PLEASE ANSWER BOTH QUESTIONS IF YOU CANNOT ANSWER BOTH DO NO

PLEASE ANSWER BOTH QUESTIONS, IF YOU CANNOT ANSWER BOTH DO NOT ANSWER

1. Modified from your lab manual: In horses, the genes for white coat color(C^W) and red coat color (C^R) are codominant. Heterozygotes (C^RC^W) have a light red coloration, called roan. You located a population of wild mustangs in a valley that had 476 red horses, 323 roan horses, and 51 white horses. First calculate p and q, where p = freq(C^R) and q = freq(C^W).

p =

q =

2. Use the Hardy-Weinburg formula to calculate the expected genotypic frequencies. Round each answer to the nearest whole number.

C^RC^R =

C^RC^W =

C^WC^W =

Could you say the population is in Hardy-Weinburg equilibrium? yes/no  

Solution

1. p² = 476; 2 pq = 323, q^{2 =} 51

Genotypec frequencies:

Frequenxy of p^{2 =} 476/850 = 0.56: Frequenxy of 2pq = 323/850 = 0.38; pq = 0.38/2 = 0.19; Frequenxy of q² = 51/850 = 0.06

Allelic frequencies: freq (CR) or p = p allele is present in Red and Roan phenotypes. So p² + 1/2 x 2pq

= 0.56 + 1/2 x 0.38

= 0.56 + 0.19 = 0.75

freq (CW) or q = 0.06 + 1/2 x 0.38

= 0.06 + 0.19

= 0.25

2.

CRCR = Also known as p x p = 0.56 x 0.56 = 0.313; CRCW = Also called pq: 0.56 X 0.19 = 0.106; CWCW = Also called qxq: 0.06 X 0.06 = 0.0036

According to Hardy - Wein berg law pp + 2 pq + qq = 1

0.313 + (2 x 0.106) + 0.0036 = 1

0.53 # 1

The Above population is not in HW equilibrium