In: Biology
Explain and demonstrate how a dihybrid test cross can be used to determine if autosomal linkage exists.
A dihybrid cross is a cross which involves two different genes. It can be used to determine whether the genes are showing independent assortment or are linked to each other.
If two organisms heterozygote for both the genes are crossed with each other, and if the number of progeny is obtained is in the ratio of 9:3:3:1, then the genes are showing independent assortment. It will follow mendelian inheritance.
But if the ratio is deviating from this value, then the genes are said to be linked and will follow non mendelian inheritance. In this case, the ratio will depend upon the distance between the genes.
Example,
1. If the genes are showing independent assortment
AaBb × AaBb
Ratio of gametes : AB, ab, Ab, aB = 1:1:1:1
Gametes | AB | Ab | aB | ab |
AB | AABB | AABb | AaBB | AaBb |
Ab | AABb | AAbb | AaBb | Aabb |
aB | AaBB | AaBb | aaBB | aaBb |
ab | AaBb | Aabb | aaBb | aabb |
Phenotypic ratio = 9 (A_B_) : 3 (A_bb) : 3 (aaB_) : 1 (aabb)
2. If the genes are linked and are 10 cM apart
AB/ab × AB/ab
Number of recombinant gametes = distance between the two gene
Number of parental gametes = 100 - number of Recombinant gametes
Gametes :
Parental gametes (90%) = AB (45%), ab (45%)
Recombinant gametes (10%) = Ab (5%), aB (5%)
Gametes | AB 45% | ab 45% | Ab 5% | aB 5% |
AB 45% | AABB 20.25 | AaBb 20.25 | AABb 2.25 | AaBb 2.25 |
ab 45% | AaBb 20.25 | aabb 20.25 | Aabb 2.25 | aaBb 2.25 |
Ab 5% | AABb 2.25 | Aabb 2.25 | AAbb 0.25 | AaBb 0.25 |
aB 5% | AaBB 2.25 | aaBb 2.25 | AaBb 0.25 | aaBB 0.25 |
Independent assortment means that the genes are located on two different chromosomes are located very far from each other on the same chromosome, more than 50 cM away. Linkage means that the genes are very close to each other on the same chromosome and the distance between them is either equal to or less than 50 cM.
Please rate high.