Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Genetics: I don\'t understand why the professor used this equation instead of ju

ID: 181763 • Letter: G

Question

Genetics: I don't understand why the professor used this equation instead of just square rooting the pp(BB) and qq(bb) given. Please explain this. and explain why this isn't in hardy weinberg equillibrium.

3. Assume you have 100 Andalusian chickens, of which 36 are black (BB), 44 are blue (Bb) and 20 are white (bb). (15 pts) a. Calculate both the allelic (p &q;, and the genotypic (p2,2pq amd frequencies for this population. Is it in Hardy-Weinberg equilibrium? b. If you decide that you no longer like the white Andalusians, what will the allelic and genotypic frequencies be after one generation of selection against the white phenotype? 2% Zoo Go Not

Explanation / Answer

Answer:

a).

white frog frequency = q^2 = 20/100 = 0.2

The frequency of allele b = q = 0.44

As, p + q = 1

The frequency of allele B = p = 0.56

The expected frequency of genotype BB (Black) = p^2 = 0.56 * 0.56 = 0.31 = 31%

The observed frequency of genotype BB = 36

The expected frequency of genotype Bb (Blue) = 2pq = 2 * 0.56 * 0.44 = 0.49 = 49

The observed frequency of genotype Bb = 44

p^2 + 2pq + q^2 = 1(Hardy-Weinberg equilibrium)

31 + 49 + 20 = 1

Based on the expected allelic frequency, the population is in Hardy-Weinberg equilibrium.

Based on the observed phenotypes…

The frequency of allele B = 36/100 =0.36 = 0.6

The frequency of allele b = 20/100 =0.2 = 0.4

p^2 = 0.36

Q^2 = 0.2

2pq = 2 * 0.6 * 0.44 = 0.53

p^2 + 2pq + q^2 =1

0.36 + 0.53 + 0.2 = 1.09 ( it is not in Hardy-Weinber equilibrium)

b). After one generation the allele number will be doubled, so their frequency as follows

The frequency of allele B = p = 2(36) + 44 / 2(100) = 0.58

p^2 = 0.58 * 0.58 = 0.34

The frequency of allele b = p = 2(20) + 44 / 2(100) = 0.42

q^2 = 0.18

2pq = 2 * 0.58 * 042 = 0.49

p^2 + 2pq + q^2 =1

0.34 + 0.49 + 0.18 = 1.01

The total frequency of populations is not equal to one, so the population is not in Hardy-Weinberg equilibrium.

It is not possible to get allele frequency by square rooting, because heterozygous population will be present that also consists of both alleles. They are also be considered for calculating allele frequency.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at drjack9650@gmail.com
Chat Now And Get Quote