Answer:
These genes are 12.3 MU apart.
Explanation:
<u>Available data</u>:
- Two diallelic genes coding for eyes number and legs position
- Normal phenotype: 2-eyed, antennae fly
- Mutant: three-eyed, fly with legs coming out of its head instead of antennas
- 1st Cross: a pure breeding wildtype with a pure breeding mutant
- F1) 100% normal dihybrid
- 2nd Cross: normal dihybrid from F1 with mutant
- F2) 500 normal flies, 50 2-eyed, head-legged flies, 76 three-eyed, antenna flies and 400 three-eyed, head-legged flies.
Let us first name the alleles
- E allele codes for two eyes
- e allele expresses three eyes
- L allele codes for legs in the body
- l allele codes for legs in the head
Normal phenotype → two eyes,antenna in the head → EELL, EeLL, EELl, EeLl
Mutant phenotype → three eyes, legs in the head → eell
1st cross:
Parentals) EELL x eell
F1) 100% EeLl normal individuals
2nd Cross:
Parentals) EeLl x eell
F2) 500 normal flies, EeLl
50 2-eyed, head-legged flies, Eell
76 three-eyed, antenna flies, eeLl
400 three-eyed, head-legged flies, eell
When two genes that assort independently are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1.
However, if these genes are linked, after the test cross, the distribution of phenotypes among the progeny differs from 1:1:1:1. Phenotypes appear in different proportions, so we can assume that these genes are linked.
In the exposed example, the phenotypic frequency is different from the expected one is genes were not linked.
We can recognize the parental gametes in the descendants because their phenotypes are the most frequent,
Parental
Recombinant
Now we need to calculate the recombination frequency in order to get the distances between the two genes.
To calculate the recombination frequency, P, we will make use of the following formula:
P = Recombinant number / Total of individuals.
P = 50 + 76 / 50 + 76 + 500 + 400
P = 126 / 1026
P = 0.123
The genetic distance, GD, will result from multiplying that frequency by 100 and expressing it in map units (MU).
The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.
GD = P x 100
GD = 0.123 x 100
GD = 12.3 MU
Two genes that are very close will have a few recombination events and are strongly bounded. The more separated two genes are, the more chances of recombination there will be. The closer they are, the fewer chances of recombination there will be. Genes that express 50% of recombination frequency or more are not linked genes. As long as they do not exceed this percentage, they are linked.
According to this information, we can assume that genes coding for eyes number and legs position are linked because their recombination frequency is 12.3% (<50%).
