Question

Phenotype brown pigmentation brown pigmentation red pigmentation (strawberry leopard) Genotype SS ss Number of leopards 112 98 SS Note: S = dominant allele for brown pigmentation s = recessive allele for red pigmentation (erythrism) Over the next several questions, we will work our way through the process to determine if this leopard population is in Hardy-Weinberg equilibrium (this will include applying the Chi-square goodness-of-fit test). (Before beginning this question, you may find it helpful to review the worked problem in your textbook. Section 18.2 - "Testing for Hardy- Weinberg Proportions"). Round your answer to three decimal places and submit your answer in the correct format (e.g. 0.062). (a) What is the frequency of the 5 (dominant) allele in this population? 0.718 (1 mark) (b) What is the frequency of the s (recessive) allele in this population? 0.282 (1 mark)

Considering the above information, how would you go about completing the following question?

Answer 1

Answer:

6).

SS 112= 224 S alleles

Ss 98= 98 S and 98 s alleles

ss 7 = 14 s alleles

Total alleles= 434

Total S alleles = 224+98= 322

Frequency of S allele = 322/434= 0.742

Total s alleles = 98+14= 112

Frequency of s allele = 112/434= 0.258

7).

Total progeny= 217

Expected values:

Number of SS = 0.742*0.742*217= 120

Number of Ss = 2*0.742*0.258*217=83

Number of ss = 0.258*258*217= 14

8).

Genotype	observed (O)	Expected (E)	(O-E)2	(O-E)2/E
SS	112	120	64	0.533
Ss	98	83	225	2.711
ss	7	14	49	3.5

Total of (o-e)2 = chi-square value= 6.744

Degrees of freedom = number of Genotypes -1

Df = 3-1=2

Critical value = 5.99

The chi-square value of 6.745 is greater than the critical value of 5.99. hence, the hypothesis is rejected and data is not fit into Hardy-weinberg equilibrium.