Genetics Chapter 4

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Which of the following statements is always true when mutations occur in genes whose products are essential to an organism's survival? A. A homozygote for a recessive lethal allele will not survive. B. Reversion of the mutation must occur for a heterozygous individual to survive. C. One normal copy of the gene will allow survival. D. The heterozygote always has a normal phenotype.

A. A homozygote for a recessive lethal allele will not survive.

Comb shape in chickens represents one of the classic examples of gene interaction. Two gene pairs interact to influence the shape of the comb. The genes for rose comb (R) and pea comb (P) together produce walnut comb. The fully homozygous recessive condition (rrpp) produces the single comb. Assume that a rose-comb chicken is crossed with a walnut-comb chicken and the following offspring are produced: 17 walnut, 16 rose, 7 pea, 6 single. What are the probable genotypes of the parents? A. Rrpp × RrPp B. RrPp x RrPp C. rrPp x RrPp D. RRpp x Rrpp E. rrpp x RrPp

A. Rrpp × RrPp

In Dexter and Kerry cattle, animals may be polled (hornless) or horned. The Dexter animals have short legs, whereas the Kerry animals have long legs. When many offspring were obtained from matings between polled Kerrys and horned Dexters, half were found to be polled Dexters and half polled Kerrys. When these two types of F1F1 cattle were mated to one another, the following F2F2 data were obtained: 3/8 polled Dexters 3/8 polled Kerrys 1/8 horned Dexters 1/8 horned Kerrys A geneticist was puzzled by these data and interviewed farmers who had bred these cattle for decades. She learned that Kerrys were true breeding. Dexters, on the other hand, were not true breeding and never produced as many offspring as Kerrys. Provide a genetic explanation for these observations. Select the four correct statements. A. The Dexter breed is heterozygous. B. The Kerry breed is heterozygous. C. The Dexter breed is homozygous dominant. D. Polled is caused by an independently assorting dominant allele, whereas horned is caused by the recessive allele to polled. E. The homozygous recessive type is lethal. F. Horned is caused by an independently assorting dominant allele, whereas polled is caused by the recessive allele to horned. G. The Kerry breed is homozygous recessive. H. The homozygous dominant type is lethal.

A. The Dexter breed is heterozygous. D. Polled is caused by an independently assorting dominant allele, whereas horned is caused by the recessive allele to polled. G. The Kerry breed is homozygous recessive. H. The homozygous dominant type is lethal.

Mutations in which the level of expression is based on environmental conditions is an example of ________. A. a conditional mutation B. expressivity C. genetic imprinting D. penetrance E. pleiotropy

A. a conditional mutation

In the mouse, gene B can produce black pigment from a colorless precursor molecule. A mouse having at least one B allele can produce black pigment, whereas the homozygous recessive mouse (bb) cannot and is albino. The agouti locus (A) can convert the black pigment to brown in the presence of at least one dominant A allele, whereas the homozygous recessive (aa) cannot convert the black pigment to brown. What color fur would a mouse have with the genotype aaBB? A. black B. albino C. brown D. patches of brown and white E. can't be determined

A. black

Fruit flies homozygous for the eyeless mutation demonstrate various gradations in phenotype. This represents an example of ________. A. expressivity B. genetic imprinting C. pleiotropy D. a conditional mutation E. penetrance

A. expressivity

Because of the mechanism of sex determination, males of many species can be neither homozygous nor heterozygous. Such males are said to be ________. A. hemizygous B. dominant C. complementary D. recessive E. None of the answers listed is correct.

A. hemizygous

The following coat colors are determined by alleles at one locus in horses: palomino = golden coat with lighter mane and tail cremello = almost white chestnut = brown The following table gives ratios obtained in matings of the above varieties: cremello × cremello = all cremello chestnut × chestnut = all chestnut cremello × chestnut = all palomino palomino × palomino 1/4 = chestnut 1/2 = palomino 1/4 = cremello Based on this data set, which of the following is the correct statement? A. palomino is the incomplete dominant form of chestnut and cremello alleles B. cremello is dominant to chestnut; chestnut is dominant to palomino C. cremello is dominant to palomino and chestnut; palomino and chestnut are codominant D. palomino is dominant to chestnut; chestnut is dominant to cremello E. cremello and chestnut are codominant; palomino is recessive to both

A. palomino is the incomplete dominant form of chestnut and cremello alleles

Complementation analysis is used to determine __________. A. whether two mutations that produce the same phenotype reside in the same or different genes B. whether two alleles can cause incomplete dominance C. whether a gene is X-linked or autosomal D. whether two alleles are codominant

A. whether two mutations that produce the same phenotype reside in the same or different genes

Which of the following symbols indicate a wild-type allele? A. d B. + C. R^1 D. D

B. +

The white-eye gene in Drosophila is recessive and sex-linked. Assume that a white-eyed female is mated to a wild-type male. What would be the phenotypes of the offspring? A. 1/4 wild-type females, 1/4 white-eyed females, 1/4 wild-type males, 1/4 white-eyed males B. 1/2 wild-type females, 1/2 white-eyed males C. 1/2 wild-type females, 1/4 wild-type males, 1/4 white-eyed males D. 1/2 wild-type males, 1/4 wild-type females, 1/4 white-eyed females E. all wild type males and females

B. 1/2 wild-type females, 1/2 white-eyed males

Assume that a dihybrid cross for two different traits (AaBb × AaBb) is made in which the gene loci are autosomal, independently assorting, and both incompletely dominant. What phenotypic ratio would you expect from such a cross? A. 1:2:1:1:2:1:1:2:1 B. 1:2:1:2:4:2:1:2:1 C. 1:2:1 D. 1:1:2:2:1:1 E. 9:3:3:1

B. 1:2:1:2:4:2:1:2:1

Which of the following statements about autosomal recessive traits is true? A. If an individual expresses the trait, then all of his or her offspring will also express the trait. B. If neither parent expresses the trait, but the offspring does, both parents must be heterozygous for the trait. C. If an individual expresses the trait, none of his or her offspring will express the trait. D. Only females can pass the trait to their offspring.

B. If neither parent expresses the trait, but the offspring does, both parents must be heterozygous for the trait.

What is the difference between incomplete dominance and codominance? A. The phenotype reveals the genotype only in codominance. B. In codominance, both phenotypes are expressed in heterozygotes. In incomplete dominance, the heterozygote shows a phenotype that is intermediate between the two homozygotes. C. In incomplete dominance, the dominant phenotype is expressed in heterozygotes. In codominance, both phenotypes are expressed in heterozygotes. D. In both types of inheritance, the heterozygotes express the phenotypes dictated by both alleles.

B. In codominance, both phenotypes are expressed in heterozygotes. In incomplete dominance, the heterozygote shows a phenotype that is intermediate between the two homozygotes.

Below is a pedigree of a fairly common human hereditary trait in which the boxes represent males and the circles represent females. Shading symbolizes the abnormal phenotype. Given that one gene pair is involved, what is/are the possible mode(s) of inheritance? A. autosomal recessive or X-linked recessive B. autosomal recessive C. autosomal recessive, X-linked recessive, or X-linked dominant D. autosomal dominant E. X-linked recessive

B. autosomal recessive

Individuals from two separate true-breeding strains of white deer mice are crossed yielding all grey offspring. White is recessive to gray color based on crossing mice from each strain with a grey mouse. Which of the following would best explain this result? A. multiple alleles B. complementation C. epistasis D. incomplete dominance E. lethal alleles

B. complementation

A human disorder resulting from a mitochondrial DNA mutation such as MERFF might show various levels of expression among children of a mother with MERFF. This is most likely the result of ________. A. genetic anticipation v C. a conditional mutation D. expressivity E. maternal effect

B. heteroplasmy

A gene in which the heterozygous condition is displayed differently in males and females is referred to as __________. A. genetic anticipation B. sex-influenced C. X-linked D. sex-limited

B. sex-influenced

A trait exhibited in one sex but not the other is referred to as ________. A. genomic imprinting B. sex-limited C. sex-linked D. sex-influenced

B. sex-limited

Pattern baldness is a sex-influenced trait, with heterozygous males exhibiting the trait. What would be the probability of the daughters exhibiting the trait from a woman with pattern baldness and a male without pattern baldness? A. 100% B. 25% C. 0% D. 50% E. not enough information to determine

C. 0%

A deficiency of the enzyme glucose-6-phosphate dehydrogenase (G6PD) is inherited as an X-linked recessive trait in humans. A phenotypically normal woman (whose father had G6PD) is married to a man with normal G6PD function. What fraction of their sons would be expected to have G6PD deficiency? A. 1/4 B. all C. 1/2 D. 0 E. not enough information available to determine

C. 1/2

What is the ratio expected for offspring of a cross between two yellow mice? A. 1 yellow: 1 agouti B. 3 yellow: 1 agouti C. 2 yellow: 1 agouti D. all yellow E. cannot be determined due to unknown genotypes

C. 2 yellow: 1 agouti

The following F2 results occur from a typical dihybrid cross: purple: A_B_ 9/16 white: aaB_ 3/16 white: A_bb 3/16 white: aabb 1/1 If a double heterozygote (AaBb) is crossed with a fully recessive organism (aabb), what phenotypic ratio is expected in the offspring? A. 1 (purple): 1 (white) B. 9 (white): 7 (purple) C. 3 (white): 1 (purple) D. 9 (purple): 7 (white) E. 3 (purple): 1 (white)

C. 3 (white): 1 (purple)

Some genes and mutations vary in their phenotypic expression. What term describes a disorder that is NOT expressed by everyone who carries a mutation? A. pleiotropy B. expressivity C. sex-linked inheritance D. incomplete penetrance

D. incomplete penetrance

Croaking in frogs is inherited by one gene with a pair of alleles −− dominant uttering (RR) and recessive muttering (rr). Eye color is inherited by epistatic interaction of two genes with the following possible phenotypes: A_B_ −− blue eyes; A_bb− purple eyes and aaB_or aabb−green eyes. The frog geneticist mated purple utterer and green mutterer with the following results: 1/4 purple, utterers1/4 purple, mutterers1/4 green, utterers1/4 green, mutterers What were the genotypes of the parents? A. AabbRr×AaBbRr B. AabbRr×AAbbRr C. AabbRr×aabbrr D. AAbbRr×AaBbRr E. AabbRr×aabbRr F. AaBbrr×AabbRr

C. AabbRr×aabbrr

Which term describes the individual in a pedigree whose phenotype was first brought to the attention of a medical researcher? A. Sibship B. Progeny C. Proband D. Allele

C. Proband

What is the most likely mode of transmission for a trait that is not expressed in parents but is expressed by one half of the sons? A. Autosomal recessive B. Autosomal dominant C. Rare X-linked recessive D. Rare X-linked dominant

C. Rare X-linked recessive

Why do multiple and lethal alleles often result in modifications of the classic Mendelian monohybrid and dihybrid ratios? Select the two correct statements. A. When the expression of a gene masks the effect of a second gene, it can result in a lethal phenotype. In this case, classic Mendelian monohybrid and dihybrid ratios will not be observed. B. Multiple alleles mean that there are more than one gene at a given locus. A diploid organism has one gene locus that may be occupied by different alleles of the same gene. This can result in many different phenotypes for traits, which may not follow typical Mendelian ratios. C. When an essential gene is mutated, it can result in a lethal phenotype. In this case, classic Mendelian monohybrid and dihybrid ratios will not be observed. D. When an essential gene is mutated, it can result in a lack of dominance. The resulting phenotype is called lethal. In the case of presence of such allele, classic Mendelian monohybrid and dihybrid ratios will not be observed. E. Multiple alleles mean that there are more than two alternatives of a gene at a given locus. A diploid organism has two homologous gene loci that may be occupied by different alleles of the same gene. This can result in many different phenotypes for traits, which may not follow typical Mendelian ratios. F. Multiple alleles mean that the joint expression of several alleles in a heterozygote shows the third (not blended) phenotype. This can result in many different phenotypes for traits, which may not follow typical Mendelian ratios.

C. When an essential gene is mutated, it can result in a lethal phenotype. In this case, classic Mendelian monohybrid and dihybrid ratios will not be observed. E. Multiple alleles mean that there are more than two alternatives of a gene at a given locus. A diploid organism has two homologous gene loci that may be occupied by different alleles of the same gene. This can result in many different phenotypes for traits, which may not follow typical Mendelian ratios.

A condition in which one gene pair masks the expression of a nonallelic gene pair is called ________. A. dominance B. codominance C. epistasis D. additive alleles E. recessiveness

C. epistasis

A mutation in a gene that results in a loss of a functional product of that gene best defines what type of mutation? A. codominance B. gain of function C. null D. incomplete dominance E. multiple allelism

C. null

The phenomenon of a gene which has multiple phenotypic effects on an individual is referred to as ________. A. penetrance B. continuous variation C. epistasis D. pleiotropy E. expressivity

D. pleiotropy

In the mouse, gene B can produce black pigment from a colorless precursor molecule. A mouse having at least one B allele can produce black pigment, whereas the homozygous recessive mouse (bb) cannot and is albino. The agouti locus (A) can convert the black pigment to brown in the presence of at least one dominant A allele, whereas the homozygous recessive (aa) cannot convert the black pigment to brown. What would be the probability of an albino mouse offspring if the parents were genotypes AaBb and AAbb? A. 1/16 B. 0 C. 3/16 D. 1/2 E. 1/4

D. 1/2

A couple each with blood type AB and normal pigmentation have a child with AB blood type and albinism. What is the probability that their next child will have the same phenotype as the first child? A. 1/16 B. 9/16 C. 3/16 D. 1/8 E. 1/4

D. 1/8

Pattern baldness is a sex-influenced trait, with heterozygous males exhibiting the trait. What would be the probability of the sons exhibiting the trait from a woman with pattern baldness and a male without pattern baldness? A. 0% B. 50% C. 25% D. 100% E. not enough information to determine

D. 100%

Mice have a set of multiple alleles of a gene for coat color. Four of those alleles are as follows: C = full color (wild) c^ch = chinchilla c^d = dilution c = albino Given that the gene locus is not sex-linked and that each allele is dominant to those lower in the list, what would be the phenotypic ratio of a cross between a mouse of full color (heterozygous for dilution) and a mouse with the chinchilla coat color (heterozygous for albino)? A. 2 full color: 2 chinchilla B. all full color C. 1 full color: 1 chinchilla: 1 dilution: 1 albino D. 2 full color:1 chinchilla:1 dilution E. 3 full color: 1 chinchilla

D. 2 full color:1 chinchilla:1 dilution

In foxes, two alleles of a single gene, P and p, may result in lethality (PP), platinum coat (Pp), or silver coat (pp). What ratio is obtained when platinum foxes are interbred? A. 1/2 platinum, 1/2 silver B. 1/3 platinum, 2/3 silver C. 3/4 platinum, 1/4 silver D. 2/3 platinum, 1/3 silver E. 1/4 platinum, 3/4 silver Is the P allele behaving dominantly or recessively in causing lethality? A. recessively B. dominantly Is the P allele behaving dominantly or recessively in causing platinum coat color? A. recessively B. dominantly

D. 2/3 platinum, 1/3 silver A. recessively B. dominantly

A couple in which one parent is blood type A and the other is blood type B, and both have normal pigmentation, have a child that is blood type O and has albinism. What is the probability that their next child will have normal pigmentation and blood type A? A. 1/8 B. 9/16 C. 1/16 D. 3/16 E. 0

D. 3/16

________ occurs when the offspring's phenotype is under the control of the mother's nuclear gene products that are present in the egg. A. Maternal inheritance B. Maternal imprinting C. Genetic anticipation D. Maternal effect E. Sex-linked inheritance

D. Maternal effect

A gain-of-function mutation __________. A. is the wild-type allelei B. s usually recessive C. results in the absence of the gene product D. results in an increased quantity of the normal gene product

D. results in an increased quantity of the normal gene product

Many of the color varieties of summer squash are determined by several interacting loci: AA or Aa gives white, aaBB or aaBb gives yellow, and aabb produces green. Assume that two fully heterozygous plants are crossed. Give the phenotype ratio of the offspring. A. 9 (white): 7 (yellow) B. 12 (yellow):3 (green): 1 (white) C. 9 (white): 3 (yellow): 4 (green) D. 1 (green): 1 (yellow): 1 (white) E. 12 (white):3 (yellow):1 (green)

E. 12 (white):3 (yellow):1 (green)

90% of children that inherit a mutated retinoblastoma gene develop the disease. This represents an example of ________. A. expressivity B. genetic imprinting C. a conditional mutation D. pleiotropy E. penetrance

E. penetrance

The MN blood group is a codominant trait as a result of which of the following? A. presence of M cell surface glycoproteins on some blood cells and N glycoproteins on others B. presence of M, M/N intermediate, and N cell surface glycoproteins on blood cells C. presence of a M/N intermediate cell surface glycoprotein on blood cells D. presence of either M, M/N intermediate, or N cell surface glycoproteins on different blood cells E. presence of both M and N cell surface glycoproteins on the same blood cell

E. presence of both M and N cell surface glycoproteins on the same blood cell

A cross was made between homozygous wild-type female Drosophila and yellow-bodied male Drosophila. All of the resulting offspring were phenotypically wild type. Offspring of the F2 generation had the following phenotypes: male/wild/96 male/yellow/99 female/wild/197 Which is correct for the mode of inheritance of this gene? A. autosomal dominant B. autosomal recessive C. Y-linked D. sex-linked dominant E. sex-linked recessive

E. sex-linked recessive

In laborador retrievers, pigment color is influenced by two genes. Gene A determines the type of pigment produced and gene B affects whether the pigment gets deposited in the hair shaft.If two heterozygous black laborator retrievers were crossed and offspring were produced in a ratio of 9 black dogs to 3 brown dogs to 4 yellow dogs, what are the genotypes of the offspring? A__B__ A__bb aaB__ aabb Genotype(s) of Black Dogs Genotype(s) of Brown Dogs Genotype(s) of Yellow Dogs Is this an example of dominant or recessive epistasis?

Genotype(s) of Black Dogs: A__B__ Genotype(s) of Brown Dogs: aaB__ Genotype(s) of Yellow Dogs: A__bb aabb Recessive epistasis

In freshwater snails, pigment color is influenced by two genes. If two heterozygous pigmented freshwater snails were crossed and offspring were produced in a ratio of 9 pigmented snails to 7 albino snails, what are the genotypes of the offspring? A__B__ aaB__ A__bb aabb Genotype(s) of Pigmented Snails Genotype(s) of Albino Snails Is this an example of dominant or recessive epistasis?

Genotype(s) of Pigmented Snails: A__B__ Genotype(s) of Albino Snails: aaB__ A__bb aabb Recessive epistasis

In sheep, coat color is influenced by two genes. Gene A influences pigment production, while gene B produces black or brown pigment.If two heterozygous white sheep were crossed and offspring were produced in a ratio of 12 white sheep to 3 black sheep to 1 brown sheep, what are the genotypes of the offspring? A__B__ A__bb aaB__ aabb Genotype(s) of White Sheep Genotype(s) of Black Sheep Genotype(s) of Brown Sheep Is this an example of dominant or recessive epistasis?

Genotype(s) of White Sheep: A__B__ A__bb Genotype(s) of Black Sheep: aaB__ Genotype(s) of Brown Sheep: aabb Dominant epistasis

In cats, orange coat color is determined by the bb allele, and black coat color is determined by the BB allele. The heterozygous condition results in a coat pattern known as tortoiseshell. This gene is XX-linked. Part A: What kinds of offspring would be expected from a cross of an orange male and a tortoiseshell female? Select the four correct kinds of offspring. A. orange female B. black female C. tortoiseshell female D. orange male E. tortoiseshell male F. black male Part B: What are the chances of getting a tortoiseshell male? A. 1/4 B. 1/8 C. 3/4 D. 1/2 E. 0

Part A: A. orange female C. tortoiseshell female D. orange male F. black male Part B: E. 0

Three autosomal recessive mutations in Drosophila, all with tan eye color (r1r1, r2r2, and r3r3), are independently isolated and subjected to complementation analysis. Part A: Of the results shown below, which, if any, are alleles of one another?Cross 1: r1×r2→F1: all wild-type eyes Cross 2: r1×r3→F1: all tan eyes A. none B. r1 and r3 C. r1 and r2 Part B: Predict the results of the cross that is not shown-that is, r2×r3. A. all tan eyes B. all wild-type eyes C. both wild-type and tan eyes

Part A: B. r1 and r3 Part B: B. all wild-type eyes

Labrador retrievers may be black, brown, or golden in color. Although each color may breed true, many different outcomes occur if numerous litters are examined from a variety of matings, where the parents are not necessarily true-breeding. The following results show some of the possibilities. black×brown→all black black×brown→1/2 black 1/2 brown black×brown→3/4 black 1/4 golden black×golden→all black black×golden→4/8 golden 3/8 black 1/8 brown black×golden→2/4 golden 1/4 black 1/4 brown brown×brown→3/4 brown 1/4 golden black×black→9/16 black 4/16 golden 3/16 brown Part A: What is the mode of inheritance that is consistent with these data? A. multiple alleles B. codominance C. epistasis D. incomplete dominance Part B: Indicate the genotype(s) of the dogs that breed true for black color Part C: Indicate the genotype(s) of the dogs that breed true for brown color Part D: Indicate the genotype(s) of the dogs that breed for golden color

Part A: C. epistasis Part B: AABB Part C: aaBB Part D: AAbb Aabb aabb

The following genotypes of two independently assorting autosomal genes determine coat color in rats: A_B_A_B_ (gray) A_bbA_bb (yellow) aaB_aaB_ (black) aabbaabb (cream) A third gene pair on a separate autosome determines whether any color will be produced. The CCCC and CcCc genotypes allow color according to the expression of the AA and BB alleles. However, the cccc genotype results in albino rats regardless of the AA and BB alleles present. Part A: What are the genotype(s) and phenotype(s) of the parents who produced the following F1 offspring? F1: 9/16 gray: 3/16 yellow: 3/16 black: 1/16 cream Select all that apply. AaBbCC(gray) AaBbCc(gray) AABbCc(gray) aaBbCc(black) aabbCc(cream) aaBbcc (albino) aaBbCC(black) aabbcc (albino) AabbCc (yellow) AAbbCc(yellow) AAbbcc(albino) Part B: What are the genotype(s) and phenotype(s) of the parents who produced the following F1F1 offspring? F1:F1: 9/16 gray: 3/16 yellow: 4/16 albino Select all that apply. AaBbCC(gray) AaBbCc (gray) AABbCc(gray) aaBbCc(black) aabbCc (cream) aaBbcc(albino) aaBbCC(black) aabbcc(albino) Aabbcc(yellow) AAbbCc(yellow) AAbbcc (albino) Part C: What are the genotype(s) and phenotype(s) of the parents who produced the following F1F1 offspring? F1:F1: 27/64 gray: 16/64 albino: 9/64 yellow: 9/64 black: 3/64 cream Select all that apply. AaBbCC(gray) AaBbCc(gray) AABbCc (gray) aaBbCc(black) aabbCc (cream) aaBbcc(albino) aaBbCC(black) aabbcc(albino) AabbCc(yellow) AAbbCc (yellow) AAbbcc(albino)

Part A: AaBbCC(gray) AaBbCc(gray) Part B: AaBbCc (gray) AABbCc(gray) Part C: AaBbCc(gray)

In cattle, coats may be solid white, solid black, or black-and-white spotted. When true-breeding solid whites are mated with true-breeding solid blacks, the F1F1 generation consists of all solid white individuals. After many F1×F1F1×F1 matings, the following ratio was observed in the F2F2 generation: 12/16 solid white 3/16 black-and-white spotted 1/16 solid black Part A: How many gene pairs are involved in the inheritance of cattle coat color? A. 1 B. 2 C. 3 D. 4 Part B: What type of inheritance mode governs coat color? A. dominant epistasis B. complementary gene interaction C. recessive epistasis D. codominance Part C: Is it possible to isolate a true-breeding strain of black-and-white spotted cattle? What genotype would they have? A. Yes. The genotype would be AABB. B. Yes. The genotype would be aaBB. c. Yes. The genotype would be aabb D. Yes. The genotype would be aaBb. E. Yes. The genotype would be AAbb. F. No. It is not possible to isolate a true-breeding strain.

Part A: B. 2 Part B: A. dominant epistasis Part C: B. Yes. The genotype would be aaBB.

In pigs, coat color may be sandy, red, or white. A geneticist spent several years mating true-breeding pigs of all different color combinations, even obtaining true-breeding lines from different parts of the country. For crosses 1 and 4 in the following table, she encountered a major problem: her computer crashed and she lost the F2F2 data. She nevertheless persevered and, using the limited data shown here, was able to predict the mode of inheritance and the number of genes involved, as well as to assign genotypes to each coat color. Attempt to duplicate her analysis, based on the available data generated from the crosses shown. sandy x sandy F1: All red F2: Data lost red x sandy F1: All red F2: 3/4 red : 1/4 sandy sandy x white F1: All sandy F2: 3/4 sandy: 1/4 white white x red F1: All red F2: Data lost Part A: What is the mode of inheritance and how many genes are involved? A. 1 gene, incomplete dominance B. 1 gene, codominance C. 1 gene with 3 alleles D. 2 genes, no epistasis E. 2 genes, epistasis Part B: What are the genotypes of the coat colors? AABB AABb AaBB AaBb AAbb Aabb aaBb aaBB aabb Red Sandy White Part C: What is the outcome of the F2 in cross 1? Answers are listed in the following order: red:sandy:white. A. 4 : 4 : 1 B. 1 : 2 : 1 C. 1 : 6 : 1 D. 9 : 6 : 1 E. 4 : 1 : 4 F. 1 : 4 : 4 Part D: What is the outcome of the F2F2 in cross 4? Answers are listed in the following order: red:sandy:white. A. 4 : 4 : 1 B. 1 : 2 : 1 C. 1 : 4 : 4 D. 1 : 6 : 1 E. 4 : 1 : 4 F. 9 : 6 : 1

Part A: E. 2 genes, epistasis Part B: Red: AABB, AABb, AaBB, AaBb Sandy: AAbb, Aabb, aaBb, aaBB White: aabb Part C: D. 9 : 6 : 1 Part D: F. 9 : 6 : 1

Pigment in mouse is produced only when the CC allele is present. Individuals of the cccc genotype have no color. If color is present, it may be determined by the AA and aa alleles. AAAA or AaAa results in agouti color, whereas aaaa results in black coats. Part A: What F1F1 and F2F2 genotypic and phenotypic ratios are obtained from a cross between AACC and aacc mice? black white agouti F1: A__C__ - F2: A__C__ = 9/16 - A__cc = 3/16 - aaC__ = 3/16 - aacc = 1/16 - Part B: In the cross between agouti female whose genotype was unknown and male of the aaccaacc genotype, what is the genotype of female parent for the following phenotypic ratio? 8 agouti and 8 colorless A. AaCc B. AACc C. AAcc D. AaCC Part C: In the cross between agouti female whose genotype was unknown and male of the aaccaacc genotype, what is the genotype of female parent for the following phenotypic ratio? 9 agouti and 10 black A. AACc B. AaCC C. AAcc D. AaCc Part D: In the cross between agouti female whose genotype was unknown and male of the aaccaacc genotype, what is the genotype of female parent for the following phenotypic ratio? 4 agouti, 5 black, and 10 colorless A. AACc B. AaCc C. AaCC D. AAcc

Part A: F1: A__C__ - agouti F2: A__C__ = 9/16 - agouti A__cc = 3/16 - white aaC__ = 3/16 - black aacc = 1/16 - white Part B: B. AACc Part C: B. AaCC Part D: B. AaCc


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