Answer:
Explanation:
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%.
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.
So, en the exposed example:
- J and K are autosomal genes
- J and K are separated by 60 M.U.
- 60 M.U. means that there is 60% of recombination.
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 equals 1% of recombination frequency. This means that every 100 meiotic products, one of them is a recombinant one.
1 M.U. -------------- 1% recombination
60 M.U. ------------ 60% recombination
30% Jk + 30% jK
100 M.U. - 60 M.U. = 40 M.U.
40M.U.--------------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%
Mendelian genetics one of the fundamental laws is The Law of Independent Assortment. The law states that parental traits are passed independently from parent to child. The recessive trait, vestigial Wings, occurs in an approximate phenotypic ratio of 1.3. In monohybrid Cross of heterozygous (Rr) parents the expected phenotypic ratio correlates with the given 1:3 result therefore l can conclude that the parents are both heterogeneous (Rr) for vestigial wings. Normal Wings-R, Vestigial Wings (Parent 1) Rr* Rr (Parent 2) R*R- RR- Normal Wings (Child 1) R*r Rr- Normal WIngs (Child 2) r * R - Rr- Normal Wings (Child 3) r*r- rr Vestigial Wings (Child 4) 1 Vestigial Wings: 3 Normal Wings
This process is mitosis. mitotic cell division is how they grow.
