Question

in dolphins, gray [G) s dominant to black (g Sleek 〔S) ls dominant to pudgy s) and friendly闩is dominant to mean lf). All loci are autosomal Pure breeding black, sleek mean dolphins were mated to pure-breeding gray, pudgy, friendly dolphins. The F1 were testcrossed. Based on the map below and a coefficient of coincidence of 0.8, determine the frequency of each phenotype in the testcross progeny. Round each answer to 5 decimal digits. Color Body Type Disposition 10cM 5cM

What is the frequency of gray, sleek testcross progeny?

What is the frequency of gray, friendly testcross progeny?

What is the frequency of black, friendly testcross progeny?

What is the frequency of black, pudgy, mean testcross progeny?

What is the frequency of gray, pudgy, friendly testcross progeny?

What is the frequency of gray, sleek, mean testcross progeny?

What is the frequency of pudgy, friendly testcross progeny?

What is the frequency of pudgy, mean testcross progeny?

Answer 1

Coefficient of coincidence is given by

The observed frequency of double recombinant/ expected frequency of double recombinant

given that COC = 0.8

Map distance between color and body type gene is 10cM

Map distance between body type and disposition is 5cM

expected frequency of double recombinant is given by

Map distance between color and body type /100 * map distance body type and disposition/100

10/100 * 5/100 = 0.005

The observed frequency of double recombinant we need to find out

Coefficient of coincidence = the observed frequency of double recombinant/ expected frequency of double recombinant

0.80 = X /0.005

0.004

The observed frequency of double recombinant = sum of the frequency of both the double recombination
Since nothing is given about the progeny numbers so we will take the double recombinant frequency as half of the total double recombinants that we got.

So the frequency will be 0.004/2 = 0.002

The frequency of gray, pudgy, friendly testcross progeny = 0.002

The frequency of black, sleek, mean testcross progeny 0.002

The distance between black and sleek is 10 cM.

A total number of the recombinant/total number of progeny = frequency of recombinant.

The frequency of recombination between black and sleek will be a sum of frequencies of gray, pudgy, friendly testcross progeny, black, sleek, mean testcross progeny, Black sleek and friendly progeny and gray, pudgy and mean progeny

Linkage distance = total number of recombinant * 100 / total number of progeny

10 = Recombination frequency between black and sleek* 100

Recombination frequency between black and sleek = 10/100 = 0.1

0.1 = sum of gray, pudgy, friendly testcross progeny, black, sleek, mean testcross progeny, Black sleek and friendly progeny and gray, pudgy and mean progeny

0.1 = 0.002 + 0.002 = Frequency of Black sleek and friendly progeny and frequency of gray, pudgy and mean progeny

0.1- 0.004 = 0.096

Frequency of Black sleek and friendly progeny = 0.096/2 = 0.048
frequency of gray, pudgy and mean progeny =0.096/2 = 0.048

The distance between Friendly and sleek is 5 cM.

A total number of the recombinant/total number of progeny = frequency of recombinant.

The frequency of recombination between friendly and sleek will be a sum of frequencies of gray, pudgy, friendly testcross progeny, black, sleek, mean testcross progeny, gray sleek and mean progeny and black, pudgy and friendly progeny

Linkage distance = total number of recombinant * 100 / total number of progeny

5 = Recombination frequency between black and sleek* 100

Recombination frequency between black and sleek = 5/100 = 0.05
0.05= sum of gray, pudgy, friendly testcross progeny, black sleek, mean testcross progeny, gray sleek and mean progeny and black, pudgy and friendly progeny

0.05= 0.002 + 0.002 + Frequency of Black sleek and friendly progeny and frequency of gray, pudgy and mean progeny

0.05 - 0.004 = 0.0.046

Frequency of Black sleek and friendly progeny = 0.046 / 2 =0.023

Frequency of gray, pudgy and mean progeny = 0.046 / 2 =0.023

As we know the frequency is always equal to one. So subtract 1 from all the frequencies that we have got we can get the parent frequency.

1 – 0.046+0.004+0.096

= 1 – 0.146

= 0.854

Frequency of Gray sleek pudgy = 0.854/2 = 0.427

Frequency of black pudgy and mean = 0.854/2 = 0.427

The frequency of gray, pudgy, friendly testcross progeny = 0.048
The frequency of black, sleek, mean testcross progeny 0.048

The frequency of gray, pudgy, friendly testcross progeny = 0.002

The frequency of black, sleek, mean testcross progeny 0.002

Frequency of Black sleek and friendly progeny = 0.023

Frequency of gray, pudgy and mean progeny = 0.023

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