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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?

Respuesta :

Answer:

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.

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