 What is the map unit distance between L and S allele ?

My working : If we take the normal ratio :9:3:3:1 , first one should be 9/16 = 0.5625 and then subtract 0.51 from it, we get 0.0525, similarly working we get 0.0525 difference for all 4 and so the distance is 0.0525 X 4 X 100 = 21

Is my working correct ?

• I did not get your working and the logic behind it.Can you explain it a bit? But I got exact $20%$ as the answer. – Satwik Pasani Jan 11 '14 at 2:56
• I found out the difference between expected and observed and then added them. Could you tell me what you did ? – biogirl Jan 11 '14 at 4:06

With the heterozygous genotype $LlSs$, there are two kinds of linkages possible.
1. Cis=$LS.ls$
2. Trans=$Ls.lS$

All the possible gametes ($LS,lS,Ls,ls$) in both cases are same, but their frequency differs based on the strength of linkage and hence, the distance on the chromosome map. In either case, the frequency of $LS$ and $ls$ will be same, and $Ls$ and $lS$ will be same. This is because, in the Cis linkage, the former pair will constitute the parental linkage and the latter will be the recombinants, but both the gametes within the pair are equally likely to be formed. The same is the case for Trans linkage where the former pair acts as the recombinants.

According to the given table, frequency of $llss$ is $0.01$. $llss$ can only be formed by the combination of two $ls$ gametes. Therefore the probability of an $ls$ gamete will be $\sqrt{(0.01)}=0.1$.
The probability of getting an $ls$ gamete is $0.1$. Therefore, the probability of $LS$ is $0.1$, $lS$ is $0.4$ and $Ls$ is $0.4$.

As $Ls$ and $lS$ are more probable, they are more likely to be linked, and hence the genotype will be in Trans linkage. Then the frequency of recombinants($LS,ls$) is $0.2$. The chromosomal distance is $20 \text{ cM}$ or $20\text{ map units}$.

I am not sure whether my line of arguments are true. If you do find any flaw, please let me know.

• Yes, 20 is right. (My instructor gave us the answer today.) – biogirl Jan 11 '14 at 13:26