Question

4. Three genes are mapped as shown below. The interference is -0.3. a fly homozygous for wildtype traits was crossed with a homozygous recessive. A female from Fi was crossed with an ebony, rough eyed, brevis bristled male. Give the expected numbers of phenotypes out of a total of 2400. (6 marks) e ro | 12 m.u. | 28 m.u. |

Answer 1

Answer-

According to the given question-

Three genes lies on the chromosome and the between gene ebony (e) and rough eye (ro) is 12 m.u. and the distance between rough eye (ro) and brevis bristed is 28 m.u.

bv 12 m.u. | 28 m.u. |

here the homozygous fly for wildtype traits was crossed with a homozygous recessive. and a femalew from F 1 generation was crossed with the ebony (e) , rough eye (ro), brevis bristed male (bv).

Lets

Suppose e= A, ro = B, bv = C (just for easy term)

here the interference = 0.3

number of progeny = 2400.

first we calculate the number of progeny when no interference was present then-

To get  a fly that is homozygous recessive i.e. a b c/ a b c  from selfing of A b c / a B C , then both gametes must be derived from a crossover between A  and B.

so first we have to calculate the frequency of the a b c gamete is
frequency of the a b c gamete = 1/2 probability (Crossing Over A–B)  $\times$  probability( no Crossing Over B– C )

= 1/2(0.12)(0.72)

= 0.04

frequency of the a b c  gamete = 0.04

So the frequency of the homozygous fly = (0.0432)² = 0.0018

now we have to cross between

A b c / a B C $\times$ a b c / a b c

here the parental fly are those in which the crossing over does not occur.

parental fly = probability(no Crossing Over A–B) $\times$ probability(no Crossing over B–C)
= (0. 88)(0.72)

parental fly = 0.64

each parents fly should be represented in equal proportions so

A b c = 0.32

b B C = 0.32

now calculate the the frequency of the a b c gamete -

frequency of the a b c gamete = 1/2 probability (Crossing Over A–B) $\times$ probability ( no Crossing Over B–C)

= 1/2(0.12)(0.72)

frequency of the a b c  gamete = 0.04

frequency of the a b c gamete = frequency of the A B C gamete = 0.04

now calculate the the frequency of the A b C gamete-

= 1/2 probability (Crossing Over B–C) $\times$ probability ( no Crossing Over A–B)

frequency of the A b C gamete = 1/2(0.28)(0.88) = 0.1232

frequency of the A b C gamete = frequency of the a B c   gamete  = 0.1232

now calculate the the frequency of the A B c gamete-  

1/2 probability (Crossing Over A–B) $\times$ probability ( Crossing Over B–C)

frequency of the A B c gamete = 1/2(0.12)(0.28) = 0.168

frequency of the A B c gamete = frequency of the a b C gamete = 0.168

here the number of progeny is 2400

so number of

A b c = 2400  $\times$ 0.32 = 768

a B C= 2400  $\times$ 0.32 = 768

a b c = 2400  $\times$ 0.04= 96

A B C = 2400  $\times$ 0.04= 96

A b C = 2400  $\times$ 0.1232 = 296

a B c = 2400  $\times$ 0.1232= 296

A B c = 2400  $\times$ 0.0168 = 40

a b C = 2400  $\times$ 0.0168 = 40

here the Interference (I)= 0.3

now we have to calculate the value based on Interference 0.3 or 30 % interference.

Interference (I) = 1 – observed Double Crossing Over / expected Double Crossing Over

0.3 = 1 –  observed Double Crossing Over / ( 0.12 ) (0.28)

0.3 (0.12) (0.28)   = (0.12) (0.28)  –  observed Double Crossing Over

observed Double Crossing Over= 0.0336 - 0.01008= 0.024

The distance between A - B = 12% = 100% [probability (Crossing over A–B) + probability (Double Crossing Over)]

so the probability (Crossing over A–B) = 0.12 – 0.024 = 0.096

The distance between the B - C = 28% = 100% [probability (Crossing over B–C) + probability (Double Crossing Over)]

probability (Crossing over B–C) = 0.28 – 0.024 = 0.256

Therefore the  probability of (parental fly ) = 1 – probability (Crossing over A–B) – probability (Crossing over B–C) – probability (observed Double Crossing Over)

= 1 – 0.096 – 0.256 – 0.024= 1 – 0.376 = 0.624
probability of (parental fly ) = 0.624

we know that

each parents should be represented in equal proportions so

A b c = 0.624 / 2 = 0. 312

a B C = 0.624 / 2 = 0. 312  

so the number of progeny from the 2400

A b c = 2400  $\times$ 0. 312  = 748

a B C= 2400  $\times$ 0. 312  = 748

a b c = 2400  $\times$ 0.096/ 2  = 115

A B C = 2400  $\times$ 0.096/ 2  = 115

A b C = 2400  $\times$ 0.256 / 2  = 307

a B c = 2400  $\times$ 0.256 / 2  = 307

A B c = 30

a b C = 30.

So here are the expected number of progeny with 0.3 level of Interference.