The frequency of each phenotype in the testcross progeny are:
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
What is phenotype?
Phenotype characters are those which are shown outside like shape, color, size, voice, hair etc.
Given data,
G= gray
g= black
S= Sleek
s= pudgy
F= friendly
f= mean
G is dominant to g
S is dominant to s
F is dominant to f
Pure-breeding gray, sleek, friendly (ggSSff) is mixed with pure-breeding black, sleek, and mean (ggSSff) (GGssFF). GgSsFf is the F1 from this mating. It has been tested with one of his parents (either male or female) The obtained gene distances for GGssFF (female) 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
As a result, the interference is 0.15, indicating that the interference is positive. In this case, one crossover diminishes the possibility of another crossover between homologous chromosomes in its vicinity.
gSsFf x GGssFF
a) GGSsFF
b) GGSsFf
c) GGssFF
d) GGssFf
e) GgSsFF
f) GgSsFf
g) GgssFF
h) GgssFf
Between "g" and "s," crossing over occurs 9 - 0.5 = 8.5 percent of the time, which indicates it does not happen 91.5 percent of the time.
Between "s" and "t," crossing over occurs 13 -0.5 = 12.5 percent of the time, implying that it does not occur 87.5 percent of the time.
a. and b. The frequency of parental = p (no crossover between either gene)
= p(no CO g—s) ' p(no CO s—f)
= (0.92)(0.88) = 0.8096
c. and d. The frequency show recombination between "g" and "s" only
= p(CO g—s) ' p(no CO s—t)
= (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 I only = p(CO s—t) ' 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 I
= p(CO g—s) ' p(CO s—fl = (0.09)(0.13)
= 0.0117 Or 1/2(0.0117)
= 0.005 each
Thus, the frequencies are 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.
Learn more about genotype, here:
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