Question

Equations:

p + q = 1

p

2

+ 2pq + q

2

= 1

Three-way cross [(P1 x P2) x P3] predicted from F1's: 1/2[F1(P1 x P3) + F1(P2 x P3)]

Double cross [(P1 x P2) x (P3 x P4)] predicted from F1's: 1/4[F1(P1 x P3) + F1(P1 x P4) + F1(P2 x P3) +

F1(P2 x P4)]

V

p

= V

A

+ V

E

+ V

D

V

G

= V

A

+ V

D

V

E

= (V

p1

+ V

p2

+ V

F1

)/3

V

A

= 2V

F2

– (V

B1

– V

B2

)

V

D

= [(V

B1

+ V

B2

) – V

F2

– (V

p1

+ V

p2

+ V

F1

)]/3

h

B

2

= V

G

/V

P

h

n

2

= V

A

/V

P

R = i



h

2

Questions:

1. The frequency of two alleles in a gene pool is 0.42 (A) and 0.58 (a). Calculate the percentage of

homozygous dominant, homozygous recessive, and heterozygous individuals assuming that the

population is in Hardy-Weinberg equilibrium (5 pts).

2. In a sample of 600 corn plants, 358 plants have yellow kernels and 242 have white kernels. The

allele Y for yellow kernels is dominant over the y allele for white kernels. Calculate the allelic

frequencies within this population, assuming that the population is in H-W equilibrium (5 pts).

3. Compute the broad sense and the narrow sense heritability estimates. (5 pts.)

V

p1

= 14.1

V

p2

= 12.2

V

F1

= 13.3

V

F2

= 40.2

V

B1

= 35.2

V

B2

= 34.6

4. Using the narrow sense heritability calculated in question 3, calculate the response to selection

when 20% of the top performing individuals are selected from the population. The phenotypic

standard deviation (



) of the population is 3. (5 pts.)

5. Five superior inbred lines were identified (A, B, C, D, E) which were intercrossed in all

combinations and the F1 progenies evaluated for productivity in a field yield trial. The results

from each single cross in the trial were:

Cross

Yield

AxB

58

AxC

63

AxD

59

AxE

67

BxC

61

BxD

64

BxE

62

CxD

61

CxE

57

DxE

63

Calculate the expected yield from the double cross hybrids [ (D x E) X (A x C) ] and

[ (A x E) X (B xD) ]. Which is the ‘better’ hybrid? (5 pts)

Answer 1

1). Here frequency of A is 0.42 and frequency of a is 0.58

So, p= 0.42, q= 0.58.

Frequency of homozygous dominant, that is, p²= (0.42)²= 0.1764= 17.64 %

Frequency of homozygous recessive, q²= (0.58)²= 0.3364= 33.64%

Frequency of heterozygous individuals, 2pq= 2×0.42×.58= 0.4872= 48.72%.

2). Here y allele is recessive and Y is dominant.

Out of 600, 242 have white kernels, that us, they have yy allele.

So, q²= 242/600×100= 40.33%= 0.403.

Therefore, q= (0.403)¹/²= 0.635

We know that p+q= 1

So, p= 1- 0.635= 0.365.

So, frequency of allele Y is 0.365 and y is 0.635.