Final Exam Study Questions
Calvin Bridges determined that the rare white-eyed males produced from the mating of true-breeding red-eyed females with white-eyed males resulted from a nondisjunction of X chromosomes in the female parent during meiosis. If such a mating produced 1234 red-eyed progeny and three white-eyed males, what is the observed rate of nondisjunction of the X chromosomes? A. 0.97% B. 0.49% C. 0.24% D. 0.12% E. This is impossible to calculate with the given information.
A. 0.97% Each nondisjunction of X chromosomes in the red-eyed female will generate gametes with two X chromosomes and with no X chromosomes with equal frequency. The white-eyed male will generate gametes with an X chromosome, bearing the white trait, or a Y chromosome, again with equal frequency. The union of these gametes generates four potential types of offspring. If an egg with two X chromosomes is fertilized by a sperm with one X chromosome, the result would be an X+X+Xw metafemale fly that is nonviable. If an egg bearing two X chromosomes is fertilized by a sperm bearing a Y chromosome, the result would be an X+X+Y female that will be phenotypically indistinguishable from the red-eyed females resulting from gametes without any nondisjunction. If an egg with no X chromosomes is fertilized with an X chromosome bearing sperm, the result will be an X Xwmale with white eyes. If an egg with no X chromosomes is fertilized with a Y-bearing sperm, the zygote will die because survival requires at least one X chromosome. So if there are three white-eyed males and 1234 red-eyed offspring, we can assume that there were (3×4) 12 zygotes resulting from nondisjunction gametes. Of these, we would have expected only six to survive. Therefore the rate of nondisjunction is 12/(1234 + 9) = 0.0097. The total number of offspring was 1243 to account for the six zygotes that would not have survived and the three white offspring which led to the discovery of the nondisjunction.
What is the width (diameter) of B-form DNA? A. 2 nm B. 2 mm C. 0.34 nm D. 3.4 nm E. 2 angstoms
A. 2 nm The diameter of B-form DNA is 2 nm. Individual base pairs are 0.34 nm apart, and one complete turn of the double helix is 3.4 nm encompassing 10 base pairs.
While anueploidy is a change in the number of chromosomes, polyploidy is: A. a change in the number of sets of chromosomes. B. a change in the structure of the chromosomes. C. an increase in the rate of recombination. D. an increase in the rate of mutation.
A. a change in the number of sets of chromosomes.
Mendel's law of independent assortment depends on the events of meiosis because: A. different chromosomes will segregate independently during anaphase I. B. different chromosomes will segregate independently during anaphase II. C. sister chromatids will segregate independently during anaphase I. D. sister chromatids will segregate independently during anaphase II. E. None of the above.
A. different chromosomes will segregate independently during anaphase I.
Meiosis generally produces all of the following EXCEPT: A. diploid cells. B. four cells instead of two. C. variability due to crossing over. D. variability due to random assortment. E. Actually, it generally produces all of these.
A. diploid cells.
Eukaryotic DNA condensation occurs in which order? A. double helix, nucleosome, 30nm fiber, loops, chromosome B. double helix, loops, nucleosome, 30nm fiber, chromosome C. double helix, 30nm fiber, nucleosome, loops, chromosome D. double helix, nucleosome, loops, 30nm fiber, chromosome
A. double helix, nucleosome, 30nm fiber, loops, chromosome
The structure of the DNA molecule is described by all of the following EXCEPT: A. each strand is assembled by hydrogen bonding. B. it is a double helix. C. the two strands of DNA are antiparallel to each other. D. the two strands interact with each other by hydrogen bonding. E. contains nitrogenous bases.
A. each strand is assembled by hydrogen bonding.
An X-linked dominant trait such as hypophosphatemia (a familial rickets) has pedigrees characterized by all of the following EXCEPT: A. fathers will pass it to their sons. B. mothers will pass it to half their sons. C. mothers will pass it to half their daughters. D. it will be present in every generation. E. Actually, all of these characterize X-linked dominant traits.
A. fathers will pass it to their sons.
Human X and Y chromosomes: A. have pseudo-autosomal regions that are homologous. B. pair during mitosis. C. are mostly the same as each other. D. do not pair up with each other. E. All of the above.
A. have pseudo-autosomal regions that are homologous.
Gregor Mendel used peas in his experiments for all of the following reasons EXCEPT: A. he could grow the plants all year in Moravia. B. he had a number of strains with obviously different traits. C. he could count a large number of experimental results. D. he could control the fertilization. E. Actually, all of these were important reasons to use peas.
A. he could grow the plants all year in Moravia.
With a Z-linked trait such as the cameo phenotype in peacocks, the trait: A. is more often observed in females than males because it is on the Z chromosome. B. is more often observed in males than females because it is on the Z chromosome. C. is never observed in females because it is on the Z chromosome. D. is observed only in females because it is on the Z chromosome. E. None of the above.
A. is more often observed in females than males because it is on the Z chromosome.
The law of independent assortment describes the behavior of traits: A. only if genes are on different chromosomes. B. only if genes are on the same chromosome. C. only if the genes affect the same traits. D. only if genes are on the same chromosome & only if the genes affect the same traits E. None of the above.
A. only if genes are on different chromosomes
A pedigree of an autosomal dominant trait, such as familial hypercholesterolemia, will likely show all of the following characteristics EXCEPT: A. skipping of a generation by the trait. B. equal numbers of males and females affected. C. unaffected individuals do not pass on the trait. D. half the offspring of an affected individual will have the trait.
A. skipping of a generation by the trait.
Considering cases of incomplete dominance: A. the heterozygote has a phenotype intermediate between the homozygotes. B. the homozygote has a phenotype intermediate between the heterozygotes. C. the heterozygote has a phenotype identical to one of the homozygotes. D. the homozygote has a phenotype identical to one of the heterozygotes. E. None of the above.
A. the heterozygote has a phenotype intermediate between the homozygotes.
A karyotype is: A. the set of chromosomes of an individual. B. 23 pairs. C. 46 chromosomes.
A. the set of chromosomes of an individual
What is the approximate average length of DNA from the beginning of one chromatosome to the beginning of the next chromatosome? A. 2000 bp B. 200 bp C. 10 bp D. 20 bp E. 700 bp
B. 200 bp The average repeating unit of DNA in the chromatosome (beads on a string) structure is about 200 bp.
Since individual humans do not need two X chromosomes (males, for example) or a Y chromosome, we can tolerate a large amount of variation in our sex chromosomes, with variants such as: A. Down's syndrome, with 3 X chromosomes. B. Turner's syndrome, with only a single X chromosome. C. Klinefelter's syndrome, with an X and 2 Y chromosomes. D. All of the above. E. None of the above.
B. Turner's syndrome, with only a single X chromosome.
In the case of a number of bird species, the male is homogametic rather than the female. This means that: A. a ZW individual is male. B. a ZW individual is female. C. a ZZ individual is female. D. a ZW individual is a hermaphrodite, having both sexes. E. None of the above.
B. a ZW individual is female.
The chemical composition of nucleotides is known to include all of the following EXCEPT: A. a sugar—ribose in the case of RNA. B. a sugar—ribose in the case of DNA. C. a phosphate. D. a nitrogenous base, adenine, cytosine, guanine, or thymine.
B. a sugar—ribose in the case of DNA
In humans, X-linked color blindness: A. affects women more often than men because it is on the X chromosome. B. affects men more often than women because it is on the X chromosome. C. is hemizygous in women. D. affects daughters if their fathers are affected by it. E. A and C.
B. affects men more often than women because it is on the X chromosome.
Unlike prokaryotic chromosomes, eukaryotic chromosomes: A. are generally circular. B. come in pairs. C. lack telomeres. D. associate with nonhistone proteins.
B. come in pairs
What is the primary level of DNA structure? A. the nucleosome B. double-stranded B-form DNA C. the mitotic chromosome D. the 30nm fiber E. the chromatosome
B. double-stranded B-form DNA Double-stranded DNA is the primary level of chromatin structure. The most common form of DNA in biological systems is B-form DNA.
Mendel's first law is described by all of the following EXCEPT: A. it is known as the law of segregation. B. it states that two genes assort independently of each other. C. it states that we get two copies of every gene. D. it states that some alleles are dominant to others. E. it states that alleles assort randomly.
B. it states that two genes assort independently of each other.
When we consider a case of codominance, such as human blood M and N factors: A. like complete dominance, there are only two phenotypes. B. like incomplete dominance, there are more than two phenotypes. C. like incomplete dominance, the heterozygote is intermediate between the homozygotes. D. like complete dominance, only one allele produces a protein. E. B and C.
B. like incomplete dominance, there are more than two phenotypes.
Inactivation of X chromosomes by mammals results in all of the following EXCEPT: A. production of Barr bodies. B. males and females producing different amounts of protein from genes on the X chromosome. C. dosage compensation. D. unpredictable patterns of expression of some genes. E. Actually, it results in all of these.
B. males and females producing different amounts of protein from genes on the X chromosome.
Since linked genes segregate together, they would be expected to: A. not show dominance. B. not assort independently. C. produce a novel phenotype by gene interaction. D. show incomplete dominance. E. None of the above.
B. not assort independently.
Traits that have differential expression in the two sexes are known as: A. sex-linked. B. sex-influenced. C. autosomal. D. sex-determined. E. None of the above.
B. sex-influenced
Recombination frequency is simply: A. the number of recombinant offspring divided by the number of nonrecombinant offspring. B. the number of recombinant offspring divided by the total number of offspring. C. the number of nonrecombinant offspring divided by the number of recombinant offspring. D. the number of nonrecombinant offspring divided by the total number of offspring.
B. the number of recombinant offspring divided by the total number of offspring
In a case of multiple alleles, such as the ABO blood groups: A. there are multiple recessive alleles. B. there are multiple dominant alleles. C. there are multiple dominant and recessive alleles. D. the genes usually code for DNA repair genes. E. None of the above.
B. there are multiple dominant alleles.
Y-linked traits, which are not well-defined other than maleness, are described by all of the following EXCEPT: A. they are present in each generation. B. they are dominant. C. they are passed by fathers to all of their sons. D. they are passed to none of the daughters. E. Actually, all of these describe Y-linked traits.
B. they are dominant.
Sister chromatids have the following features EXCEPT that: A. both parts have telomeres. B. they are homologs to each other. C. they remain attached to each other at their centromere. D. they will separate during anaphase. E. their DNA sequences are identical.
B. they are homologs to each other
What is the protein composition of the nucleosome? A. H2A, H2B, H3, and H4 B. two copies each of H2A, H2B, H3, and H4 C. H1, H2, H3, and H4 D. H1 and two copies each of H2A, H2B, H3, and H4 E. two copies each of H1, H2, H3, and H4
B. two copies each of H2A, H2B, H3, and H4 The nucleosome consists of double-stranded DNA wrapped nearly two times around a protein core consisting of two molecules each of histone proteins H2A, H2B, H3, and H4.
Differences between prokaryotes and eukaryotes include all of the following EXCEPT: A. presence or absence of the nucleus. B. use of circular chromosomes by eukaryotes. C. small size of prokaryotes. D. absence of organelles in prokaryotes. E. packaging via supercoiling in prokaryotes.
B. use of circular chromosomes by eukaryotes
If nondisjunction of X-chromosomes occurs at a rate of 2% in a certain population of Drosophila, approximately how many white-eyed males would you expect to see from 1000 fertilized eggs resulting from breeding a true breeding red-eyed female with a white- eyed male? A. 20 B. 10 C. 5 D. 2 E. This is impossible to calculate with the given information.
C. 5 Each nondisjunction of X chromosomes in the red-eyed female will generate gametes with two X chromosomes and with no X chromosomes with equal frequency. The white-eyed male will generate gametes with an X chromosome, bearing the white trait, or a Y chromosome, again with equal frequency. The union of these gametes generates four potential types of offspring. If an egg with two X chromosomes is fertilized by a sperm with one X chromosome, the result would be an X+X+Xw metafemale fly that is nonviable. If an egg bearing two X chromosomes is fertilized by a sperm bearing a Y chromosome, the result would be an X+X+Y female that will be phenotypically indistinguishable from the red-eyed females resulting from gametes without any nondisjunction. If an egg with no X chromosomes is fertilized with an X chromosome bearing sperm, the result will be an Xw male with white eyes. If an egg with no X chromosomes is fertilized with a Y-bearing sperm, the zygote will die because survival requires at least one X chromosome. Therefore, if the nondisjunction rate is 2%, this means that 20 of the 1000 offspring were the result of nondisjunction-bearing gametes, of which 1/4 would be white-eyed males.
If the white-eye trait were X-linked dominant rather than X-linked recessive, what result would you expect when crossing a heterozygous white-eyed female with a red-eyed male? A. all red-eyed females with half of the males having white and half of the males having red eyes B. all red-eyed progeny C. In each sex, half of the offspring will have red and half of the offspring will have white eyes. D. all red-eyed females and all white-eyed males E. all red-eyed males with half of the females having white and half of the females having red eyes
C. In each sex, half of the offspring will have red and half of the offspring will have white eyes. In this case, the male's X chromosome, carrying the recessive red-eye trait, would have no bearing on the phenotype of the progeny. Whether the progeny are male or female, their eye color depends entirely on which X chromosome they inherit from their mother. Since their mother is heterozygous, half of her offspring will inherit the dominant white-eye trait and half will inherit the recessive red-eye trait.
In a cross between a white-eyed female with two X chromosomes and one Y chromosome (XwXwY) with a red-eyed male, the majority of the males have white eyes and the majority of the females have red eyes. However, rare red-eyed males and white-eyed females are present as well. What is the most likely chromosomal constitution of the red-eyed males and white-eyed females resulting from this cross? A. XwY males and XwX+ females B. X+XwY females and Xw males C. X+Y males and XwXwY females D. X+XwY males and Xw females E. X+Y males and XwXw females
C. X+Y males and XwXwY females XXY females are particularly prone to nondisjunction of the X chromosomes during meiosis. These nondisjunction events can generate both X+Y red-eyed males (resulting from a Y-bearing egg being fertilized by an X- bearing sperm) and XwXwY white-eyed females (resulting from an egg bearing two X chromosomes being fertilized by a Y-bearing sperm).
If the white-eye trait were Y-linked dominant, rather than X-linked recessive, what result would you expect when crossing a red-eyed female with a white-eyed male? A. all red-eyed males with half of the females having white and half of the females having red eyes B. In each sex, half of the offspring will have red and half of the offspring will have white eyes. C. all red-eyed females and all white-eyed males D. all red-eyed progeny E. all white-eyed females with half of the males having white and half of the males having red eyes
C. all red-eyed females and all white-eyed males If the white-eye trait were Y-linked, only flies inheriting a Y chromosome would be expected to possess white eyes. Under normal circumstances, male Drosophila are the only flies having a Y chromosome; if the trait is dominant, all flies inheriting the trait will have white eyes.
The chi-square test, which is used to discern whether data are close enough to expectations to be chance variation, can be used to test for independent assortment. Doing this requires: A. a data set with both recombinant and nonrecombinant individuals. B. calculating the number of expected individuals, assuming linkage. C. calculating the number of expected individuals, assuming independent assortment.
C. calculating the number of expected individuals, assuming independent assortment.
When two dominant alleles or two recessive alleles are linked to each other it is known as: A. linkage analysis. B. repulsion. C. coupling. D. matching.
C. coupling.
Linked genes do not segregate together 100% of the time, due to the phenomenon known as: A. linkage disequilibrium. B. repulsion. C. crossing over. D. gene conversion. E. heterologous recombination.
C. crossing over.
The test cross: A. is used to determine the genotype of the recessive individual. B. uses a homozygous dominant in the cross. C. gives results either 100% dominant or 1:1 dominant: recessive. D. A and C. E. None of the above.
C. gives results either 100% dominant or 1:1 dominant: recessive.
What drives the formation of the 30 nm fiber of chromatin structure? A. cosmic forces B. acid-base reactions in the DNA C. histone protein interactions D. spindle microtubules E. coordination of zinc ions
C. histone protein interactions Nucleosomes fold on themselves to create the 30 nm fiber based on histone interactions. In particular, histone H1 is important in creating the 30 nm fiber.
Inversions: A. include paracentric inversions, which include the centromere. B. include pericentric inversions, which do not include the centromere. C. suppress crossing over in the inverted areas. D. usually lead to dicentric bridges. E. All of the above.
C. suppress crossing over in the inverted areas.
What level of chromatin structure is most common in interphase DNA? A. chromatosomes in a "beads-on-a-string" arrangement. B. unpackaged B-form DNA C. tethered 30 nm loops on the nuclear scaffold creating a 250 nm fiber D. spiraled 250 nm fibers creating a coil of about 700 nm across E. the 30 nm fiber
C. tethered 30 nm loops on the nuclear scaffold creating a 250 nm fiber Interphase DNA consists of 30 nm loops of DNA with a length of approximately 300 nm tethered to the nuclear scaffold creating a fiber with a width of approximately 250 nm.
If we apply a branch diagram to a dihybrid cross between organisms heterozygous for both genes, we can calculate: A. the chance of an individual being recessive for both is ¼. B. the chance of an individual being recessive for both is 1/8. C. the chance of an individual being recessive for both is 1/16. D. the chance of an individual being recessive for both is 9/16. E. None of the above is correct.
C. the chance of an individual being recessive for both is 1/16.
Bacterial chromosomes are characterized by all of the following EXCEPT: A. they are associated with proteins. B. they are confined to specific regions of the cell. C. they are positively supercoiled. D. they are generally circular. E. there is typically only one per cell.
C. they are positively supercoiled.
A heterozygote is an individual: A. with two different genes for a trait. B. with one locus that has two traits. C. with two different alleles of a gene. D. whose phenotype is recessive. E. None of the above.
C. with two different alleles of a gene.
Chargaff's rules state that in molecules of DNA: A. the amount of adenine is equal to the amount of guanine. B. the amount of cytosine is equal to the amount of thymine. C. the amount of uracil is equal to the amount of adenine. D. (adenine + guanine)/(cytosine + thymine) = 1.
D. (adenine + guanine)/(cytosine + thymine) = 1.
The Hershey-Chase experiment, sometimes known as the Waring Blender experiment, demonstrated that: A. bacteria used DNA as the molecule of heredity. B. bacteria did not use proteins as the molecule of heredity. C. viruses used proteins as the molecule of heredity. D. A and B.
D. A and B
When two nucleotides are joined to each other: A. a phosphodiester bond is formed. B. a covalent bond is formed between the 3' alcohol group of the first nucleotide and the 5' phosphate of the second. C. a covalent bond is formed between the 5' alcohol group of the first nucleotide and the 3' phosphate of the second. D. A and B.
D. A and B
Polyploidy includes: A. autopolyploidy, which doubles the chromosome set as a result of mitotic failure. B. autopolyploidy, which doubles the chromosome set as a result of meiotic failure. C. allopolyploidy, which doubles the chromosome set as a result of meiotic failure. D. A and B.
D. A and B.
The fact that a large range of continuous variation is seen in some traits, such as human height, can be attributed to: A. polygenic inheritance. B. environmental effects. C. anticipation. D. A and B. E. All of the above.
D. A and B.
An X-linked recessive trait, such as the most common form of hemophilia, hemophilia A, will: A. not be passed from father to son. B. be more common in males than in females. C. be passed by affected women to all of their sons D. All of the above.
D. All of the above
The three-point testcross is key to genetic mapping because: A. it allows calculation of recombination frequencies. B. it allows determination of the rate of double crossovers. C. it allows determination of interference, the phenomenon of one crossover suppressing nearby crossovers. D. All of the above
D. All of the above
Viable human aneuploids include: A. trisomy 1, Wittgart's syndrome. B. trisomy 12, Palau syndrome. C. trisomy 17, Michael's syndrome. D. X monosomy, Turner's syndrome.
D. X monosomy, Turner's syndrome
The results of which of the following crosses allowed T. H. Morgan to conclude that the white-eye trait in Drosophila was not autosomal recessive? A. crosses between true-breeding red-eyed females and white-eyed males B. crosses between true-breeding red-eyed females and red-eyed males C. crosses between true-breeding white-eyed females and white-eyed males D. crosses between true-breeding white-eyed females and red-eyed males E. None of the above crosses would give a different result than what would be expected if the white-eye trait was autosomal recessive.
D. crosses between true-breeding white-eyed females and red-eyed males Crosses between true breeding white-eyed females and red-eyed males would result in 50% red-eyed females and 50% white-eyed males because the white eye trait is linked to the X-chromosome and males only inherit their X chromosome from their mother. If the white eye trait were autosomal recessive, this cross would have yielded all red-eyed F1 progeny. All of the other crosses described would yield identical results whether or not the white-eyed trait was X-linked.
The cell cycle is comprised of all of the following EXCEPT: A. interphase. B. mitosis. C. S phase. D. crossing over. E. checkpoints.
D. crossing over.
The Meselsohn-Stahl experiment: A. demonstrated the conservative nature of DNA replication. B. relied on density gradient centrifugation to separate old and new molecules based on their use of different isotopes of carbon. C. demonstrated the dispersive nature of DNA replication. D. demonstrated that replicated double helices consisted of one old and one new strand.
D. demonstrated that replicated double helices consisted of one old and one new strand.
Reproduction by prokaryotic cells requires all of the following EXCEPT: A. replication of the chromosomes. B. division of the cell. C. separation of the chromosomes. D. dissolution of the nucleus. E. Actually, all of these are required.
D. dissolution of the nucleus.
Unequal crossing over often leads to: A. duplications. B. deletions. C. translocations. D. duplications & deletions
D. duplications & deletions
When genes violate Mendel's second law, they are said to: A. fail to assort independently. B. be linked. C. be coupled. D. fail to assort independently & be linked
D. fail to assort independently & be linked
In order to serve as the molecule of heredity, a molecule must possess all of the following traits EXCEPT: A. the ability to specify the phenotype. B. a high degree of complexity. C. high-fidelity reproduction. D. high variability.
D. high variability.
The dihybrid cross: A. was used by Mendel to develop the law of segregation. B. uses two parents, both of which are heterozygous at a single locus. C. shows a 3:1 phenotypic ratio in the next generation. D. is a cross that follows two characters. E. None of the above.
D. is a cross that follows two characters.
Recessive epistasis is described by all of the following EXCEPT: A. it is observed in coat color in Labrador retrievers. B. it has two genes interacting to produce three phenotypes instead of four. C. it involves one gene controlling the expression of another. D. it gives the modified Mendelian ratio of 12:3:1. E. Actually, all of these describe recessive epistasis.
D. it gives the modified Mendelian ratio of 12:3:1.
Okazaki fragments are produced because: A. DNA polymerase functions only in the 3' to 5' direction. B. DNA strands are parallel to each other. C. leading and lagging strands are synthesized at different times. D. lagging strands are copied discontinuously.
D. lagging strands are copied discontinuously.
M phase of the cell cycle includes: A. mitosis. B. cytokinesis. C. meiosis. D. mitosis and cytokinesis E. Mitosis, cytokinesis, and meiosis are all parts of the M phase
D. mitosis and cytokinesis
Supercoiling of circular chromosomes, such as bacterial chromosomes, is usually: A. positive, because it compacts the chromosome more efficiently. B. positive, because it makes unwinding the DNA easier. C. negative, because it compacts the chromosome more efficiently. D. negative, because it makes unwinding the DNA easier.
D. negative, because it makes unwinding the DNA easier.
A pedigree that demonstrates an autosomal recessive trait, such as Tay-Sachs disease, will likely show all of the following characteristics EXCEPT: A. mating between cousins. B. skipping of a generation by the trait. C. equal numbers of males and females affected. D. on average, half the offspring of carriers showing the trait. E. Actually, all of these are expected.
D. on average, half the offspring of carriers showing the trait.
In a monohybrid cross: A. only a single character is studied. B. parents differ from each other in only one trait. C. only one parent is hybrid; the other is pure- breeding. D. only a single character is studied & parents differ from each other in only one trait. E. All of the above.
D. only a single character is studied & parents differ from each other in only one trait.
White eye color in Drosophila was discovered to be sex linked by Thomas Hunt Morgan, who observed that: A. reciprocal crosses produced different results, which is not true for most traits. B. the inheritance pattern depended on the sex of the individuals. C. the white-eyed flies were color blind, in contrast to the red-eyed flies. D. reciprocal crosses produced different results, which is not true for most traits & the inheritance pattern depended on the sex of the individuals E. All of the above.
D. reciprocal crosses produced different results which is not true for most traits & the inheritance pattern depended on the sex of the individuals
Sex-limited traits are different from sex-influenced traits in that: A. sex-limited traits are inherited on the X- chromosome. B. sex-influenced traits are inherited on the X- chromosome. C. sex-limited traits have higher penetrance than sex-influenced traits. D. sex-influenced traits have higher penetrance than sex-limited traits. E. None of the above.
D. sex-influenced traits have higher penetrance than sex-limited traits.
Since linked genes violate Mendel's second law, a typical dihybrid cross with them would show: A. a 9:3:3:1 ratio. B. a 1:1:1:1 ratio. C. a 3:1 ratio. D. something other than a 9:3:3:1 ratio.
D. something other than a 9:3:3:1 ratio.
What level of chromatin structure is most common in mitotic chromosomes? A. chromatosomes in a "beads-on-a-string" arrangement. B. unpacked B-form DNA C. the 30 nm fiber D. spiraled 250 nm fibers creating a coil of about 700 nm across E. tethered 30 nm loops on the nuclear scaffold creating a 250 nm fiber
D. spiraled 250 nm fibers creating a coil of about 700 nm across Mitotic chromosomes consist of tight helical coiling of the 250 nm fiber present in interphase chromatin.
Chromosome morphology types include: A. telocentric, in which the telomere is near one end of the chromosome. B. metacentric, in which the centromere is near one end of the chromosome. C. paracentric, in which the telomere is somewhat off center. D. telocentric, in which the centromere is near one end of the chromosome.
D. telocentric, in which the centromere is near one end of the chromosome.
Eukaryotic chromosomes include centromeres, described by all of the following EXCEPT: A. they associate with kinetochores. B. they are where the mitotic spindle attaches. C. they are responsible for daughter cells receiving identical chromosomes. D. they have highly conserved sequences. E. identifiable as constricted regions along the length of chromosomes.
D. they have highly conserved sequences.
Mutations that affect chromosome structure include all of the following EXCEPT: A. duplication. B. deletion. C. inversion. D. translation.
D. translation.
When there is no crossing over, meiosis produces four nonrecombinant haploid gametes. A single crossover produces: A. four recombinant diploid gametes. B. four recombinant haploid gametes. C. two recombinant diploid gametes and two nonrecombinant diploid gametes. D. two recombinant haploid gametes and two nonrecombinant haploid gametes.
D. two recombinant haploid gametes and two nonrecombinant haploid gametes.
What is the average length of linker DNA? A. 2000 bp B. 4500 bp C. 200 bp D. 10 bp E. 45 bp
E. 45 bp Although the length of linker DNA is variable in a chromosome (in fact, this may be an important way to regulate gene expression), the average length of linker DNA (the DNA between chromatosomes) is 45 bp.
Linear chromosomes have telomeres at their tips, which are described by all of the following EXCEPT: A. they shorten after replication due to the use of primers. B. they can be extended by the reverse transcriptase telomerase. C. their length is a function of age. D. they have conserved sequences within a species. E. Actually, all of these are accurate.
E. Actually, all of these are accurate.
Aneuploidy includes all of the following EXCEPT: A. nullisomy. B. monosomy. C. trisomy. D. tetrasomy. E. Actually, all of these are forms of anueploidy.
E. Actually, all of these are forms of anueploidy.
DNA replication requires all of the following EXCEPT: A. a template, that is, a strand or strands to be copied. B. the monomers that are assembled into the new strand, including dCTP. C. the enzymes that carry out the process, including primase. D. the enzymes that carry out the process, including topoisomerase. E. Actually, all of these are required for DNA replication.
E. Actually, all of these are required for DNA replication.
Eukaryotic chromosomes differ from bacterial chromosomes in all of the following ways EXCEPT: A. they are usually linear. B. they are wrapped around histones. C. they have a centromere. D. they have telomeres. E. Actually, they differ in all of these ways.
E. Actually, they differ in all of these ways.
Which of the following statements is TRUE? A. The length of a mitotic chromosome is less than the width of the average cell nucleus. B. Chromosomes never exist as naked, B-form DNA in a cell. C. The chromosome is made up of DNA and protein. D. The total length of naked DNA in any given chromosome far exceeds the width of an average cell nucleus. E. All of the above statements are true.
E. All of the above statements are true.
A series of transformation experiments, often called Avery-Griffiths, demonstrated that: A. harmless bacteria could become pathogenic by picking up material from pathogenic strains. B. proteins did not serve as the molecule of heredity. C. RNA did not serve as the molecule of heredity. D. DNA served as the molecule of heredity. E. All of the above.
E. All of the above.
Genetic differences between prokaryotes and eukaryotes include: A. the presence of a nucleus in eukaryotes. B. the use of histones by eukaryotes to condense DNA. C. the presence of chromosomes in pairs in eukaryotes. D. A and C. E. All of the above.
E. All of the above.
Investigating the sex-linked trait white eyes in fruit flies, the unusual observation was made that only about 90% of the offspring followed the expected patterns. Morgan's student Bridges explained the new observations: A. by proposing a failure in meiosis. B. by developing the theory of nondisjunction. C. by analyzing the behavior of chromosomes. D. A and B. E. All of the above.
E. All of the above.
The events of meiosis explain Mendel's law of segregation in that: A. meiosis is a reductional division. B. meiosis produces cells that are haploid. C. fertilization produces a new diploid cell from two haploid gametes. D. meiosis produces cells that are haploid & fertilization produces a new diploid cell from two haploid gametes. E. All of the above.
E. All of the above.
Translocations: A. include reciprocal translocations, in which exchange of chromosomal parts occurs. B. include nonreciprocal translocations, in which one chromosome donates material to another. C. suppress crossing over in the case of reciprocal exchanges. D. can involve loss of material, as in Robertsonian translocations. E. All of the above.
E. All of the above.
Types of DNA sequences include: A. unique sequences, which comprise up to one- half of all protein coding sequences. B. moderately repetitive sequences, which include rRNA genes. C. moderately repetitive sequences, which include interspersed sequences, making up one-fourth of the human genome. D. highly repetitive sequences, also known as satellite DNA due to its distinct density. E. All of the above.
E. All of the above.
Lethal alleles: A. produce a 3:2 phenotypic ratio instead of a 3:1 ratio. B. produce a 2:1 phenotypic ratio instead of a 3:1 ratio. C. can persist in the population if recessive. D. A and C. E. B and C.
E. B and C.
Since DNA is a double helix, and the strands are held together by complementary base pairing: A. each strand has the same information as the other strand. B. each strand has the information necessary to reconstruct the other strand. C. there is redundancy in the storage of the hereditary information. D. A and C. E. B and C.
E. B and C.
What is the function of histone H1? A. Histone H1 is necessary for the formation of B-form DNA. B. Histone H1 is one of the major proteins of the nuclear scaffold tethering folds of 30 nm fibers. C. Histone H1 is one of the core histone proteins making up the nucleosome. D. Histone H1 coats the linker DNA between nucleosomes. E. Histone H1clamps the DNA joining and leaving the nucleosome to the core protein octomer.
E. Histone H1clamps the DNA joining and leaving the nucleosome to the core protein octomer. Histone H1 clamps DNA that is entering and leaving the nucleosome to form the chromatosome. Histone H1 is not a core nucleosome protein.
Since the white-eye trait is X-linked recessive, what result would you expect when crossing a heterozygous red-eyed female with a white-eyed male? A. all red-eyed males with half of the females having white and half of the females having red eyes B. all red-eyed progeny C. all red-eyed females with half of the males having white and half of the males having red eyes D. all red-eyed females and all white-eyed males E. In each sex, half of the offspring will have red and half of the offspring will have white eyes.
E. In each sex, half of the offspring will have red and half of the offspring will have white eyes. In this case, the male's X chromosome, carrying the recessive white-eye trait, would have no bearing on the phenotype of the progeny. Whether the progeny are male or female, their eye color depends entirely on which X chromosome they inherit from their mother. Since their mother is heterozygous, half of her offspring will inherit the dominant red-eye trait and half will inherit the recessive white-eye trait.
In the replication fork, leading and lagging strand synthesis are different in that: A. the lagging strand is synthesized 3' to 5'. B. the leading strand does not use a primer. C. the leading strand is synthesized as a number of smaller pieces. D. the lagging strand does not use DNA ligase. E. None of the above is accurate.
E. None of the above is accurate.
Crossing over results in linkage being incomplete and takes place during: A. prophase of mitosis. B. telophase of mitosis. C. prophase of meiosis II. D. telophase of meiosis I. E. None of the above.
E. None of the above.
Meiosis differs from mitosis in that: A. meiosis II includes crossing over. B. meiosis I includes separation of sister chromatids. C. meiosis II includes separation of homologs. D. mitosis includes crossing over. E. None of the above.
E. None of the above.
In the case of many organisms, sex is determined by specific sex chromosomes. In the case of most mammals, females are: A. XX. B. heterogametic. C. homogametic. D. XX and heterogametic E. XX and homogametic
E. XX and homogametic
In the animation, the rare white-eyed males generated from offspring of true-breeding red-eyed females and white-eyed males were the result of a nondisjunction of X chromosomes in meiosis I. Could white-eyed males be generated in this cross by a nondisjunction of X chromosomes in meiosis II? Why or why not? A. No, if a nondisjunction of X chromosomes occurred in meiosis II all progeny would have red eyes. B. No, if a nondisjunction of X chromosomes occurred in meiosis II only white-eyed females could be generated. C. No, if a nondisjunction of X chromosomes occurred in meiosis II, none of the resulting offspring would be viable. D. Yes, because a nondisjunction of X chromosomes in meiosis II will result in both white-eyed males and white-eyed females. E. Yes, because a nondisjunction of X chromosomes in meiosis II will generate both gametes with two normal X chromosomes and gametes with no X chromosomes.
E. Yes, because a nondisjunction of X chromosomes in meiosis II will generate both gametes with two normal X chromosomes and gametes with no X chromosomes. In the mating described, a nondisjunction of X chromosomes in meiosis II would be expected to generate two gametes with one X chromosome each, one gamete with two X chromosomes, and one gamete with no X chromosomes. If an egg bearing no X chromosomes was fertilized by a sperm bearing an X chromosome, an Xw white-eyed male will result.
Since the white-eye trait is X-linked recessive, what result would you expect when crossing a heterozygous red-eyed female with a red-eyed male? A. all red-eyed progeny B. all red-eyed females and all white-eyed males C. all red-eyed males with half of the females having white and half of the females having red eyes D. In both sexes, half of the offspring will have red and half of the offspring will have white eyes. E. all red-eyed females with half of the males having white and half of the males having red eyes
E. all red-eyed females with half of the males having white and half of the males having red eyes This is, in fact, the intercross performed by T. H. Morgan after first crossing a red-eyed female with a white-eyed male. When these F1 were intercrossed, they yielded all red-eyed females with 50% of the males having white and 50% of the males having red eyes.
If the white-eye trait were X-linked dominant, rather than X-linked recessive, what result would you expect when crossing a heterozygous white-eyed female with a white-eyed male? A. all red-eyed males with half of the females having white and half of the females having red eyes B. In each sex, half of the offspring will have red and half of the offspring will have white eyes. C. all red-eyed females with half of the males having white and half of the males having red eyes D. all red-eyed progeny E. all white-eyed females with half of the males having white and half of the males having red eyes
E. all white-eyed females with half of the males having white and half of the males having red eyes If the white-eye trait were X-linked dominant, all of the female progeny would have white eyes because they would inherit the white-eye trait from their father's X chromosome. In contrast, the males would inherit their single X chromosome from their mother, with half receiving the white-eye trait and half receiving the red-eye trait.
