Question

GENETICS

Out of 1000 there are 150 that exhibit the recessive bandit eye phenotype which gives them the appearance of wearing a robber mask.

Determine the following

What is p? 0.61

What is q? 0.387

What is the null hypothesis? The values for p and q are not significantly different than the observed frequencies.

You then test the remaining animals and find the following numbers for the genotypes

RR 525    Rr 325 rr 150

What are your new allelic expected allelic frequencies based on this new information (p², 2pq, q²)? <<<Need This Answer >>> The exact values for the new p-squared, 2pq, and q-squared.

What is the X² value ? 0.026

How many degrees of freedom are there? One

What is the p-value? 0.8719

Is the population in HWE? Yes, fail to reject null hypothesis.

Just need the answer for the new P-Squared, 2pq, and q-squared. Thank you.

Answer 1

Answer:

Part 1:

• Number of recessive phenotypes = 150
• Total number of individuals = 1000
• Therefore, frequency of recessive phenotype (q²) = 150/1000 = 0.150
• Frequency of recessive allele (q) = sqrt(0.150) = 0.3873
• Frequency of dominant allele (p) = 1 - q = 1 -0.3873 = 0.6127

Part 2:

Null Hypothesis H₀:There is no difference between the observed and expected values; therefore the population is in Hardy Weinberg equilibrium.

Alternative Hypothesis H_a: There is statistically significant difference between the observed and expected values; therefore the population is not in Hardy Weinberg equilibrium.

• Observed Number of individuals with genotype RR = 525
• Observed Number of individuals with genotype Rr = 325
• Observed Number of individuals with genotype rr = 150
• Total number of individuals = 525 + 325 + 150 = 1000
  Frequency of individuals who are homozygous recessive rr (q²) =150/1000 = 0.150
• Frequency of recessive allele (q) = sqrt(0.150) = 0.3873
• Frequency of dominant allele (p) = 1 - q = 1 -0.3873 = 0.6127
• Total number of alleles = 1000*2 = 2000
• Total number of R alleles = 2*525 + 325 = 1375
• Total number of r alleles = 2*150 + 325 = 625
• If the population is in Hardy Weinberg equilibrium then:
  • Expected frequency of R allele (p) = 1375 / 2000 = 0.6875
  • Expected frequency of r allele (q) = 625/2000 = 0.3125
  • Expected number of individuals with RR genotype = p²*1000 = (0.6875)²*1000 = 472.65
  • Expected number of individuals with rr genotype = q²*1000 = (0.3125)²*1000 = 97.65
  • Expected number of individuals with Rr genotype = 2pq*1000 =
  • =2*(0.6875*0.3125)*1000 = 429.7

Chi Square test:

Genotype	Observed Number (O)	Expected Number (E)	O-E	(O-E)2/E
RR	525	472.7	52.35	5.798
Rr	325	429.7	-104.7	25.511
rr	150	97.7	52.35	28.065
			Chi Square (sum)	59.374

Degree of freedom = Number of categories - 1 = 3 - 1 = 2

P-value corresponding to a Chi square value of 59.374 with 2 degree of freedom is: P-Value < 0.00001. The result is significant at p < 0.05.

Therefore, Null Hypothesis is rejected. The population is not in Hardy Weinberg Equilibrium