GENETICS Out of 1000 there are 150 that exhibit the recessive bandit eye phenoty
ID: 199279 • Letter: G
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.
Explanation / Answer
Answer:
Part 1:
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 Ha: There is statistically significant difference between the observed and expected values; therefore the population is not in Hardy Weinberg equilibrium.
Chi Square test:
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
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) 59.374Related Questions
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