A pure breeding wildtype male fish has the following phenotypes: red fins, yello

ID: 56271 • Letter: A

Question

A pure breeding wildtype male fish has the following phenotypes: red fins, yellow eyes and frilly fins. It is crossed with a female fish with the following mutant phenotypes: blue fins, brown eyes, and straight fins. In the F1 generation, all female fish are normal, while all males have straight fins. A F1 female is crossed with multiply homozygous ( or possibly hemizygous) recessive male fish, producing the following progeny:

a. Which of the genes, if any, are X-linked?

b. Which of the mutant allele, if any, are dominant?

c. Carry out a chi-square test to determine if the genes assort independently by filling in the above table. State the degree of freedom and critical value of your test statistics. What do you conclude from your Chi-square test?

d. in the column provided (#), label whether each phenotype combination is parental (PAR) or recombinant (REC).

e. Determine the gene order by carrying out two-point crosses. Draw a map of the linked genes showing the distances between genes.

PLEASE HELP ME AND SHOW WORK SO I UNDERSTAND HOW TO DO IT THANK YOU!

| Expected no. | (Obs-Exp)2/Exp | PAR/REC (#) Phenotype | Observed No. Normal Brown eyes Brown eyes, 25 straight fins Blue fins Blue fins, brown eyes Blue fins, brown eyes, straight fins Straight fins122 Blue fins, straight fins Total 127 23 26 130 125 600

Explanation / Answer

When a complete heterozygous female is crossed to complete homozygous male then 8 phenotypes with same population are produced.

According to the given data, the fins character (frilly or straight fins) is the X-linked character. This is because when the parents are crossed in the F1 generation all the females are normal, while males represent the straight fin character. This happens only when the genes for a character are located on X-chromosome.

There is no dominant mutant allele. This is because the expected phenotype and obtained phenotypes are similar.

Degree of freedom is calculated as follows:

Where “n” is number of phenotypes produced by trihybrid cross. In this case 8 phenotypes are produced. Hence n=8.

If the chi-square value is less than 0.05 then the hypothesis is accepted, while if it is greater than 0.05 it is rejected. In the given case, the obtained chi-square value is 0.32. hence, the hypothesis is accepted.

Phenotype

Observed No.

Expected No.

(Obs-Exp)2/Exp

PAR/REC

Normal

127

36.05

PAR

Brown eyes

-36.05

REC

Brown eyes, straight fins

-33.33

REC

Blue fins

-32.01

REC

Blue fins, brown eyes

130

40.33

REC

Blue fins, brown eyes, straight fins

125

33.33

PAR

Straight fins

122

29.45

REC

Blue fins, straight fins

-37.45

REC

Total

600

0.32= 2