Answer:
1) option a is correct. (260+270)/1545
<em>(Note about this option: In the statement, it is written as </em><em>(260+270) 1545. </em><em>Lacks the division symbol, </em><em>/ </em><em>)</em>
2) From these results one can conclude that the two genes are linked (option a)
Explanation:
To know if two genes are linked, we must observe the progeny distribution. If heterozygous individuals, whose genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1. If we observe a different distribution, phenotypes appearing in different proportions, we can assume linked genes in the double heterozygote parent.
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. Phenotypes with the highest frequencies represent the parental gametes.
To calculate the recombination frequency, we will use the following formula: P = Recombinant number / Total of individuals.
In the present example:
Parental) YySs x yyss
Gametes) YS parental type ys, ys, ys, ys
ys parental type
Ys recombinant type
yS recombinante type
The total number of individuals in the offspring: 1545
Phenotypic class Number of offspring
- 500 Y-S- (parental)
- 515 yyss (parental)
- 260 Y-ss (recombinant)
- 270 yyS- (recombinant)
According to this information, the phenotypic frequencies of the progeny differ from the phenotypic ratio 1:1:1:1. We can assume then, that t<u>hese genes are linked. </u>
The recombination frequency is:
P = Recombinant number / Total of individuals
P = 260 + 270 / 1545
P = 530/1545
P = 0.343
The genetic distance between genes is 0.343 x 100= 34.4 MU.
