Question

Gene P has two alleles - P and p; gene Q has two alleles - Q and q.

If an PQ/pq female is crossed with a pq/pq male, what percent of the offspring would have genotype pq/pq if...

A) the genes were completely linked (no crossing-over at all; recombination frequency = 0; 1 mark)

B) the genes are 10 map units apart (1 mark)

C) the genes are 24 map units apart (1 mark)

For either part (B) or (C) show your work (1 mark). You cannot attach an image, use fractions or decimals and write it out.

Just fill in the requested information.

Do your results support independent assortment (yes/no)?

What is your chi-square value?

Are your results less than or greater than p = 0.05?

Correctly written cross:

Answer 1

Answer: The gene P has two alleles P and p and gene Q has two alleles Q and q. Now an PQ/pq female is crossed with a pq/pq male.

A. If the two genes are completely linked, then the results of the cross is as follows:

Parents: PQ/pq pq/pq

Gametes: PQ pq pq

Results: PQ/pq and pq/pq and the phenotypic ratio is 1:1. It shows that 100% parental combinations is formed and suggest that the gene P and Q is completely linked with each other. Percent of offsprings having genotype pq/pq is %.

If the genes are incompletely linked, then there is an occurence of crossing over, so besides the parental generation, recombinants are also found. Then the results can be interpreted as follows:

Parents:   PQ/pq   pq/pq

Gametes: PQ Pq pQ pq pq

Pesults: PQ/pq (parental combination); Pq/pq (recombinants); pQ/pq (recombinants); pq/pq (parental combination).

B. If the genes are 10 map units apart, then the total percentage of parental combination is %.

Percent of offspring having genotype pq/pq is %.

C. If the genes are 24 map units apart, then the toatl percentage of parental combinations=%. Percent of offspring having the genotype pq/pq is %.

In case of A, the results do not support independent assortment as the chi-square value is zero, so the P value is very high (P > 0.05), hence the hypothesis fails to reject.

Case B:

Phenotypes	Observed number (o)	Expected number (e)	(o-e)	(o-e)2	(o-e)2/e
Parental 1	45		20	400	16
Parental 2	45		20	400	16
Recombinant 1	5		-20	400	16
Recombinant 2	5		-20	400	16
Total	100

Chi-square value=

As the chi-square value is way too large, so the correspondent P value is much less than 0.05, so the second result is rejected. as there will be no independent assortment of genes.

Case C:

Phenotypes	Observed number (o)	Expected number (e)	(o-e)	(o-e)2	(o-e)2/e
Parental 1	38		13	169	6.76
Parental 2	38		13	169	6.76
Recombinant 1	12		-13	169	6.76
Recombinant 2	12		-13	169	6.76
Total	100

Chi-square value=

As the chi-square value is way too large, so the correspondent P value is much less than 0.05, hence the hypothesis is rejected.