Question

the Internet can contain vinuses. Unless you need to edit it's safer to stay in Protected view. IL LES 1) A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits.(Remember that extra digits is a dominant trait.) What fraction of this couple's children would be expected to have extra digits? 2) In cocker spaniels, black color (B) is dominant over red (b), and solid color (S) is dominant over spotted (s). If the genes are unlinked, and the offspring of BBss and boss individuals are mated with each other, what fraction of the next generation will be black and spotted? 3) A pea plant with red flowers is test crossed, and one-half of the resulting progeny have red flowers, while the other half has white flowers. What was the genotype of the test crossed parent? 4) In a species of wild lupine flower, the smooth seed allele (s) is completely dominant over wrinkled seed allele (s). If the alleles for height are incompletely dominant, such that Tristall, Tt is intermediate, and tt is short, what progeny will result from crossing a smooth-seeded, short (Sstt) plant with a wrinkled-seeded, tall (STT) plant? 5) Coat color in Labrador retrievers is controlled by two genes. The first gene (B) yields either black retrievers when the dominant allele is expressed or brown retrievers when the genotype is homozygous recessive. A second gene (E) acts as an epistatic inhibitor, such that when animals are homozygous recessive for E, they exhibit a yellow coat color. If two retrievers are crossed [BBEe x Bbee), please describe all the possible genotypes and phenotypes.

Answer 1

1. Ans is 1/2 fraction of this couple's children would be expected to have extra digits.

Reason: Extra digit is a dominant trait. Consider the dominant allele as F and recessive f. The genotype of daughter and the mother is ff. Genotype of daughter is ff which is only possible when both parents contain atleast one recessive allele f. Hence, genotype of man is Ff. On mating Ff * Ff half of the progeny isFF and other half will be ff.

2) All of the offsprings (4/4) will be black and spotted .

Reason: BBss * bbss

Gametes: Bs * bs

Offspring: Bbss  

Black & spotted

3) Testcrossed genotype is Heterozygous dominant.

Reason: consider R is dominant and r is recessive allele. Testcrossing is done to determine the unknown genotype of individual with dominant phenotype by crossing it with it's recessive counterpart.

Here, R_(Red, but unknown genotype) * rr

Offspring : half of them Red (Rr)

Other half is white(rr) : this is only possible when both parents have aatleast one r .

Hence the genotype of testcrossed is Rr (heterozygous dominant)  

4)Smooth seeded with intermediate height.

SStt* ssTT = all offsprings SsTt

As S is complete dominant seeds must be smooth and T is incompletely dominant they are of intermediate height.

5)following are the possible genotype and phenorypes

(a) BBEe Black

(b) BBee yellow color

(c) BbEe Black

(d) Bbee Yellow color