BIOL110 Exam 3 Mastering Bio Questions

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What is the chance that any daughter that the couple will be color blind with a widow's peak?

Suppose the couple had a daughter with normal color vision and a widow's peak. What is the chance that she is a heterozygous for both genes?

The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.What is the probability that individual III-1 is Ww?

What is the probability that individual III-1 is Ww?

In a particular plant, leaf color is controlled by gene locus D. Plants with at least one allele D have dark green leaves, and plants with the homozygous recessive dd genotype have light green leaves. A true-breeding, dark-leaved plant is crossed with a light-leaved one, and the F1 offspring is allowed to self-pollinate. The predicted outcome of the F2 is diagrammed in the Punnett square shown in the figure, where 1, 2, 3, and 4 represent the genotypes corresponding to each box within the square.Which of the boxes marked 1-4 correspond to plants that will be true-breeding?

1 and 4

What is the chance that any son the couple has will be color blind with a straight hairline?

1/2

Red-green color blindness is due to an X-linked recessive allele in humans. A widow's peak (a hairline that comes to a peak in the middle of the forehead) is due to an autosomal dominant allele. Consider the following family history: A man with a widow's peak and normal color vision marries a color-blind woman with a straight hairline. The man's father had a straight hairline, as did both of the woman's parents. Use the family history to make predictions about the couple's children. Drag the correct label to the appropriate location in the table. Not all labels will be used. If the couple has a child, what is the chance that it will be a son with a widow's peak?

1/4 For the hairline gene, the man's genotype is Ww and the woman's is ww.For the color vision gene, the man's genotype is X N Y and the woman's is X n X n. Because the genes are on different chromosomes, they assort independently. Also, because one gene is sex-linked, it exhibits a different inheritance pattern in males and females. You should use the multiplication rule to calculate the chances of two events (e.g., widow's peak and colorblindness) occurring together in a specific combination, paying attention to whether the offspring is male or female.

If two genes are found on different chromosomes, or if they are far enough apart on the same chromosome that the chance of a crossover between them is very high, the genes are considered to be unlinked. Unlinked genes follow Mendel's law of independent assortment. If, however, two genes tend to "travel together" because they are near one another on the same chromosome, they are said to be linked. Linked genes do not follow Mendel's law of independent assortment. A wild-type tomato plant (Plant 1) is homozygous dominant for three traits: solid leaves (MM), normal height (DD), and smooth skin (PP). Another tomato plant (Plant 2) is homozygous recessive for the same three traits: mottled leaves (mm), dwarf height (dd), and peach skin (pp). In a cross between these two plants (MMDDPP x mmddpp), all offspring in the F1 generation are wild type and heterozygous for all three traits (MmDdPp). Now suppose you perform a testcross on one of the F1 plants (MmDdPp x mmddpp). The F2 generation can include plants with these eight possible phenotypes: solid, normal, smooth solid, normal, peach solid, dwarf, smooth solid, dwarf, peach mottled, normal, smooth mottled, normal, peach mottled, dwarf, smooth mottled, dwarf, peach Assuming that the three genes undergo independent assortment, predict the phenotypic ratio of the offspring in the F2 generation.

1:1:1:1:1:1:1:1 The law of independent assortment states that each pair of alleles segregates independently of all other pairs of alleles during gamete formation. The three Punnett squares below show the predicted phenotypic ratios for each trait in the F2 offspring. MmmMmmmmMmmm1/2 Mm (solid)1/2 mm (mottled) DddmDddddDddd1/2 Dd (normal)1/2 dd (dwarf) PppPppppPppp1/2 Pp (smooth)1/2 pp (peach) Thus, you would predict that all eight possible phenotypes from the cross MmDdPp x mmddpp are equally likely, resulting in a phenotypic ratio of 1:1:1:1:1:1:1:1.

2 and 3

In addition to A and a, the "agouti" gene has a third allele, AY . Here is some information about the inheritance of the AY allele. The AY allele is dominant to both A and a. The homozygous genotype (AYAY ) results in lethality before birth. The heterozygous genotypes (AYA or AYa) result in yellow fur color, regardless of which alleles are present for the B/b gene. (This effect exhibited by the AY allele is known as epistasis--when the expression of one gene masks the expression of a second gene.) Suppose you mate two mice with the genotypes AYaBb x AYaBb . Considering only the live-born offspring, what would be the expected frequency of mice with yellow fur?

2/3 Because the presence of the AY allele is epistatic to (masks expression of) the B/b gene, the B/b gene does not need to be taken into consideration in this problem. For the AYa x AYa cross, 1/4 of the offspring would have the AYAY genotype, which is lethal before birth. For the live-born offspring, 2/3 would be AYa, and thus have yellow fur.

Skin color in a certain species of fish is inherited via a single gene with four different alleles. How many different types of gametes would be possible in this system? 4 16 2 8

What is the likelihood that the progeny of IV-3 and IV-4 will have the trait?

50%

In the same mouse species, a fourth unlinked gene (gene P/p) also affects fur color. For mice that are either homozygous dominant (PP) or heterozygous (Pp), the organism's fur color is dictated by the other three genes (A/a, B/b, and C/c). For mice that are homozygous recessive (pp), large patches of the organism's fur are white. This condition is called piebaldism. In a cross between two mice that are heterozygous for agouti, black, color, and piebaldism, what is the probability that offspring will have solid black fur along with large patches of white fur?

9/256 Because each gene segregates independently, you need to determine the probability of each genotype independently and then multiply the four probabilities together. The probability of offspring with solid color (aa) is 1/4; the probability of offspring with black fur (BBor Bb) is 3/4; the probability of colored fur (CCor Cc) is 3/4; and the probability of piebald, or white patches (pp), is 1/4. The combined probability is1/4 x 3/4 x 3/4 x 1/4 = 9/256.

Here are four individual strands of DNA: 1- 5'AATTCCGGCCTAACTT3' 2- 5'AAGTTAGGCCGGAATT3' 3- 5'TTAAGGCCGGATTGAA3' 4- 5'TTCAATCCGGCCTTAA3' Which strands would pair with one another? A. 1 and 2, 3 and 4 B. 1 and 4, 2 and 3 C. 1 and 3, 2 and 4 D. 1 and 3 only

A. 1 and 2, 3 and 4

Below are four DNA sequences. Which pair(s) encodes alleles of the same gene? 1-attcgctaacgca 2-accggcaacattc 3-attccctaactca 4-acccgcatcattc A. 1 and 3, 2 and 4 B. 2 and 3 C. 1 and 3 D. 1 and 2 E. 2 and 4

A. 1 and 3, 2 and 4

Skin color in a certain species of fish is inherited via a single gene with four different alleles. One fish of this type has alleles 1 and 3 (S1S3) and its mate has alleles 2 and 4 (S2S4). If each allele confers a unit of color darkness such that S1 has one unit, S2 has two units, and so on, then what proportion of their offspring would be expected to have five units of color? A. 1/2 B. 1/8 C. 0 D. 1/4

A. 1/2

Refer to the following information to answer the question below.A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive.They have a daughter who is a dwarf with normal color vision. What is the probability that she is heterozygous for both genes? A. 100% B. 50% C. 0% D. 25% E. 75%

A. 100%

In this example, crossing over could occur between ______. A. 2 and 3 B. 1 and 2 C. 3 and 4

A. 2 and 3

In rabbits, the homozygous genotype LCLC has normal legs, LCLc results in deformed legs, and LcLc results in very short legs. The genotype FBFB produces black fur, FBFb brown fur, and FbFb white fur. If a cross is made between brown rabbits with deformed legs and white rabbits with deformed legs, what percentage of the offspring would be expected to have deformed legs and white fur? A. 25% B. 100% C. 50% D. 33%

A. 25%

A. 26.0 map units

A. 38.0%

If one parent who has Huntington's disease is heterozygous for the trait and the other parent is not affected, what are the chances of their offspring inheriting the trait? A. 50% B. 100% C. 0 D. 25%

A. 50%

The recessive alleles for cinnibar eyes (c) and vestigial wings (v) identify two autosomal genes on the third chromosome of Drosophila melanogaster. When females, heterozygous at these two genes (Cc Vv), were crossed with cinnibar-eyed, vestigial winged males, the following classes and numbers of progeny (n=1000) were obtained: wild-type eye color, wild-type wings 42 Wild-type eye color, vestigial wings 464 Cinnibar eye color, vestigial wings 38 Cinnibar eye color, wild-type wings 456 A. 8 map units B. 4 map units C. 92 map units D. 46 map units

A. 8 map units

In Drosophila, two genes on the fourth chromosome, bent wings (b) and shaven bristles (s) are completely linked. If a fly with the genotype bbss is crossed to a fly with the genotype BbSs, what type(s) of gametes will their offspring produce. A. BS and bs B. BS, Bs, bS, and bs C. BS D. bs

A. BS and bs

You are examining a trait which shows three distinct phenotypes, with the phenotype of heterozygotes expressing both alleles seen in the homozygotes. This is an example of: A. Codominance. B. Complete dominance. C. Incomplete dominance. D. Pleiotrophy.

A. Codominance.

This image shows a pair of homologous chromosomes with three linked genes indicated (genes A, B, and C). Which of the following statements is true with respect to these genes? A. Crossing-over can produce recombinant gametes with the genotypes abC and ABC. B. The parental arrangement of alleles is abC and ABC. C. Crossing-over can produce recombinant gametes with the genotypes aBC and AbC. D. Independent assortment can produce gametes with the genotypes aBC, abC, AbC, ABC.

A. Crossing-over can produce recombinant gametes with the genotypes abC and ABC.

Hutchinson-Gilford progeria is an exceedingly rare human genetic disorder in which there is very early senility and death, usually from coronary artery disease, at an average age of approximately 13. Patients, who look very old even as children, do not live to reproduce. Which of the following represents the most likely assumption? A. The disorder may be due to mutation in a single protein-coding gene. B. Successive generations of a family will continue to have more and more cases over time. C. The disease is autosomal dominant. D. Each patient will have had at least one affected family member in a previous generation. E. All cases must occur in relatives; therefore, there must be only one mutant allele.

A. The disorder may be due to mutation in a single protein-coding gene.

In humans, what determines the sex of offspring and why? A. The male gamete determines sex because each male gamete can contribute either an X or a Y chromosome. B. The chromosome contribution from both parents determines sex because the offspring uses all the parents' chromosomes. C. The female gamete determines sex because only the female gamete provides cytoplasm to the zygote. D. The female gamete determines sex because only the female gametes can have one of two functional sex chromosomes. E. The male determines sex because the sperm can fertilize either a female egg or a male egg.

A. The male gamete determines sex because each male gamete can contribute either an X or a Y chromosome.

Which of these descriptions of the behavior of chromosomes during meiosis explains Mendel's law of segregation? A. The two alleles for each gene separate as homologous chromosomes move apart during anaphase I. B. The arrangement of each pair of homologous chromosomes on the metaphase plate during metaphase I is random with respect to the arrangements of other pairs. C. Sister chromatids separate during anaphase II.

A. The two alleles for each gene separate as homologous chromosomes move apart during anaphase I. The chromosomal basis of Mendel's law of segregation state that the two alleles for each gene separate during gamete formation.

You are studying two genes that are located on the same chromosome. They are located 55 map units apart. How will the alleles of these genes behave in a dihybrid cross? A. They will appear to segregate independently, as if they are located on different chromosomes. B. Can't say without knowing the recombination frequency. C. They will stay together.

A. They will appear to segregate independently, as if they are located on different chromosomes.

Consider the following family history: Bob has a genetic condition that affects his skin. Bob's wife, Eleanor, has normal skin. No one in Eleanor's family has ever had the skin condition. Bob and Eleanor have a large family. Of their eleven children, all six of their sons have normal skin, but all five of their daughters have the same skin condition as Bob. Based on Bob and Eleanor's family history, what inheritance pattern does the skin condition most likely follow? A. X-linked dominant B. autosomal recessive C. Y-linked D. autosomal dominant E. X-linked recessive

A. X-linked dominant If the skin condition is caused by an X-linked dominant allele, a father would pass the allele on to all of his daughters, who would all have the skin condition. In contrast, the father would not pass the allele on to any of his sons because the sons would receive the father's Y chromosome, not his X chromosome. As a result, none of the sons would inherit the skin condition.

Mice with the genotypes BBCc and bbcc are crossed. The alleles for brown coat (b) and no pigment (c) are recessive, and black coat (B) is dominant. What color of mice would NOT be produced by this mating? A. brown B. white C. black

A. brown

In general, the frequency with which crossing over occurs between two linked genes depends on what? A. how far apart they are on the chromosome B. the phase of meiosis in which the crossing over occurs C. whether the genes are on the X or some other chromosome D. the characters the genes code for E. whether the genes are dominant or recessive

A. how far apart they are on the chromosome The farther apart two genes are, the greater the probability that a crossover will occur between them, and therefore the greater the recombination frequency.

In cattle, roan coat color (mixed red and white hairs) occurs in the heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. Which of the following crosses would produce offspring in the ratio of 1 red:2 roan:1 white? A. roan × roan B. red × roan C. white × roan D. red × white

A. roan × roan

A homozygous tomato plant with red fruit and yellow flowers was crossed with a homozygous tomato plant with golden fruit and white flowers. The F1 all had red fruit and yellow flowers. The F1 were testcrossed by crossing them to homozygous recessive individuals and the following offspring were obtained:Red fruit and yellow flowers-41Red fruit and white flowers-7Golden fruit and yellow flowers-8Golden fruit and white flowers-44How many map units separate these genes? A. 17.6 B. 15 C. 17.1 D. 18.1

B. 15

What ratio of F1 generation phenotypes would be expected from a test cross of flies differing in two characters (females are heterozygous for both genes, and males are homozygous recessive for both genes) if the two genes assort independently of each other? A. 8:2:2:1 B. 1:1:1:1 C. 9:3:3:1 D. 3:3:1:1 E. 5:5:1:1

B. 1:1:1:1

The following question refers to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. What is the likelihood that the progeny of IV-3 and IV-4 will have the trait? A. 75% B. 50% C. 0% D. 100% E. 25%

B. 50%

Now, suppose that the three tomato genes from Part A did not assort independently, but instead were linked to one another on the same chromosome. Would you expect the phenotypic ratio in the offspring to change? If so, how? Which statement best predicts the results of the cross MmDdPp x mmddpp assuming that all three genes are linked? A. Only the parental phenotypes could possibly occur in the offspring. B. All eight possible phenotypes could occur, but a greater proportion of the offspring would have the parental phenotypes. C. All eight possible phenotypes would occur in equal numbers in the offspring (1:1:1:1:1:1:1:1). D. Lack of independent assortment means that you cannot predict the frequencies of phenotypes in the offspring.

B. All eight possible phenotypes could occur, but a greater proportion of the offspring would have the parental phenotypes. Because all three genes are linked, it is more likely that the parental allele combinations would stay together rather than be recombined through a crossover event. That is why a greater proportion of the offspring would have parental phenotypes. Nevertheless, some crossing over would likely occur, which is why a small proportion of the offspring would have recombinant phenotypes.

A woman who has blood type A positive has a daughter who is type O positive and a son who is type B negative. Rh positive is a trait that shows simple dominance over Rh negative. Which of the following is a possible phenotype for the father? A. AB negative B. B positive C. A negative D. O negative

B. B positive

Folk singer Woody Guthrie died of Huntington's disease, an autosomal dominant disorder. Which statement below must be true? A. His sons will develop Huntington's disease but not his daughters. B. It is very likely that at least one of Woody Guthrie's parents also have had the allele for Huntington's disease. C. All of his children will develop Huntington's disease. D. His daughters will die of Huntington's disease but not his sons. E. There is not enough information to answer the question.

B. It is very likely that at least one of Woody Guthrie's parents also have had the allele for Huntington's disease. Unless the disease is caused by a new mutation, which is quite rare, individuals with a dominant condition must have inherited the dominant allele from one of their parents. As it happens, Guthrie's mother also died of Huntington's disease.

__________ is an example of a character that is controlled by more than one gene (a quantitative character) and demonstrates a continuum of phenotypes. A. Pleiotropy B. Polygenic inheritance C. Epistasis D. Codominance E. Incomplete dominance

B. Polygenic inheritance

Which of these descriptions of the behavior of chromosomes during meiosis explains Mendel's law of independent assortment? A. The two alleles for each gene separate as homologous chromosomes move apart during anaphase I. B. The arrangement of each pair of homologous chromosomes on the metaphase plate during metaphase I is random with respect to the arrangements of other pairs. C. Sister chromatids separate during anaphase II.

B. The arrangement of each pair of homologous chromosomes on the metaphase plate during metaphase I is random with respect to the arrangements of other pairs. Alleles of genes on nonhomologous chromosomes assort independently during gamete formation

Of the following statements, which is/are usually true if the genes for two different characters are linked? A. The alleles for each character won't segregate. B. The genes are on the same chromosome and do not assort independently. C. The alleles for the linked genes assort independently from one another. D. A proportional number of parental phenotypes and recombinant phenotypes are represented among offspring. E. All of the above

B. The genes are on the same chromosome and do not assort independently.

Which of the following statements about independent assortment and segregation is correct? A. The law of segregation is accounted for by anaphase of mitosis. B. The law of independent assortment requires describing two or more genes relative to one another. C. The law of segregation requires describing two or more genes relative to one another. D. The law of independent assortment is accounted for by observations of prophase I.

B. The law of independent assortment requires describing two or more genes relative to one another.

What is the reason that closely linked genes are typically inherited together? A. The number of genes in a cell is greater than the number of chromosomes. B. They are located close together on the same chromosome. C. Alleles are paired together during meiosis. D. Genes align that way during metaphase I of meiosis.

B. They are located close together on the same chromosome.

Duchenne muscular dystrophy is a serious condition caused by a recessive allele of a gene on the human X chromosome. The patients have muscles that weaken over time because they have absent or decreased dystrophin, a muscle protein. They rarely live past their 20s. How likely is it for a woman to have this condition? A. Women can never have this condition. B. Very rarely: it is rare that an affected male would mate with a carrier female. C. One-fourth of the daughters of an affected father and a carrier mother could have this condition. D. One-half of the daughters of an affected man would have this condition. E. Only if a woman is XXX could she have this condition.

B. Very rarely: it is rare that an affected male would mate with a carrier female.

Can we determine mom's chromosomes if son is affected with baldness? A. X B X B B. X B X b C. X b X b D. it could be X B X B or X b X b E. we cannot say for sure

B. X B X b Since the mother has an affected son who inherited his X from her, she must have an X b to pass on to him. But she has another son who inherited an X from her but is not affected, so she must also have an X B to pass on.

The locations for three genes have been added to this cartoon. For clarity, the locations have only been noted on one chromatid. The lowest frequency of crossovers would occur between genes ____ and ____. A. b and c B. a and b C. a and c

B. a and b

SRY is best described as _____. A. an autosomal gene that is required for the expression of genes on the Y chromosome B. a gene region present on the Y chromosome that triggers male development C. a gene present on the X chromosome that triggers female development D. an autosomal gene that is required for the expression of genes on the X chromosome

B. a gene region present on the Y chromosome that triggers male development

SRY is best described in which of the following ways? A. an autosomal gene that is required for the expression of genes on the Y chromosome B. a gene region present on the Y chromosome that triggers male development C. a gene required for development, and males or females lacking the gene do not survive past early childhood D. an autosomal gene that is required for the expression of genes on the X chromosome E. a gene present on the X chromosome that triggers female development

B. a gene region present on the Y chromosome that triggers male development

All female mammals have one active X chromosome per cell instead of two. What causes this? A. attachment of methyl (CH3) groups to the X chromosome that will remain active B. activation of the XIST gene on the X chromosome that will become the Barr body C. activation of the BARR gene on one X chromosome, which then becomes inactive D. inactivation of the XIST gene on the X chromosome derived from the male parent E. crossing over between the XIST gene on one X chromosome and a related gene on an autosome

B. activation of the XIST gene on the X chromosome that will become the Barr body

The figure below shows the pedigree for a family. Dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis. Males are represented by squares, females by circles. From this pedigree, this trait seems to be inherited _____. A. as a result of epistasis B. as an autosomal dominant C. as an autosomal recessive D. from mothers

B. as an autosomal dominant

Radish flowers may be red, purple, or white. A cross between a red-flowered plant and a white-flowered plant yields all-purple offspring. The flower color trait in radishes is an example of which of the following? A. a multiple allelic system B. incomplete dominance C. codominance D. sex linkage

B. incomplete dominance

DNA strands are called antiparallel because of _______________. A. semiconservative replication B. the two strands having opposite polarities C. the alignment of bases required for protein synthesis D. the specific base-pairing rules

B. the two strands having opposite polarities

________________________ is an example of a recessive inherited trait. A. Huntington's disease B.Cystic fibrosis C. Brachydactyly. D. The AB blood type

B.Cystic fibrosis

In the same mouse species, a third unlinked gene (gene C/c) also has an epistatic effect on fur color. The presence of the dominant allele C (for color), allows the A/a and B/b genes to be expressed normally. The presence of two recessive alleles (cc), on the other hand, prevents any pigment from being formed, resulting in an albino (white) mouse. Match the phenotypes on the labels at left to the genotypes listed below. Labels can be used once, more than once, or not at all.

Because the C/c gene is epistatic to both the A/a and B/b genes, any offspring with the cc allele combination will be albino. Otherwise, the A/a and B/b genes are expressed normally.

In water snakes, body color is controlled by two alleles that display incomplete dominance. Snakes with the BB genotype are brown, those with the Bb genotype are gray, and individuals with the bb genotype are green. If a brown snake and a gray snake are crossed, what ratio of phenotypes should be expected in the offspring? A. 1:1, gray:green B. 1:2:1, brown:gray:green C. 1:1, brown:gray D.1:1:1, brown:gray:green1:1, brown:green

C. 1:1, brown:gray

The gene controlling ear color in an organism known as a gizmo has two alleles that exhibit incomplete dominance: CR, which codes for red ears; and CY, which codes for yellow ears. Individuals that are homozygous for the CR allele have red ears, whereas the heterozygous genotype produces orange ears. Those that are homozygous for the CY allele have yellow ears. If two individuals with orange ears are crossed, what ratio of phenotypes should be expected in the offspring? A. 9:3:1, red:yellow:orange ears B. 3:1, red:yellow ears C. 1:2:1, red:orange:yellow ears D. 1:2:1, red:yellow:orange ears E. 1:3, red:yellow ears

C. 1:2:1, red:orange:yellow ears

A DNA sequence reads 3' AGGCTTA 5'. Therefore, the complementary DNA strand must read _______________.' A. 3' AGGCTTA 5' B. 3' TCCGAAT 5' C. 3' TAAGCCT 5' D. 3' UAAGCCU 5'

C. 3' TAAGCCT 5'

The following enzymes are involved in DNA replication on the lagging strand: 1-ligase 2-primase 3-helicase 4-polymerase In order for normal DNA replication to occur, these enzymes must work in the following order: A. 2,3,4,1 B. 4,2,1,3 C. 3,2,4,1 D. 1,2,3,4

C. 3,2,4,1

If the recombination frequency between P and O is 7.4, and between N and O it is 7.9, what is the likely order of these genes on the chromosome if the distance between N and P is 15.7? A. None of these choices B. N-O-PO-P-N C. Either P-O-N or N-O-P D. O-N-P

C. Either P-O-N or N-O-P

Pull out a piece of paper and pencil because this question will take some thought. First, work out the predicted offspring of a mating between a white-eyed female and a red-eyed male Drosophila. Next, work out the predicted offspring of a heterozygous red-eyed female and a white-eyed male. What is the result of both crosses? A. In the first case, all females will have white eyes and all males red eyes. In the second case, this is reversed. B. In the first case, all males will have white eyes. In the second case, this will be reversed. C. In the first case, all females will have red eyes and all males will have white eyes. In the second case, there will be equal numbers of white and red eyes. D. In both cases, the result is the same. E. In the first case, there will be equal numbers of white and red eyes. In the second case, only females will have red eyes.

C. In the first case, all females will have red eyes and all males will have white eyes. In the second case, there will be equal numbers of white and red eyes.

Use the following information to answer the question. In a Drosophila experiment, a cross was made between homozygous wild-type females and yellow-bodied males. All of the resulting F1s were phenotypically wild type. However, adult flies of the F2 generation (resulting from matings of the F1s) had the characteristics shown in the figure. How is the mutant allele for yellow body inherited? A. It is not X-linked. B. It is Y-linked. C. It is X-linked. D. It is inherited by X inactivation.

C. It is X-linked.

Hemophilia (recessive) male? A. X h Y 0 B. X H Y H C. X H Y 0 D. X H Y h E. we cannot say for sure

C. X H Y 0 Since hemophilia is sex-linked, I must have only one copy of the gene, on my X chromosome. And since I don't have the recessive disorder, I must have the dominant form of the gene.

Red-green color blindness is a sex-linked recessive trait in humans. Two people with normal color vision have a color-blind son. What are the genotypes of the parents? A. XNXN and XNY B. XnXn and XnY C. XNXn and XNY D. XNXN and XnY

C. XNXn and XNY

Pedigrees reveal that a trait shows a dominant pattern of inheritance if that trait generally _______. A. skipped generations B. was lethal C. did not skip generations D. was only inherited by males

C. did not skip generations

A gene that affects the expression of a second gene is an example of __________________. A. pleiotropy B. codominance C. epistasis D. polygenic inheritance E. incomplete dominance

C. epistasis

When genes on a particular chromosome do not assort independently in genetic crosses, the genes are _______________. A. recombinants B. wild type C. linked D. homozygous alleles E. not linked

C. linked

Males are more often affected by sex-linked traits than females because _____. A. X chromosomes in males generally have more mutations than X chromosomes in females B. male hormones such as testosterone often alter the effects of mutations on the X chromosome C. males are hemizygous for the X chromosome D. female hormones such as estrogen often compensate for the effects of mutations on the X chromosome

C. males are hemizygous for the X chromosome

The effects of ______________________ can be almost completely overcome by regulating the diet of the individual from infancy. A. cystic fibrosis B. sickle-cell disease C. phenylketonuria (PKU) D. Huntington's disease

C. phenylketonuria (PKU)

If a male affected by a recessive sex-linked trait reproduces with a female who is homozygous wild type for that trait, what is the chance that any of their offspring will have the trait? A. 50% B. 75% C. 100% D. 0 E. 25%

D. 0

Odds that homozygous recessive bald mom and normal dad have a son that is bald? A. 0% B. 25% C. 50% D. 100% E. the answer cannot be determined from the information given

D. 100% If a woman is bald, she must have two copies of the recessive gene, so she must pass one on to her children. Since her son would get his Y chromosome from his father and his X chromosome from his mother, he would inherit her baldness gene. That means that all of her sons will be bald.

In mice, agouti fur is a dominant trait resulting in individual hairs having a light band of pigment on an otherwise dark hair shaft. A mouse with agouti fur is shown here, along with a mouse with solid color fur, which is the recessive phenotype (A = agouti; a = solid color). A separate gene, which is not linked to the agouti gene, can result in either a dominant black pigment or a recessive brown pigment (B = black; b = brown). A litter of mice from the mating of two agouti black parents includes offspring with the following fur colors:solid color, blacksolid color, brown (sometimes called chocolate)agouti blackagouti brown (sometimes called cinnamon) What would be the expected frequency of agouti brown offspring in the litter? A. 9/16 B. 1/2 C. 1/8 D. 3/16 E. 1/4 F. not enough information given

D. 3/16 Because the two traits are determined by unlinked genes, they assort independently. As a result, you need to use the multiplication rule to calculate the probability of agouti brown offspring (A_ bb) from AaBb parents. The probability of A_offspring is 3/4, and the probability of bb offspring is 1/4. The combined probability is therefore 3/4 x 1/4 = 3/16.

You cross a true-breeding red-flowered snapdragon with a true-breeding white-flowered one. All of the F1 are pink. What does this say about the parental traits? A. Red is completely dominant. B. Both red and white are pleiotropic. C. Pink is dominant, and red and white are recessive. D. Red shows incomplete dominance over white.

D. Red shows incomplete dominance over white. Red shows incomplete dominance over white, and the F1 is therefore pink.

How would one explain a testcross involving F1 dihybrid flies in which more parental-type offspring than recombinant-type offspring are produced? A. Recombination did not occur in the cell during meiosis. B. Both of the characters are controlled by more than one gene. C. The two genes are linked but on different chromosomes. D. The two genes are closely linked on the same chromosome.

D. The two genes are closely linked on the same chromosome.

Gene S controls the sharpness of spines in a type of cactus. Cacti with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cacti have dull spines. At the same time, a second gene, N, determines whether or not cacti have spines. Homozygous recessive nn cacti have no spines at all. The relationship between genes S and N is an example of which of the following inheritance patterns? A. codominance B. pleiotropy C. incomplete dominance D. epistasis

D. epistasis

A mutation in the DNA sequence of a gene that codes for a protein: A. is always harmful to the organism, impairing the function of the protein. B. cannot be repaired. C. is a very common occurrence, happening in about one out of every 100 nucleotides. D. is only heritable if it occurs in a gamete producing cell.

D. is only heritable if it occurs in a gamete producing cell.

What is daughter if mom carries sex-linked traits? A. X B X B B. X b X b C. X B X b D. it could be X B X B or X B X b E. we cannot say for sure

D. it could be X B X B or X B X b Since she is not affected, she must have at least one X B , which she inherited from her father. But her mother is heterozygous, so we cannot say for certain whether she is a carrier or not.

In cats, black fur color is caused by an X-linked allele; the other allele at this locus causes orange color. The heterozygote is tortoiseshell. What kinds of offspring would you expect from the cross of a black female and an orange male? A. tortoiseshell females; tortoiseshell males B. orange females; black males C. black females; orange males D. tortoiseshell females; black males

D. tortoiseshell females; black males

How do the events of meiosis I promote the production of new combinations of alleles? A. via both independent assortment and crossing over between sister chromatids B. via independent assortment alone C. via crossing over between homologous chromosomes only D. via both crossing over between homologous chromosomes and independent assortment E. via crossing over between sister chromatids only

D. via both crossing over between homologous chromosomes and independent assortment

The recessive alleles for yellow body (y) and cut wings (c) identify two autosomal genes on the second chromosome of Drosophila melanogaster. When females, heterozygous at these two genes (YyCc), were crossed with yellow-bodied, cut-winged males, the following classes and numbers of progeny (out of 1000) were obtained: wild type body color, wild type wings 120wild type body color, cut wings360yellow body color, cut wings140yellow body color, wild type wings380 Which of the following crosses could have given rise to the females used in the cross described above? Assume that all individuals are true breeding. A. wild type body color, wild type wings X yellow body color, cut wings B. wild type body color, wild type wings X wild type body color, wild type wings C. yellow body color, cut wings X yellow body color, cut wings D. wild type body color, cut wings X yellow body color, wild type wings

D. wild type body color, cut wings X yellow body color, wild type wings

Here is a cartoon of homologous chromosomes. Sister chromatids are represented by _____ and nonsister chromatids are represented by ________. A. 2 and 3 B. 1 and 2; 3 and 4 C. 1 and 3; 2 and 4 D. 3 and 4 E. 1 and 2, 3 and 4; 1 and 3, 1 and 4, 2 and 3, 2 and 4

E. 1 and 2, 3 and 4; 1 and 3, 1 and 4, 2 and 3, 2 and 4

What possible blood types could offspring have if their two parents have the blood types A and AB, respectively? A. AB only B .A, B, AB, and O C. A and O only D. A and B only E. A, B, and AB only

E. A, B, and AB only

Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolize the amino acid phenylalanine. This amino acid is not naturally produced by humans. Therefore, the most efficient and effective treatment is which of the following? A. Transfuse the patients with blood from unaffected donors. B. Feed them the substrate that can be metabolized into this amino acid. C. Feed the patients the missing enzymes in a regular cycle, such as twice per week. D. Feed the patients an excess of the missing product. E. Regulate the diet of the affected persons to severely limit the uptake of the amino acid.

E. Regulate the diet of the affected persons to severely limit the uptake of the amino acid.

Height in humans generally shows a normal (bell-shaped) distribution. What type of inheritance most likely determines height? A. a combination of epistasis and environmental factors B. a combination of multiple alleles and codominance C. a combination of complete dominance and environmental factors D. incomplete dominance E. a combination of polygenic inheritance and environmental factors

E. a combination of polygenic inheritance and environmental factors Several genes (polygenic inheritance) control height in humans, giving an overall normal distribution. Environmental factors such as nutrition smooth out the curve.

Use the following pedigree (the figure) for a family in which dark-shaded symbols represent individuals with one of the two major types of colon cancer. Numbers under the symbols are the individual's age at the time of diagnosis. From this pedigree, how does this trait seem to be inherited? A. as an incomplete dominant B. from mothers C. as a result of epistasis D. as an autosomal recessive E. as an autosomal dominant

E. as an autosomal dominant

Refer to the following information to answer the question below. A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive. What proportion of their sons would be color-blind and of normal height? A. one out of four B. all C. none D. three out of four E. half

E. half

Hemophilia mother carrier? A.X B X B B. X b X b C. X B X b D. X B Y E. it could be X B X B or X B X b

E. it could be X B X B or X B X b We know that my mother passed on a dominant allele to me, so she must have an X B . But whether she has one or two copies of the X B , we cannot say for sure from the information shown here.

Refer to the following information to answer the question below. A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive.How many of their daughters might be expected to be color-blind dwarfs? A. half B. one out of four C. three out of four D. all E. none

E. none

The inheritance of eye color in Drosophila is controlled by genes on each of the fly's four chromosome pairs. One eye-color gene is on the fly's X chromosome, so the trait is inherited in a sex-linked manner. For this sex-linked trait, the wild-type (brick red) allele is dominant over the mutant vermilion (bright red) allele. A homozygous wild-type female fly is mated with a vermilion male fly. XVXV×XvY Predict the eye colors of F1 and F2 generations. (Assume that the F1 flies are allowed to interbreed to produce the F2 generation.)

F1 Females: all wild typeF2 Females: all wild typeF1 Males: all wild typesF2 Females: 1/2 wild type, 1/2 vermilion Both F1 females and F1 males will have the wild-type eye color, but because the trait is sex-linked, F1 females will be heterozygous for the trait. As a result, F2 males have a 1/2 chance of inheriting each allele, and thus of having that eye color. F2 females, on the other hand, will all have the wild-type eye color because they inherit the dominant allele on the X chromosome from their fathers (the F1 males).

The pedigrees below show the inheritance of three separate, rare autosomal conditions in different families. For each pedigree, decide if the condition is better explained as recessive or dominant. Drag the correct label to the appropriate location. Labels can be used once, more than once, or not at all.

If an individual has a genetic condition that neither parent has, then that condition must be recessive. Dominant conditions require that every affected individual have at least one affected parent.In situations where the inheritance mode of a rare condition cannot be definitely determined, the most likely mode is the one that requires the fewest unrelated individuals to have the condition-causing allele.

Suppose that you perform the cross discussed in Part B: MmDdPp x mmddpp. You plant 1000 tomato seeds resulting from the cross, and get the following results: Drag the labels onto the chromosome diagram to identify the locations of and distances between the genes. Gene m has already been placed on the linkage map. See picture attached to previous question for data

The recombination frequencies between genes can be used to construct a linkage map, as you have just done. The closer two genes are to each other on the same chromosome, the less likely they are to be separated by a crossover event, resulting in a lower recombination frequency. To calculate the distance between m and p, you would add the map distances between m and d and between d and p (12 cM + 5 cM = 17 cM). You probably noticed that the recombination frequency between m and p is slightly less than that sum. This is because of double crossover events--the times that crossovers occur both between m and d and between d and p. The second crossover effectively "cancels out" the first, reducing the number of recombinants between m and p that are observed, while contributing to the number of recombinants between each of the other two pairs of genes. Therefore, adding the smaller map distances to calculate larger distances avoids inaccuracies due to double-crossover events that produce non-recombinant genotypes.

The pedigree from Part C is shown below. Individuals III-3 and III-4 are expecting their first child when they become aware that they both have a family history of this recessive condition. As their genetic counselor, you can calculate the probability that they are carriers and that their child will be affected with the condition. Complete each statement by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all.

You can calculate probabilities in pedigrees by considering the requirements for a specific outcome. In this case, the outcome that individual IV-1 will be affected has four requirements: Individual III-3 is a carrier (probability = 2/3). Individual III-4 is a carrier (probability = 1/2). Individual III-3 passes the r allele to his child (probability = 1/2, assuming III-3 is a carrier, which is accounted for in requirement 1). Individual III-4 passes the r allele to her child (probability = 1/2, assuming III-4 is a carrier, which is accounted for in requirement 2). Because requirements 1 AND 2 AND 3 AND 4 must be met for the outcome in question (individual IV-1 to be affected), you calculate the answer by applying the multiplication rule (as implied by "AND"): The probability that IV-1 will be affected (rr) = 2/3 x 1/2 x 1/2 x 1/2 = 1/12.

Pedigree 2 from Part A is shown below. Recall that this pedigree shows the inheritance of a rare, autosomal dominant condition. Fill in the genotypes for the indicated individuals in the pedigree by dragging the best label to the appropriate location. Labels can be used once, more than once, or not at all.

You can deduce the genotype of an individual in a pedigree based on two types of information: the individual's phenotype (affected or unaffected) and the phenotypes of his or her parents and/or children.For autosomal dominant conditions: Unaffected individuals are homozygous for the recessive, wild-type allele. Affected individuals with only one affected parent are heterozygous. Affected individuals with any unaffected children are heterozygous. Affected individuals with two affected parents may be homozygous dominant or heterozygous.

Pedigree 3 from Part A is shown below. Recall that this pedigree shows the inheritance of a rare, autosomal recessive condition.Note that individual II-3 has no family history of this rare condition. Fill in the genotypes for the indicated individuals in the pedigree by dragging the best label to the appropriate location. Labels can be used once, more than once, or not at all.

You can deduce the genotype of an individual in a pedigree based on two types of information: the individual's phenotype (affected or unaffected) and the phenotypes of his or her parents and/or children.For autosomal recessive conditions: Affected individuals are always homozygous recessive. Unaffected children of an affected parent are always carriers (heterozygous). Both parents of affected individuals must have at least one recessive allele. If both parents are carriers, their unaffected children may be carriers or homozygous for the dominant, wild-type allele. For rare conditions, you can assume that individuals marrying into a family do not carry the recessive allele if there is no evidence that they are carriers.

You are working for a fly geneticist who sends you to a produce market to find new mutations. You rummage through the dumpsters and capture 100 flies. You note that 100% of the flies have red eyes and 90% have long body bristles (10% have short body bristles). Based on these results, you can conclude that __________________. a. red eyes are dominant and short body bristles are recessive b. short body bristles and long body bristles are located on different chromosomes c. red eyes and short body bristles are located on different chromosomes d. in this produce market, red eyes and long bristles are wild-type traits e. red eyes and long body bristles are inherited in a dominant manner

d. in this produce market, red eyes and long bristles are wild-type traits

An individual that is heterozygous for an allele and that exhibits a phenotype intermediate between that of the two homozygous alternatives is an example of _____. a.) polygenic inheritance b.) pleiotropy c.) codominance d.) incomplete dominance e.) epistasis

d.) incomplete dominance

The following question refer to the pedigree chart in the figure below for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. What is the genotype of individual II-5?

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