Question

1) Rabbit’s ears can be either short (dominant) or floppy (recessive). If a population of rabbits in Hardy-Weinberg equilibrium has 250 short eared rabbits and 100 floppy eared rabbits, what is the allele frequency of the “short” allele?

A.0.714

B.0.465

C.0.845

D.0.535

2)You are observing the suit (clubs, diamonds, hearts, or spades) of a randomly drawn card from a deck. What is the degree(s) of freedom if you were to perform the Chi-square test on the expected and observed outcome of the suits of the cards?

A.1

B.0

C. 3

D.4

3)You perform a Chi-square test for gene X and find the p-value to be between 0.05 and 0.01. What can you say with 95% certainty about this population?

A.This population is in Hardy-Weinberg equilibrium at the X locus: gene X is influenced by at least one force of evolution.

B.This population is in Hardy-Weinberg equilibrium at the X locus: gene X is not influenced by any forces of evolution.

C.This population is not in Hardy-Weinberg equilibrium at the X locus: gene X is influenced by at least one force of evolution.

D.This population is not in Hardy-Weinberg equilibrium at the X locus: gene X is not influenced by any forces of evolution.

Please do all three questions.

Answer 1

1) Short-ear rabbit( Aa or AA) = 250

Floppy ear rabbit (aa) = 100

Total = 350

Frequency of floppy ear rabbit (aa) = q2 = 100/350 = 0.2857

So, frequency of floppy ear allele = q = sqrt (0.2857) = 0.5345

Since the population is in HWE, so p+q = 1. So,

p= 1-q = 1- 0.5345 = 0.465

So the correct option is B

2) The degree of freedom = variable – 1

Number of variables here = 4

So degree of freedom = 4-1 = 3

The correct option is C.

3) 0.05 < p < 0.01 represents that only 1-5% of the observed genotypic differences are by chance while rest of the differences are due to some evolutionary forces favoring one of the genotypes. Thus, the population is not in Hardy Weinberg equilibrium.