which of the following types of mutation?
The mutation described here involves the substitution of a guanine for a cytosine to generate the sequence UGA, which is a stop codon.
Generation of an early stop codon is known as a nonsense mutation
Following the binding of a loaded tRNA to its codon during translation, which of the following steps occurs next?
Following the binding of a loaded tRNA to a codon, the growing peptide chain is transferred from the tRNA occupying the P site to the tRNA occupying the A site via a peptidyl transfer reaction (i.e., a peptide bond is formed between the last amino acid in the chain and the new amino acid on the tRNA in the A site
Polysaccharides can be used for many different functions. Which of the following is/are polysaccharides that are used primarily for glucose storage?
Item I is true: starch is the polysaccharide used by plants to store glucose (choice B can be eliminated). Item II is true: glycogen is the polysaccharide that animals use to store glucose (choice A can be eliminated). Item III is false: cellulose is a polymer of glucose, but is used primarily for plant structure (choice D can be eliminated and choice C is correct).
During prophase II of meiosis:
Cells undergoing meiosis are haploid after the first cytokinesis event. In prophase II, the sister chromatids have not separated yet. therefore there are two chromatids per chromosome and the cell is haploid
A scientist has had a stationary-phase culture of E. coli at 4°C for several weeks. If she takes a small amount of this culture and puts it into 100 mL of new media (shaking at 37°C overnight), which of the following will be observed?
Bacteria are very hardy cells. While the original culture was in stationary phase, all it takes is more food to get the cells growing again. Initially, the new culture will undergo a lag phase as the bacteria replicate their genomes and get ready to divide, but then the cells will start undergoing binary fission and the culture density will increase exponentially. This is termed the log phase.
Which of the following provides the best description of diffusion?
A solute moves passively down its concentration gradient in a thermodynamically favorable way, increasing entropy.
Polydactyly (extra fingers or toes) is a congenital physical anomaly that can be caused by recessive or dominant alleles. It is found more frequently in blacks than in whites, and more frequently in men than in women. A study on autosomal recessive polydactyly in black women showed the incidence of this condition to be 10 in 1000 births. What is the frequency of the allele causing polydactyly in this population?
If 10 in 1000 births have polydactyly, then the condition is found at a frequency of 1% in this population. In the Hardy-Weinberg equation for genotype frequency
(p2 + 2pq + q2 = 1), q2 is the frequency of the autosomal recessive genotype. Since the question states that polydactyly (at least in this study) is an autosomal recessive disorder, q2 = .01, so q (the frequency of the recessive allele)
A researcher isolates Gram-negative bacilli bacteria. This organism has a:
Gram-negative bacteria have a thin cell wall and an outer membrane which leads to a light pink staining. The bacterial cell wall is made of peptidoglycan; note that the plant cell wall is made of cellulose and the fungal cell wall is made of chitin. Bacilli means rod-shaped, cocci means round, and spirochetes or spirilla mean spiral-shaped.
A researcher dissects testes from a mutant mouse and isolates individual gametes. Flow cytometry analysis shows that there are three populations of cells. Population I has 20 chromosomes, Population II has 19 chromosomes, and Population III has 21 chromosomes. Which of the following is most likely?
Nondisjunction during meiosis I leads to two cells with too many chromosomes (n + 1) and two cells with too few (n - 1). Thus, nondisjunction in meiosis I would only produce two populations of cells (choice C is not correct). Nondisjunction during meiosis II leads to two cells with the correct number of chromosomes (n), one cell with an extra chromosome (n + 1) and one cell that is missing a chromosome (n - 1). This could account for the three populations of cells isolated by the researcher (choice D is correct, note that aneuploidy is just an abnormal number of chromosomes). There is no information to support the fact that meiosis is not being completed (choice A is wrong), or that there is limited mitosis of spermatogonia (this would lead to fewer gametes, but not a change in chromosome number; choice B is wrong).
The viral genome integrates into the host genome during the lysogenic cycle. After this:
In the lysogenic cycle, the viral genome inserts into the host genome. The host genome continues to be transcribed and translated, but the viral genome is silent due to viral-encoded repressor proteins. Both the host and viral genomes are copied during DNA replication. The viral genome is activated and excised when the host cell is stressed, and will enter the lytic or productive cycle. If the virus waits until the host cell is dead, it will certainly not be able to reproduce. Finally, excision of the viral genome is relatively imprecise. Some of the host genome can also be excised and packaged into viral particles. This new bit of DNA can be transferred to the next host on subsequent infection
Which of the following describes a trafficking pathway through the Golgi complex?
The portion of the Golgi complex closest to the endoplasmic reticulum is its cis stack so this is the first step in trafficking. From there, proteins travel to the medial stack and finally the trans stack, the Golgi component farthest from the endoplasmic reticulum. At this point, the protein can be released via exocytosis or anchored to the cell membrane, depending on its function
Eye color is a sex-linked trait in the fruit fly Drosophila melanogaster. A pure-breeding red-eyed female is mated with a pure-breeding white-eyed male. All offspring have red eyes. If an F1 female was backcrossed, which of the following would be observed in the F2?
If all the F1 flies have red-eyes, you know the red eye trait is dominant. Let's assign R = red and r = white. The parental (P) cross is female XRXR × male XrY. The F1 flies would be XRXr (red-eyed females that carry the white allele but don't express it) and XRY (red-eyed males). The next cross in the question is between a XRXr female and the XrY male from the parental generation (remember that a backcross is when an individual is crossed to a previous generation). The resultant F2 flies would be 25% XRXr (red-eyed females), 25% XrXr (white-eyed females), 25% XRY (red-eyed males) and 25% XrY (white-eyed males). You can see that regardless of sex, half the offspring will have white eyes and half will have red eyes
Mitochondria share what characteristic with prokaryotic cells?
Both mitochondria and prokaryotes have a single, distinct chromosome (choice D is correct). Mitochondria are a type of membrane-bound organelle, but do not contain such structures (choice A is wrong). Maternal inheritance refers to the inheritance of the mitochondria and its genome exclusively from the mother (as are all organelles); such a pattern does not exist in prokaryotes (choice B is wrong). Both mitochondria and prokaryotes have ribosomes (choice C is wrong).
The lytic and productive viral cycles are similar in that they both:
involve a virus using host cellular machinery to replicate the viral genome and capsid.
In lab mice, agouti coat color is dominant over chinchilla fur, and black eyes are dominant over pink eyes. If a pure-breeding male agouti black-eyed mouse is crossed with a pure-breeding female chinchilla pink-eyed mouse, then one of the male pups is backcrossed, what is the probability of having a black-eyed chinchilla female F2?
Let's assign alleles as A = agouti, a = chinchilla, B = black, and b = pink. The original cross is AABB × aabb. All F1s are AaBb and a male from this group is then backcrossed (must be to the female parent, so AaBb × aabb). The probability of getting chinchilla fur (aa) is 0.5. The probability of getting black eyes (Bb) is 0.5. The probability of having a female mouse is 0.5. Therefore, overall the answer is (0.5)(0.5)(0.5) = 0.125
A virologist dips his pipette tip into a plaque on a bacterial plate, then into an actively growing culture of E. coli cells. A new strain of E. coli results. Which of the following best explains what occurred?
A plaque is a clear area on a plate otherwise covered in bacterial cells. It can be caused by addition of a toxin, antibiotic, or virus; each of these can kill bacteria and would generate a clear area. Since whatever was in the plaque did not outright kill the actively growing culture, it is not likely a toxin or antibiotic (choice C is wrong). Most likely, the plaque was a region of dead bacteria due to viral infection, and contains active virus. Infection of new cultures with active virus can sometimes result in new strains, if the virus transfers some DNA from its previous host; this process is known as transduction (choice A is the most likely scenario). Since a plaque is typically an area of dead bacteria, choice D is wrong; dead bacteria cannot undergo conjugation. Random genomic mutations are possible but less likely than transduction (choice B is wrong).
Which of the following is true regarding viral entry into a host cell?
All viruses are very specific regarding their host cell. This specificity comes from the requirement of a virus to bind a host cell receptor to allow viral entry (choice A is wrong). Attachment can also be called adsorption, and penetration can also be called eclipse (choice B is wrong). Prokaryotic viruses lack an envelope and therefore cannot undergo membrane fusion with their host. Instead, they rely on injection of viral contents (choice C is wrong). Animal viruses can enter their host cell via membrane-fusion (where the envelope of the virus fuses with the host plasma membrane) or endocytosis (choice D is correct). In both cases, receptor binding is still required, and the viral genome is uncoated after entry into the host. A few animal viruses insert their genome without being taken up entirely, but this is rare.
If a Hfr cell mates with an F- bacterium, which of the following is true?
The Hfr strain has the F factor in the genome, male (F+) bacteria contain the F factor as a plasmid, while the female bacteria (F-) has no F factor. Both male (F+) and Hfr strains can mate with female strains (choice D is wrong). The Hfr cell keeps a copy of its genome (choice B is wrong) and passes a copy to the female cell, through the conjugation bridge (or sex pilus). The female cell can receive a portion of the Hfr chromosome, or a copy of the whole thing (if the cells stay connected for a longer time). The F factor is the final gene transferred during conjugation; if the F factor doesn't make it over to the female cell, she will stay female (but may acquire new genetic traits). If the F factor is transferred, it can end up in a plasmid (the cell would then be male, F+) or the genome (the cell would then be Hfr). While choice A is possible, choice C is more likely.
Black coat color is dominant to brown coat color in Labrador retrievers. A second gene (E or e) controls expression of the fur pigment gene, where the dominant allele allows pigment expression and the recessive allele prevents pigment expression. Labradors lacking black or brown pigment are referred to as "golden". If a dihybrid (heterozygous) male is bred to a homozygous recessive female, which of the following would be expected?
Let's assign B = black and b = brown. For the second gene, E allows pigment expression and e does not (this leads to the golden labs). The cross in the question stem is BbEe (a black lab) × bbee (a golden lab), so choices A and C can be eliminated. The genotypic ratio of the puppies would be 25% BbEe (black lab), 25% Bbee (golden lab), 25% bbEe (brown lab) and 25% bbee (golden lab). Overall, 25% of the puppies would have black fur, 25% would have brown fur, and 50% would have golden fur (due to pigment expression being turned off).
Which of the following has a vesicle fuse with the plasma membrane in order to release material from the cell?
Exocytosis is the process by which the cell uses vesicles to move material out of the cell (choice D is correct). Phagocytosis and pinocytosis are large-scale and small-scale uptake, respectively (choices A and C are wrong), while clathrin-coated pits are involved in receptor-mediated endocytosis (choice B is wrong).
Which of the following is a FALSE statement about (+) RNA viruses and (-) RNA viruses?
(+) RNA viral genomes can serve as templates for transcription, but (-) RNA genomes are complementary to a transcript template (choice C is true and can be eliminated). As a consequence, both must encode an RNA-dependent RNA polymerase to replicate the genome (choice A is true of (+) and (-) RNA viruses and can be eliminated), but a (-) RNA virus must also carry a copy of this protein, to generate functional mRNA from the genome (choice B is false and the correct answer choice). Since a (+) RNA genome can serve as a template for protein production (i.e. mRNA), if a (+) RNA viral genome was injected into a host cell, it would be immediately infective (choice D is true and can be eliminated).
A geneticist testcrosses a dihybrid (heterozygous for two genes) mouse and notices the double dominant and double recessive phenotypes are present at 8% and 10% frequency, respectively. Which of the following is the best conclusion from these data?
Let's assign the dihybrid mouse a genotype of AaBb. The testcross would be AaBb × aabb and the expected ratios of offspring would be 25% AB/ab, 25% Ab/ab, 25% aB/ab and 25% ab/ab (in linkage notation). The question stem says AB/ab is present at only 8% and ab/ab is present at only 10%. These two combinations must represent the recombinant offspring given their low percentages, Ab/ab and aB/ab must be the parental combination of alleles, and the genotype of the dihybrid parent must have been Ab/aB. In other words, the two genes are linked and the dominant allele of one gene is linked to the recessive allele of the other. To determine how far apart the genes are, we can calculate Rf, where Rf = (# recombinants / total) x 100% = [(8% + 10%) / 100%] x 100% = 18% recombination frequency = 18 cM = 18 map units. Overall, the best answer is choice A.
Red-green colorblindness is an X-linked recessive trait in humans. If a carrier female mates with a normal male, what is the probability they will have a colorblind son?
If we assign D = normal and d = colorblind, the cross in the question stem is XDXd × XDY. The probability of receiving the colorblind allele from the mother is 0.5. The probability of receiving the Y chromosome from the father is 0.5. Therefore, the probability they will have a colorblind son is (0.5)(0.5) = 0.25 (choice C is correct).
Alcohol flush reaction is due to an accumulation of acetaldehyde after alcohol consumption and results in a red face and neck. It is due to a dominant missense polymorphism that encodes the enzyme acetaldehyde dehydrogenase (ALDH2); this allele, ALDH2*2, occurs at a frequency of 0.3 in the human population. What is the proportion of individuals in this population that have a red face after drinking alcohol?
The question stem says ALDH2*2 is dominant so we will assign it A; the normal allele is recessive and we will assign it a. In the equation for allele frequency (p + q = 1), p = A = 0.3, so q = a = 1 - 0.3 = 0.7). Since this trait is dominant, both homozygous dominants (AA) and heterozygotes (Aa) will express the phenotype. From the equation for genotype frequency (p2 + 2pq + q2 = 1), the proportion of AA in the population is p2 and the proportion of heterozygotes is 2pq. Therefore the answer is (0.3)2 + 2(0.3)(0.7) = 0.09 + 0.42 = 0.51, or 51% (choice D is correct). Alternatively 1 - q2 can be used in place of the above
Solution A is 1 M glucose. Solution B is 1 M NaCl. Which of the following is true?
The boiling point elevation of Solution B is twice the boiling point elevation of Solution A.
A chemoheterotrophic bacteria that is a leucine auxotroph:
Chemotrophs get their energy from chemicals, not light (choice D is wrong). Heterotrophs rely on organic molecules made by other organisms; they cannot make them from CO2 (choice B is wrong). A leucine auxotroph cannot synthesize the amino acid leucine; to survive, leucine must be provided for the organism in its medium (choice A is wrong and choice C is correct).
In Kartagener's syndrome, defective dynein is produced causing a paralysis of microtubule-based movement of flagellae and cilia. One could expect to find all of the following outcomes EXCEPT:
A. failure to ovulate in women.
B. male infertility.
C. chronic lung infections.
D. ectopic pregnancy in women.
In Kartagener's syndrome, defective dynein is produced causing a paralysis of microtubule-based movement of flagellae and cilia. One could expect to find all of the following outcomes EXCEPT failure to ovulate in women.
Large proteins, most notably albumin, are dissolved in the plasma and serve an important role in regulation of plasma volume. Reducing the amount of albumin to below-normal levels would most likely have which of the following effects?
Large proteins, most notably albumin, are dissolved in the plasma and serve an important role in regulation of plasma volume. Reducing the amount of albumin to below-normal levels would most likely have the effect of the movement of water from the bloodstream into the tissues with resulting swelling, due to reduced osmotic pressure.
The loss of protein from the plasma would cause the osmotic pressure of the blood to go down. It would not affect the hydrostatic pressure (blood pressure). The reduced osmotic pressure of the blood will have the effect of allowing water to leave the bloodstream and enter the tissues, where it will cause the tissues to swell. (You can also think of it as the tissues' relative osmotic pressure increasing, thus they have a greater tendency to draw water out of the blood.)
Vertebrates have developed various renal structures for osmoregulation based on their habitats. Bony fish that live in seawater drink large amounts of seawater and use cells in gills to pump excess salt out of the body. This is in response to:
Vertebrates have developed various renal structures for osmoregulation based on their habitats. Bony fish that live in seawater drink large amounts of seawater and use cells in gills to pump excess salt out of the body. This is in response to a need to maintain their tissues in a hypoosmotic state.
A physiologist observes a dark, dense subcellular structure via electron microscopy that spans the space between two adjacent cell membranes and extends to the underlying cytoskeleton. Upon further investigation, she discovers that two cells with this structure are electrically isolated but that a tracer dye can travel between the cells through the pericellular space. What is the identity of this structure?
The dark structure spanning the cell membranes is a desmosome. Desmosomes are a type of anchoring junction which connect adjacent cells and extend through the cell membrane to associate with components of the cytoskeleton. Unlike gap junctions (also known as communicating junctions), desmosomes do not link the cytoplasms of the two cells (the cells are not electrically coupled). Tight junctions form connections with adjacent cells but prevent pericellular diffusion of molecules, including that of a tracer dye.
How does the van't Hoff factor of NaCl compare to that of NH3?
The i of NaCl is 2 while the i of NH3 is 1. The van't Hoff or ionizability factor is determined by the number of ions produced when a single molecule dissolves in solution. Since NaCl is ionic, it breaks into Na+ and Cl- giving an i = 2. Since NH3 is non-ionic, it does not break down and thus has an i = 1.
Kinesin plays a significant role in mitosis and helps in the spindle apparatus assemble. In which direction does kinesin travel and in which phase of mitosis does its activity begin?
Kinesin would travel away from the microtubule organizing center (MTOC) and begin activity in the prophase when microtubule assembly begins. The spindle apparatus, composed of microtubules, assembles with the minus ends toward the MTOC and the plus ends away from the MTOC. Kinesin travels in the plus direction and would be visible in the prophase as this is when the spindle apparatus begins forming.
Where in the adult human is both mitosis and meiosis occurring in parallel?
In the adult human, both mitosis and meiosis occur in parallel in the seminiferous tubules.
Meiosis is preceded by mitosis. Mitotic production of gametic cells is completed in the human female at the time of birth. However, adult males have a constant cycle of spermatogonia production (via mitosis) leading to meiosis in the seminiferous tubules. Thus both processes occur in that location. Meiosis does not occur in bone marrow or hypodermis.
An inhibitor of β-tubulin polymerization is applied to actively replicating cells. Which of the following processes would be most impacted by this inhibitor?
An inhibitor of β-tubulin polymerization is applied to actively replicating cells. Mitosis would be most impacted by this inhibitor.
β-tubulin polymerization is important in the formation of microtubules. Cytokinesis relies on actin polymerization, not β-tubulin, so would not be significantly impacted. Of the other three processes, all of which rely on microtubules, the most impacted would be mitosis. Formation of the mitotic spindle and movement and separation of replicated chromosomes rely on polymerization and depolymerization of microtubules. Transport of vesicles occurs along microtubules, but these are not actively polymerizing and depolymerizing, as are the spindle fibers during mitosis. The flagella is composed of microtubules in a 9 + 2 arrangement, but is also not dependent on polymerization and depolymerization.
A researcher observes cells from a multicellular organism under the microscope. The cells appear to have cell walls, but lack chloroplasts. He identifies the cells as belonging to Kingdom Fungi. Is his identification correct?
No, without knowing the composition of the cell walls, he cannot distinguish between Fungi and Plantae.
Barr bodies are inactivated X chromosomes, more densely coiled than typical chromosomes. They are typically formed during embryonic development of females; one of the two X chromosomes becomes a Barr body and the other remains active. This prevents the overexpression of proteins coded for by the X chromosome. What is the fate of the Barr body during meiosis?
Barr bodies undergo reactivation before prophase I of meiosis. Oogonia entering into meiosis must have a full complement of chromosomes to ensure the successful production of an ovum. To this end, the Barr body is reactivated prior to meiosis. Replication does not occur in a chromosome-selective fashion; the other X would replicate but this would not prevent nondisjunction. Both X chromosomes must be replicated and subsequently divided. Polar bodies result from unequal apportioning of the cytoplasm during cytokinesis, not the presence of a Barr body (which would not exist in a haploid). Failure to enter into meiosis may ultimately lead to apoptosis but if Barr bodies caused this, no oogonia would undergo meiosis.
A small subpopulation of beetles with a slightly advantageous modification of pincer structure was found to be wiped out after a locally isolated severe wind storm. A biologist studying this event would most likely attribute the loss of the advantageous gene as due to:
The storm is a random event unrelated to the apparent fitness of the beetle in its normal environment. Genetic drift is the random change over time in the allele frequency within a population, such as that caused by the storm in the loss of allele(s) for the altered pincer structure.
A cell has three copies of a gene. All of the following are possible causes of this EXCEPT:
B. non-disjunction during meiosis telophase I.
C. non-disjunction during meiosis anaphase I.
D. non-disjunction during meiosis anaphase II.
A cell has three copies of a gene. All of the following are possible causes of this EXCEPT: non-disjunction during meiosis telophase I. Non-disjunction is a cause of trisomy (a cell having three copies of a gene). During anaphase in meiosis I and II, homologous chromosomes and sister chromatids are separated, respectively. If this separation fails to happen, one cell gets an extra chromosome and the other cell is deficient a chromosome. Translocation is where a piece of one chromosome breaks off and attaches to a different chromosome. This can also lead to three copies of a gene in a cell. Non-disjunction does not occur during telophase, which involves reformation of daughter cells, not separation of chromosomes.
A researcher working on an isolated island discovers what she thinks is a new species of bird. It has a longer beak than most birds on the island and different color plumage. Its song is different from the songbirds on the island. Which of the following tests would be most helpful in determining if the bird is a new species or not?
The defining characteristic of distinct species is the inability to interbreed. If all attempts to breed the new bird with existing birds fail, the new bird can be considered a separate species. There can be differences or similarities in DNA sequence, physical characteristics, and protein composition, however these factors do not distinguish between different species ("sequence the bird's genome," "document all of the physical differences between the new bird and the existing island birds," and "isolate proteins from the bird's eggs, separate them on a western blot, and compare them to proteins isolated from other island bird's eggs" are wrong).
Which of the following is found in meiosis but not mitosis?
A. Paired homologous chromosomes
B. Cell division
C. Separation of sister chromatids
D. A spindle consisting of microtubules
Paired homologous chromosomes are found in meiosis but not mitosis. Spindles and cell division are clearly found in both mitosis and meiosis. Separation of sister chromatids occurs during anaphase of mitosis and anaphase II of meiosis. However, only in meiosis do the homologous chromosomes pair up. This occurs during prophase I of meiosis.
Which of the following is most likely an example of pleiotropism?
A point mutation in a single gene leads to both kidney and lung malformations in utero.
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