Question

Two large populations of horses are being systematically crossed (mares from one population bred to stallions of the other and vice versa). Coat color is not a factor in determining which animals are selected and which individual matings are made (random matings). Frequencies of coat color genes at the C locus for population 1 are.85 for Cand.15 for c. Frequencies for Care.6 and care.4 for population 2. Given these values, what are the gene and genotypic frequencies of the F1?

a. p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06
b. p=0.725.q = 0.275; P=0.06. H=0.56, Q=0.51
c. p=0.4.q = 0.6: P=0.12. H=0.56, Q=0.32
d. p=0.725.q = 0.275: P=0.34. H=0.57. Q=0.09

Answer 1

Answer:

a. p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06

Explanation:

Let state our given parameters from the question:

Frequencies of coat color genes at the C locus for population 1 are .85 for C

This implies that the Allelic frequency C for population p1 =0.85

Frequencies of coat color genes at the c locus for population 1 are .15 for c

This implies that the Allelic frequency c for population q1 = 0.15

Frequencies for Care .6 i.e p2= 0.6

Frequencies for care .4 i.e, let that be q2= 0.4

The table below shows a diagrammatic representation of the above expression:

Alllelic Frequency                      C                                          c

Population 1                        (p1)   0.85                              (q1)   0.15

Population 2                       (p2)   0.6                               (q2)   0.4

Now, from above: let think of the table as a punnet square and then cross it together;

                                            (p1)  = 0.85                              (q1) =  0.15

p2 = 0.6                               p1p2                                       p2q1

                                            = 0.6 × 0.85                           = 0.15 × 0.6

                                            = 0.51 (P)                                = 0.09 (H)              

                                                                                                   

q2 = 0.4                               p1q2                                       q1p2

                                            = 0.85 × 0.4                           = 0.4 × 0.15

                                            =0.34 (H)                                = 0.06 (Q)

From the above table, the heterozygous are represented by (H)

∴ Frequency of heterozygous can be calculated as:

= 0.09 + 0.34

= 0.43

Thus, we can conclude that the progeny F1 genotypic frequencies are:

P= 0.51

H= 0.43

Q= 0.06

Now, let us calculate the allelic frequencies, p and q in F1

p = P + 1/2 × (H)

= 0.51 + (1/2 × 0.43)

= 0.51 + 0.215

= 0.725

q = Q + 1/2 × (H)

= 0.06 + (1/2 × 0.43)

= 0.06 × 0.215

= 0.275

Hence, p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06 , This makes option a the correct answer.