EXAM 2 GENETICS CH 4 MG Q'S

In complementary gene interactions, two genes work in tandem to produce a particular phenotype. Functional copies of both genes are required to produce the phenotype. What phenotypic ratios would be expected among the F2 progeny of a dihybrid cross involving dominant and recessive alleles for two such complementary genes?

9:7, Correct. Nine sixteenths of the progeny would possess a dominant allele of both genes (A-B-) and therefore express the phenotype associated with the gene interaction. Seven sixteenths of the progeny would be homozygous recessive for one or both of the genes.

Which of the following is the most likely example of codominance?

A pure-breeding plant with red flowers is crossed to a pure-breeding plant with white flowers. All their progeny have flowers with some red patches and some white patches.

What is meant by variable expressivity in genetics?

A single genotype produces a range of phenotypes that differ in the degree or magnitude of expression of the particular trait specified by the gene.

A recessive allele that results in a complete loss of gene function is known as which of the following types of mutation?

Amorphic

In 1905, Lucien Cuenot observed that crosses between two yellow mice consistently produced progeny ratios of two yellow : one agouti (wild-type). What is the explanation for the observed 2:1 phenotype ratio?

The yellow allele is a recessive lethal. Correct. This is a modified 1:2:1 ratio. Individuals that are homozygous for the yellow allele (AY/ AY) die during embryogenesis and therefore are not represented among the progeny. Yellow mice are heterozygous for the yellow and agouti alleles. Agouti mice are homozygous for the agouti allele. See Figure 4.8.

Is this an example of dominant or recessive epistasis?

This is an example of recessive epistasis, where two recessive alleles mask expression of an allele at a different locus. In this case, the presence of recessive alleles at the B locus (bb) prevented pigment deposition into the hair shaft, causing dogs to be yellow regardless of the genes they carried for pigment color.

In a cross between two strains that are true breeding for purple and white flowers, the F2 phenotypic ratio would be ¼ purple, ¼ white, and ½ lavender if the flower color phenotype exhibits incomplete dominance.

True The heterozygous offspring exhibit a phenotype that is intermediate between the parents.

Choose the offspring phenotypes you would expect from the following cross: R 1 R 1 ×× R 2 R 2

all pink

What types of mutants commonly result from regulatory mutations that increase the rate of transcription of a gene?

Hypermorphic

hoose the offspring phenotypes you would expect from the following cross: R 1 R 2 ×× R 2 R 2

1/2 pink, 1/2 white The genotypic ratio for this cross is 1/2 R 1 R2 and 1/2 R 2 R 2, making the phenotypic ratio 1/2 pink and 1/2 white.

Incomplete dominance is shown in the flower color of snapdragons, as shown in this figure. Choose the offspring phenotypes you would expect from the following cross: R 1 R 1 ×× R 1 R 2 R1 R1= red R1 R2= pink R2 R2= white

1/2 red, 1/2 pink The genotypic ratio for this cross is 1/2 R 1 R 1 and 1/2 R 1 R 2, making the phenotypic ratio 1/2 red and 1/2 pink.

Choose the offspring phenotypes you would expect from the following cross: R 1 R 2 ×× R 1 R 2

1/4 red, 1/2 pink, 1/4 white The genotypic ratio for this cross is 1/4 R 1 R 1, 1/2 R 1 R 2, and 1/4 R 2 R 2, making the phenotypic ratio 1/4 red, 1/2 pink, and 1/4 white.

If two alleles of a gene (T1 & T2) are incompletely dominant, then selfing individuals that are heterozygous for these two alleles will produce what phenotypic ratios among the progeny?

1:2:1

A woman with type A blood (whose father was type O) has children with a man that has type O blood. Both individuals are heterozygous for the MN antigen. Recall that MN blood group antigens are independent of the ABO locus, and that the alleles are codominant. Determine the proportion of various phenotypes of offspring that this couple may have.

Genotype of mother = AO father = OO AO OO= AO (1/2) OO(1/2) both hetero for MN- MMxMN- (1/4) MN (1/2) NN (1/4) type A with N= 1/2 x 1/4= 1/8 type A with M and N antigens= 1/2 x 1/2= 1/4 etc

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?

Genotype(s) of Black Dogs A_B_ Genotype(s) of Brown Dogs aaB_ Genotype(s) of Yellow Dogs A_bb aabb The 9:3:4 ratio obtained is a modification of the 9:3:3:1 ratio observed in a dihybrid cross of heterozygous individuals. In this case, dogs with black or brown coats are produced only when a dominant allele is present at the B loci and yellow dogs result from the presence of recessive alleles, bb. Of the pigmented dogs, black dogs are produced 75% of the time, while dogs with brown coats are produced 25% of the time. This 3:1 ratio suggests that the black allele is dominant to brown. Therefore, black dogs can be assigned the A_B_ genotype, brown dogs the aaB_ genotype, and yellow dogs the A_bb and aabb genotypes.

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?

Genotype(s) of White Sheep A_B_ A_bb Genotype(s) of Black Sheep aaB_ Genotype(s) of Brown Sheep aaabb The 12:3:1 ratio obtained is a modification of the 9:3:3:1 ratio observed in a dihybrid cross of heterozygous individuals. In this case, white sheep are produced 75% of the time, while sheep with pigmented coats are produced 25% of the time. The 3:1 ratio observed suggests that a dominant allele is responsible for suppressing pigment color. As gene A is responsible for inhibiting pigment production, sheep with an A_ genotype will be white (A_B_ and A_bb).As for the remaining sheep, 75% are black and 25% are brown, suggesting that black is dominant over brown. Therefore, black sheep can be assigned the aaB_ genotype while brown sheep can be assigned the aabb genotype.

What is the name given to the phenomenon in which some individuals with a particular genotype fail to display the corresponding phenotype?

Incomplete penetrance

Is this an example of dominant or recessive epistasis?

dominant epistasis This is an example of dominant epistasis, where one copy of an allele, in this case A, can mask the expression of another gene. Any sheep possessing the A allele (A_B_ and A_bb) had a white coat regardless of the gene present for pigment at the B locus.

Merle is a marking pattern in canine coats that appears as a marbling of color in a solid coat. The merle gene displays incomplete dominance. CSCS individuals display a solid coat CMCS individuals are merle and show patches of fur with a lighter, mottled effect CMCM individuals are "double merle" and show patches of fur with much more pronounced lightening Another gene, FGF5, also influences coat phenotype and determines whether dogs display long or short coats, with short coats being dominant to long coats. Determine the proportion of offspring phenotypes that would result when two merle dogs mate, if one dog is true-breeding for the long-coat trait and the other dog is true-breeding for the short-coat trait.

double merle with short coat:1/4 double merle with long coat:0 merle with short coat:1/2 merle with long coat:0 solid with short coat:1/4 solid with long coat:0 The key to solving this problem is to consider the two genes separately.First, focus on the merle gene. By definition, individuals displaying the merle trait must be heterozygotes (CMCS ). Because the merle gene displays incomplete dominance, the outcome of the CMCS x CMCS cross would be 1/4 double merle, 1/2 merle and 1/4 solid.Now consider the gene for coat length: a cross between dogs that are true breeding for short coat length and long coat length (LL x ll) would yield 100% Ll offspring, all with short coats. As a result, this cross would not produce any offspring with long coats.

Determine the proportion of offspring phenotypes that would result when two merle dogs mate, if both dogs are heterozygous (Ll) for the gene that regulates coat length.

double merle with short coat:3/16 double merle with long coat:1/16 merle with short coat:3/8 merle with long coat:1/8 solid with short coat:3/16 solid with long coat:1/16 From Part A, you determined that the results of the CMCS x CMCS cross would be 1/4 double merle, 1/2 merle and 1/4 solid. Now consider the gene for coat length: a cross between dogs heterozygous for coat length (Ll x Ll) would yield offspring that are 1/4 LL, 1/2 Ll and 1/4 ll. Therefore 3/4 of offspring will have short coats (1/4 LL + 1/2 Ll) and 1/4 of offspring will have long coats (ll).You can then use the product law to determine the chance of these two independent events occurring simultaneously. You would expect the following results: 1/4 x 3/4 = 3/16 offspring would be double merle with short coats 1/4 x 1/4= 1/16 offspring would be double merle with long coats 1/2 x 3/4= 3/8 offspring would be merle with short coats 1/2 x 1/4= 1/8 offspring would be merle with long coats 1/4 x 3/4=3/16 offspring would be solid with short coats 1/4 x 1/4=1/16 offspring would be solid with long coats

Epistasis is the interaction between genes such that one gene influences or interferes with the expression of another gene, leading to a specific phenotype. Epistatic genes can be dominant or recessive. Use your knowledge of epistasis to determine the genotypes of offspring in the following crosses: 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?

genotypes of pigmented snails = A_B_ Genotype(s) of Albino Snails= A_bb aaB_ aabb The 9:7 ratio obtained is a modification of the 9:3:3:1 ratio observed in a dihybrid cross of heterozygous individuals. In this case, pigmented snails are produced only when dominant alleles at both loci are present (A_B_) and albino snails result from any of the other possible genotypes (aaB_, A_bb, aabb).

Consider a recessive mutation that results in a complete loss of gene function. If heterozygous individuals (mutant/wild-type) are phenotypically wild-type, then the wild-type allele would be described as being __________.

haplosufficient; Correct. Haplosufficiency describes the case in which one copy of a wild-type allele is sufficient to produce the wild-type phenotype.

Interestingly, double merles are subject to a variety of health problems, including hearing loss and vision deficiencies. Therefore, it is not recommended that breeders breed for the double merle phenotype.If you were in the business of breeding dogs, which of the following crosses would you avoid in an effort to not produce double merle puppies

merle x merle If breeders selectively mate dogs in an effort to reduce production of double merle puppies, they should avoid mating two merle dogs together. The CMCS x CMCS cross would produce double merle offspring (CMCM ) 25% of the time. Breeders should also avoid mating two double merle dogs together (CMCM x CMCM ) because this combination would produce double merle puppies 100% of the time.

Is this an example of dominant or recessive epistasis?

recessive epistasis In recessive epistasis, two recessive alleles mask expression of an allele at a different locus. The 9:7 ratio observed in this example is actually a special case of epistasis called duplicative recessive epistasis, meaning the presence of either aa or bb was sufficient to mask expression of the other gene. In this case, if snails had two copies of either allele a or allele b, pigment production was suppressed, resulting in albino individuals.

How many different phenotypes are possible in a one gene/three allele system that displays codominance to each other?

six, Correct. With three alleles, a1, a2, a3, each unique combination of two alleles results in a distinct phenotype. Possible combinations are:a1/a1a1/a2a1/a3a2/a2a2/a3a3/a3