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In mice the allele for black coat color (B) is dominant to the allele for white...

In mice the allele for black coat color (B) is dominant to the allele for white coat color (b). The allele for long tail (T) is dominant to the allele for short tail (t).

For the same cross: BbTt x bbTt

a. Using the Probability Method illustrated in lecture, break the complex two-gene cross into two simple single-gene crosses (note that the Probability Method can be used if it is known that the alleles of the different genes Assort Independently)

b. Show the expected genotypic and phenotypic ratios for each of the simple single-gene crosses.

c. Using this information, show the calculations for determining the expected number of genotypes and the expected number of phenotypes among the offspring of the BbTt x bbTt cross.

d. What is the expected frequency of BbTt offspring from the cross? (Show the calculation using the Product Rule).

e. What is the expected frequency of white coat, long tail offspring? (Show the calculation using the Product Rule).

f. Using Branching Diagrams show the Full expected genotypic and phenotypic ratios among offspring of the cross. Include the calculation of the frequency of each genotype or phenotype in the branching diagram.

Expert Solution

The allele for black coat color (B) is dominant to the allele for white coat color (b) and the allele for long tail (T) is dominant to the allele for short tail (t).

(a) Mice with black coat color and long tail (BbTt), this complex two- gene cross can be broken into two simple single cross using probability method. In order for a mice to have the genotype BbTt, the mice must receive Bb alleles and it must also receive Tt alleles, both these events must take place.The two events are independent because the genes assort independently. The two simple single gene cross are as follows:

Coat color: Bb Bb Tail length: Tt Tt

Results: BB, Bb, Bb, bb Results: TT, Tt, Tt, tt

Probability of Bb=1/2, probability of Tt=1/2.

(b) The phenotypic and genotypic ratios for coat color are 3(black):1(white) and 1:2:1 respectively; the phenotypic and genotypic ratios of tail length are 3(long):1(short) and 1:2:1 respectively.

(c) The cross between BbTt and bbTt is as follows:

parents: BbTt bbTt

gametes: BT Bt bT bt bT bt

results; BbTT, BbTt, bbTT, bbTt, BbTt, Bbtt, bbTt, bbtt

Number of genotypes=6 ; number of phenotypes=4 (black and long, black and short, white and long, white and short)

(d) From the dihybrid cross the probability of BbTt offspring=

the probability of BbTt offspring is the product of individual probabilities of Bb and Tt (according to product rule) from the two simple single gene cross.

probability of BbTt offspring=

(e) The genotypes of white coat and long tail will be either bbTT or bbTt.

Probability of bb=1/4 and probability of TT and Tt are 1/4 and 1/2 respectively

Expected frequency of offspring with white coat and long tail= probability of bbTT + probability of bbTt=

gladiator answered 2 years ago

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