In a certain diploid plant, three loci A/a, B/b, and C/c are linked as follows:

A/a is linked to B/b by 40 m. u.

B/b is linked to C/c by 30 m. u.

A/a is linked to C/c by 70 m. u.

One plant is available to you (call it the parental plant). It has the constitution A b c / a B C.

(a) Assuming no interference between the regions, if the parental plant is crossed with the a b c/a b c plant, how many progeny will have the phenotype of A B c if there are 1000 total progeny?

(b) Assuming 20 percent interference between the regions, if the parental plant is crossed with thea b c/a b c plant, how many progeny will have the phenotype of A B c if there are 1000 total progeny?

(c) Assuming 20 percent interference between the regions, if the parental plant is crossed with thea b c/a b c plant, how many progeny will have the phenotype of A B C if there are 1000 total progeny?

a) 60 b) 48 c) 152

Explanation:

According to the given question.

selfing of A b c/a B C, crossover between A and B-

first we have to calculate the frequency of the a b c gamete is:

1/2 p (CO A-B) * p (no CO B-C) = 1/2 (0.40) * (0.70) = 0.14

Frequency of homozygous plant will be 0.14*0.14 = 0.0196.

The cross between A b c/a B C * a b c/a b c. we know that the parental are those who who did not get Crossing over

so parental = p (no CO A-B) * p (no CO B-C) = 0.60 * 0.70 = 0.42

we know that each parental should be represented equally: i ... e.

A b c = 1/2 (0.42) = 0.21

and a B C = 1/2 (0.42) = 0.21.

So the frequency of the a b c gamete is = 1/2 p (CO A-B) * p (no CO B-C) = 1/2 (0.40) (0.70) = 0.14 which is equal to the frequency of ABC.

The frequency of the A b C gamete is = 1/2 p (CO B-C) * p (no CO A-B) = 1/2 (0.30) (0.60) = 0.09, which is equal to the frequency of a B c.

the frequency of the A B c gamete is = 1/2 p (CO A-B) * p (CO B-C) = 1/2 (0.40) (0.30) = 0.06, which is equal to the frequency of a b C.

for 1,000 progeny, the expected results is-

A b c 210 = a B C 210, A B C 140 = a b c 140, A b C 90 = a B c 90, A B c 60 = a b C 60.

Interference = 1 - observed DCO/expected DCO

0.2 = 1 - observed DCO / (0.40) (0.30)

observed DCO = (0.40) (0.30) - (0.20) (0.40) (0.30)

=0.12 - 0.024

observed DCO = 0.096.

The A-B distance = 40% = 100% [p (CO A-B) + p (DCO) ]

p (CO A-B) = 0.40 - 0.096 = 0.304

and B-C distance = 30% = 100% [p (CO B-C) + p (DCO) ]

p (CO B-C) = 0.30 - 0.096 = 0.204.

p (parental) = 1 - p (CO A-B) - p (CO B-C) - p (observed DCO)

= 1 - 0.304 - 0.204 - 0.096

= 1-0.604

p (parental) = 0.396

for 1,000 progeny, the expected results are-

A b c 198 = a B C 198, A B C 152 = a b c 152, A b C 102 = a B c 102, A B c 48 = a b C 48.

So the answer-

(a) assuming no interference phenotype of A B c = 60

(b) assuming 20% interference phenotype of A B c = 48.

(c) assuming 20% interference phenotype of A B C = 152.

a) 60.00

b) 48.00

c) 152.00