Question

In recent years, some leopard populations have suffered from a loss of habitat. You are part of a research team that is studying the effects that decreased range has on these populations. Your team observed that a portion of one population relocated to the desert and merged with another small leopard population after its own habitat was threatened by deforestation. You have been assigned to lead the study of this new population and look at its gene pool make-up. In particular the pigmentation locus. The following is the breakdown of their phenotypes and genotypes in this leopard population. The total number of leopards in this population is 217.

Phenotype	Genotype	Number of leopards
brown pigmentation	SS	112
brown pigmentation	Ss	98
red pigmentation (strawberry leopard)	ss	7

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).

Part 1: 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 S (dominant) allele in this population?  

(b) What is the frequency of the s (recessive) allele in this population?

Part 2: Calculate the genotypic frequencies that you would expect if the population is in Hardy-Weinberg equilibrium.

Round your answer to three decimal places and submit your answer in the correct format (e.g. 0.062).

(a) The frequency of the SS genotype is:

(b) The frequency of the Ss genotype is

(c) The frequency of the ss genotype is:

Part 3: Using the genotypic frequencies you calculated above in Question 6, determine the expected number of leopards with each of the following genotypes:

Round your answers to the nearest whole number (because you cannot have part of a leopard!).

(a) Expected number of leopards with genotype SS =  (1 mark)

(b) Expected number of leopards with genotype Ss =  (1 mark)

(c) Expected number of leopards with genotype ss =  (1 mark)

Part 4: Based on the expected numbers you calculated above in Question 7 for each genotype, and the observed numbers of leopards given in Question 5 with these genotypes, calculate the Chi-square statistic to determine if this population is in Hardy-Weinberg equilibrium.

Chi-square value is:

Part 5: The degrees of freedom for a Chi-square test of Hardy-Weinberg equilibrium will equal the number of expected genotypic classes minus the number of associated alleles.

Therefore, the degrees of freedom for this Hardy-Weinberg test is:

Part 6: Using the following table of critical values of the Chi-square distribution, what is the probability associated with your calculated Chi-square value for this Hardy-Weinberg test? (1 mark)

The probability is:

Part 7: Answer the following questions:

(a) Based on your Chi-square goodness-of-fit test, do you think this population is in Hardy-Weinberg equilibrium? Explain your reasoning based on your probability results.

(b) Briefly explain if the conclusion you reached in part (a) was expected for this population (i.e., are all the Hardy-Weinberg assumptions valid? If not, which one(s) have been violated?) Explain your reasoning based on your probability results.

Answer 1

Part - 1:

Total No. of an allele in the leopard's population = 217

Total No. of S allele in the population = 112 + 49 = 161

Total No. of s allele in the population = 7 + 49 = 56

Therefore, (a) frequency of the S (dominant) allele in the population = No. of S allele/ Total No. of allele in the population

= 161/217 = 0.742

(b) Frequency of the s (recessive) allele in the population = No. of s allele/ Total no. of allele in the population

= 56/217 = 0.258

Part - 2:

brown pigmentation	SS	112
brown pigmentation	Ss	98
red pigmentation	ss	7

(a) frequency of the SS genotype is = No. of pouplation with SS genotype/Total No. of genotype

= 112/217 = 0.512

(b) frequency of the Ss genotype is = No. of pouplation with Ss genotype/Total No. of genotype

= 98/217 = 0.452

(c) frequency of the ss genotype is = No. of pouplation with ss genotype/Total No. of genotype

= 7/ 217 = 0.032