Answer:
- J K / j k = 20%
- j k / j k = 20%
- J k / j k = 30%
- j K / j k = 30%
- To find the recombination frequency, it's essential to understand that
1% of recombinations equals 1 map unit = 1cm. Furthermore, the
maximum recombination frequency can never exceed 50%. Each map unit signifies the distance between gene pairs, wherein for every 100 meiotic outputs, a single one results in a recombinant outcome.
In the given example:
- J and K are autosomal genes
- J and K are positioned 60 M.U. apart
- 60 M.U. indicates a 60% recombination probability.
Cross) J K / j k x j k / j k
Gametes) JK Parental jk, jk, jk, jk
jk Parental
Jk Recombinant
jK Recombinant
One map unit translates to 1% of recombination frequency. This indicates that out of every 100 meiotic products, one is recombinant.
1 M.U. is equivalent to 1% recombination
60 M.U. corresponds to 60% recombination
30% Jk + 30% jK
100 M.U. - 60 M.U. = 40 M.U.
40 M.U. equates to a 40% parental (not recombinant)
20% JK + 20% jk
Punnet Square) JK jk Jk jK
jk JK/jk jk/jk Jk/jk jK/jk
J K / j k = 20%
j k / j k = 20%
J k / j k = 30%
j K / j k
= 30%
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