Answer:
The recombination frequency between genes E and B is 0.1 = 10%
Explanation:
<u>Available data</u>:
Cross: BbEeHh x bbeehh
F1) 1000 individuals in total
- 36 Bb ee hh
- 400 Bb ee Hh
- 4 Bb Ee Hh
- 426 bb Ee hh
- 46 Bb Ee hh
- 38 bb Ee Hh
- 2 bb ee hh
To know if two genes are linked, we must observe the progeny distribution. In a tri-hybrid cross, If individuals, whose genes assort independently, are test crossed, they produce a progeny with equal phenotypic frequencies 1:1:1:1:1:1:1:1. If we observe a different distribution, that is that phenotypes appear in different proportions, we can assume that genes are linked in the heterozygote parent.
In the present example, the phenotypic ratio shows different proportions than the expected ones if they were not linked.
We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. And simple recombinant gametes produced by the cross, which frequencies are intermediate.
Parentals:
- 400 Bb ee Hh
- 426 bb Ee hh
Double recombinants:
Simple recombinant
- 48 bb ee Hh
- 36 Bb ee hh
- 46 Bb Ee hh
- 38 bb Ee Hh
Comparing the parental and the double recombinant we will realize that they only change in the position of the alleles E/e. This suggests that the position of the gene E is in the middle of the other two genes, B and H, because in a double recombinant only the central gene changes position in the chromatid.
B------E------H
Now we will call Region I to the area between B and E and Region II to the area between E and H.
Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between B and E genes, and P2 to the recombination frequency between E and H.
P1 = (R + DR) / N
P2 = (R + DR)/ N
Where: R is the number of recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals. To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.
So:
• P1 = (R + DR) / N = 48 + 46 + 4 + 2 / 1000 = 100/1000 = 0.1
• P2 = (R + DR)/ N = 36 + 38 + 4 + 2 / 1000 = 80/1000= 0.08
Now, to calculate the recombination frequency between the two extreme genes, B and H, we can just perform addition or a sum:
P1 + P2 = Pt
0.1 + 0.08 = Pt
0.18 = Pt
The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU). The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.
• P1 = 0.1 x 100 = 10%
• P2 = 0.08 x 100 = 8%
• Pt = 0.18 x 100 = 18 %
