Question

Consider a population of 1000 birds in Florida. Gene 1 is in Hardy-Weinberg equilibrium, and the frequency of the N allele is 0.2. Also, there are 30 BB and 80 bb individuals. Answer the following:

a. What is the frequency of the R allele?

b. What are the frequencies of the RR, RN, and NN genotypes?

c. How many individuals have the RN genotype?

d. How many individuals have the Bb genotype?

e. What are the frequencies of the B and b alleles?

f. If I selected a bird at random from this population, what is the probability that the bird will have a BB genotype?

Answer 1

Total population = 1000

Gene 1 is in Hardy-Weinberg equilibrium means follow Hardy-Weinberg equation = p^2 + q^2 + 2pq = 1

Frequency of N allele = 0.2 ( can be p or q, let it consider q)

Also given there are 30 BB and 80bb individuals.

Ans 1) frequency of R allele =

As there are 2 genes one is B/b and one is N and R. So N and R are alleleic pair. And we have been given frequency of N = 0.2

By using Hardy-Weinberg equation = p+q = 1

N + R = 1 ( as p and q are 2 Alleles the same way N and R are 2 Alleles).

R = 1 - N

R = 1 - 0.2

R = 0.8

Ans 2) frequencies of RR, RN and NN are :

RR = R^2  

R^2 = 0.8^2 -------- ( as R = 0.8)

R^2 = 0.64

NN = N^2

N^2 = 0.2^2 -------- ( as N=0.2)

N^2 = 0.04

RN = 2RN ( as per Hardy-Weinberg equilibrium heterozygotes = 2pq = 2RN)

2RN = 2 * 0.2 * 0.8

2RN = 0.32

RN = 0.32/2

RN = 0.16

Ans 3) Individuals having RN Genotype =

Frequency of 2RN * total population

= 0.32 * 1000

= 320

Ans 4) individuals with Bb Genotype =

Frequency of bb = 80/1000 ------ bb individuals = 80

= 0.08

b = √b^2

b = √0.8

b = 0.282= 0.28

Frequency of BB = 30/1000

BB = 0.03

B = √B^2

B = √0.03

B = 0.173 = 0.17

Frequency of Bb = 2 * B * b

2Bb = 2 * 0.17 * 0.28

2Bb = 0.0952

So, frequency of Bb = 0.0952

Ans 5) frequency of B and b Alleles =

Frequency of B = 0.17 ( calculated in and 4)

Frequency of b = 0.28 ( calculated in Ans 4)

Ans 6) if a bird selected at random what is the Probability that it has the Genotype = BB

BB Genotype birds = 30

Total bird population = 1000

Probability = 30/1000

Probability = 3/100 or 0.03

Thank you.?