Answer and Explanation:
Available data:
- Allele "B" expresses brown, and "b" expresses black the wing
- Allele "N" expresses normal wings, and allele "n" expresses wingless.
- "P" generation: cross two true-breeding flies: brown and wingless with black with normal wings.
- All of the flies in the F1 generation are brown and have normal wings.
(a) What are the genotypes in P generation? BBnn x bbNN
BBnn= brown and wingless
bbNN=black and normal wings
(b) What are the genotypes in F1 generation? The F1 genotype is 100% Heterozygote, BbNn
(c) You now take an F1 female and cross her to a true-breeding black, wingless male: BbNn x bbnn
and you count:
- Total number of individuals, N= 1600
- 85 brown winged flies, B-N-
- 728 black winged flies, bbN-
- 712 brown wingless flies, B-nn
- 75 black wingless flies, bbnn
These results suggest the genes are on the same chromosome, why?
To know if two genes are linked in the same chromosome, we must observe the progeny distribution. If individuals, whos genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1, in the example this would be 1/4 BbNn, 1/4 Bbnn, 1/4 bbNn, 1/4 bbnn, or 400 individuals per genotype.
But if instead of this distribution, we observe a different one, that is that phenotypes appear in different proportions, we can assume that genes are linked in the same chromosome of the double heterozygote parent. In the present example, the proportion of the F2 has a different distribution, so that is how we know they are on the same chromosome, very close to each other. The F2 is distributed like this: 85 B-N-, 728 bbN-, 712 B-nn, 75 bbnn.
(d) What is the genetic distance between the color and wing genes?
We need to know that 1% of recombination frequency = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.
The map unit is the distance between a pair of genes for which every 100 meiotic products one of them results in a recombinant one.
In the present example, the genotype, in linked gene format, of the double heterozygote parent is Bn/bN.
In this way, we might verify which are the recombinant gametes produced by the di-hybrid, and we will be able to recognize them by looking at the phenotypes with lower frequencies in the progeny: 85 B-N- and 75 bbnn.
To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals.
P = Recombinant number / Total of individuals.
P= 85 + 75 / 85 + 75 + 728 + 712
P= 160/ 1600
P= 0.1
The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU).
Genetic distance= 0.1 x 100 = 10 MU.
(e) A series of fruit fly matings shows that the recombination frequency between the gene for wing size and the a third gene (the gene for antenna length is 5%. List all possible recombination frequencies between the gene for color and the gene for antenna length and draw the possible chromosome map(s).
There are two possibilities:
- Antena--------wing------------color
5 MU 10 MU
Where the genetic distance between the gene for color and the gene for antenna length is 15MU
- Color----------Antena---------Wing
5 MU 5 MU
Where the genetic distance between the gene for color and the gene for antenna length is 5MU