In beetles, a gene coding for body color and a gene coding for bristle length are located on the same chromosome. Black...

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In beetles, a gene coding for body color and a gene coding for bristle length are located on the same chromosome. Black...

In beetles, a gene coding for body color and a gene coding for bristle length are located on the same chromosome. Black (B) is dominant over grey body (b) and long bristles (s) are recessive to short bristles (S). A beetle that is heterozygous for black body and short bristles is mated with one that has a grey body and long bristles.

What are the genotypes of the parents? ____________ ____________
Assume that the B allele is linked with the S allele and the b allele is linked to the s allele and that these genes are located 12 map units apart on the chromosome. In the above cross, if you counted 2846 offspring and crossing over occurred during meiosis, what are the expected numbers (theoretical) of each genotype of offspring?

genotype: number:

_____________________               _____________________
_____________________               _____________________
_____________________               _____________________
_____________________           _____________________

science biology

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Answer 1

Answer #1

Representation:

1. Trait 1: Body color

Black= Dominant = BB/Bb
Grey = Recessive = bb

2. Trait 2: Bristle size

Short Bristle= Dominant =SS/Ss
Long bristle = recessive = ss

Parents:

Beetle with black body and short bristle (heterozygous): Genotype = BbSs
Beetle with grey body and long bristle: Genotype = bbss

Cross:

BbSs x bbss

	bs	Bhenotybe	numbers
BS	BbSs	Black-short bristle	4/16= 1/4
Bs	Bbss	Black-long bristle	4/16= 1/4
bS	bbSs	Grey-short bristle	4/16= 1/4
bs	bbss	Grey-long bristle	4/16= 1/4

Ratio = 1: 1:1:1 (in case of indepent assortment.

b.

Since, the genes are linked, the ratio will not be 1:1:1:1.

Recombination frequency (RF or θ) indicates the frequency of chromosomal crossover between two genes.
RF measures genetic linkage and the loci of gene on a chromosome. Recombination frequency (measured in %) determines the distance of two gene loci (measured in centi-Morgan (cM) or as map unit (MU).

Thus, recombination frequencies can be used to determine the map distances of two gene loci.

If RF = 1%, map distance = 1 cM.

Given:

Map distance= 12 MU
Total = 2846
RF = 12%

R.F.= (Number of recombinants/ Number of recombinants + Number of non-recombinants) x 100

Or,

12= (Number of recombinants/ 2846) x100 = 341.52

Number of recombinants = 341.52

Number of parental = 2846 - 341.52 =2504. 48

Parental genotypes: BbSs and bbss

Recombinant genotypes: Bbss and bbSs

Genotype	Number
BbSs	341.52/2 = 170.76
bbss	341.52/2 = 170.76
Bbss	2504.48/2 = 1252.24
bbSs	2504.48/2 = 1252.24

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Answer 2

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