Question

In dolphins, gray (G) is dominant to black (g), Sleek (S) is dominant to pudgy (s) and friendly (F) is dominant to mean (f). 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 0.64, determine the frequency of each phenotype in the testcross progeny. Round each answer to 5 decimal digits.

Body Type Disposition Color 3cM 17cM What is the frequency of gray, sleek testcross progeny? Answer:

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

Based on the given data,

• G= gray
• g= black
• S= Sleek
• s= pudgy
• F= friendly
• f= mean

1. G is dominant to g
2. S is dominant to s
3. F is dominant to f

Pure-breeding black, sleek , mean (ggSSff) are crossed with pure-breeding gray, pudgy, friendly (GGssFF). The F1 from this mating is: GgSsFf. It is testcrossed with it either of his parents (either male or female) GGssFF (female), the obtained gene distances, are:

• Color to body type = 9cm
• Body type to disposition = 13cm
• Color to disposition is = 22cm

The interference of this cross is calculated by the formula = 1- coefficient of coincidence.

1-0.85 = 0.15

Thus, the interference is 0.15; hence it is a Positive interference. In this one crossover reduces the occurrence of likelihood another crossover in its neighborhood between homologous chromosomes.    

a) GGSsFF

b) GGSsFf

c) GGssFF

d) GGssFf

e) GgSsFF

f) GgSsFf

g) GgssFF

h) GgssFf

Crossing over occurs 9 - 0.5 = 8.5 percent of the time between “g” and “s,” which means it does not occur 91.5 percent of the time. Crossing over occurs 13 -0.5 = 12.5 percent of the time between “s” and “f,” which means that it does not occur 87.5 percent of the time.

a. and b. The frequency of parentals = p (no crossover between either gene)

= p(no CO g–s) ´ p(no CO s–f) = (0.92)(0.88)                        

= 0.8096

or 1/2(0.8096) = 0.4048 each

c. and d. The frequency that will show recombination between “g” and “s” only

= p(CO g–s) ´ p(no CO s–f) = (0.09)(0.88) = 0.0762

or 1/2(0.0762) = 0.0381 each

e. and f. The frequency that will show recombination between s and f only

= p(CO s–f) ´ p(no CO g–s) = (0.13)(0.92) = 0.1196

or 1/2(0.1196) = 0.0598 each

g. and h. The frequency that will show recombination between g and s and s and f

= p(CO g–s) ´ p(CO s–f) = (0.09)(0.13) = 0.0117

Or 1/2(0.0117) = 0.005 each

Thus,

a) GGSsFF = 0.4048

b) GGSsFf = 0.4048

c) GGssFF = 0.0381

d) GGssFf = 0.0381

e) GgSsFF = 0.0598

f) GgSsFf = 0.0598

g) GgssFF = 0.005

h) GgssFf = 0.005