LECTURE 4: LINKAGE AND RECOMBINATION

Thomas Hunt Morgan and his colleagues discovered the variation that arises due to sexual reproduction.  Morgan worked on Drosophila.

Why Drosophila to study linkage and recombination?

• Small genome size with only 4 pairs of chromosomes
• Genome size = 1.2 x 10^8 bp (approx 14,000 genes)
• Short life cycle (2 weeks)
• Controlled cross possible
• Many offspring from one cross
• Rich source of mutation

There are some simple notations that need to be considered while studying linkage and recombination:

• Letters describe mutant phenotype, for example (sw) for short wing
• Plus sign (+) is used to denote wild-type allele (sw+)
• Wild type alleles are more common in nature
• Lower case letter is used to show recessive wild type sw+ allele
• Upper case is used to denote mutant allele dominant to wild type allele.

Unlinked versus Linked Genetic Inheritance:

A cross has been shown above between a wild type drosophila (female) and a short winged male. Since wild type is dominant, the F1 generation shows a wild - type phenotype. 

-In Mendelian inheritance, we discussed about test cross, the monohybrid test cross ratio is 1:1 and the dihybrid test cross ratio is 1 : 1 : 1 : 1

-Test cross is also used to find linked genes.

- In the following cross, we have considered that the genes are unlinked

-Let's see the expected and observed value of the following test cross.

-- According to Mendel's test cross experiment, the expected ratio should be 250 : 250 : 250 : 250, however, the observed ratio was quite different (415 : 410 : 90 : 85). So the observed data does not fit the expected data and this means that the genes are not unlinked. 

=> Now take another situation where we consider that the genes are linked. 

-The above picture shows the types of gametes being formed from b/sw+ b/sw

-The figure above shows the test cross, however, genes are linked.

- The figure above shows that the genes are totally linked (means closed enough that no crossing over is possible) and therefore no recombination is done. Therefore, we can only expect parental genotypes.

The above picture shows that if F1 is crossed with a recessive parent (test cross), the ratio would differ depending on whether alleles are unlinked, partially linked or fully linked.

The above picture shows the expected and observed number of progeny formed after a test cross. The picture also shows how to calculate the recombination frequency. The recombinant frequency calculated is the distance between two genes. 

Well, the distance between two genes have already been given, which is 20cM. It means that the recombinant frequency is 20%. Since there are two recombinants so each will have 10% frequency. And each parent will have (100-20 =80/2 = 40%). 

So, AB/ab = 40%

Ab/ab = 10%

aB/ab = 10%

ab/ab = 40%