In: Biology
explain and demonstrate how a dihybrid test cross can be used to determine if autosomal linkage exists
Answer: A dihybrid cross is a cross between parents differing in two pairs of contrasting character.
A test cross is when the parent of an unknown genotype is crossed with a recessive genotype.
Suppose a dihybrid test cross is made as follows
P1 genotype RrYy P2 genotype rryy
P1 phenotype round and yellow seeds P2 phenotype wrinkled and green seeds
Gamates RY,Ry,rY,ry ry
Punnett square
gamates |
RY |
Ry |
rY |
ry |
ry |
RrYy Round,yellow parental |
Rryy Round, green recombinant |
rrYy wrinkled,Yellow recombinant |
ryry green, wrinkled parental |
If the genes are not linked then they could be very far from each other on the same chromosome or they are present on different chromosomes.
Hence the phenotypic ratio obtained in this case will be 1:1:1:1
In a test cross with a F1 of dihybrid cross the ratio obtained will be always 1:1:1:1 the genes for the two characters are showing independent assortment.
If the genes are linked then the ratio obtained would be different than the above ideal test cross ratio. The ratio obtained in case of linkage will be showing more number of parental phenotypes and correspondingly there will be less of recombinants. Suppose in above case if the ratio obtained is 5:1:1:5 that means the genes for the two traits are linked and they are inherited together. In the gamate formation the alleles RY and ry will tend to remain together.
Resulting in more number of parental combinations.
If the distance between the genes is very less then finding the recombinants proportion of such genes would be very very less. Hence we can say that the distance between the genes determine their recombination frequency,more the distance more is the recombination frequency.Less the distance (more linkage) less is the possibility of finding recombinants
Therefore in above case if the genes are linked then the parental combinations will increase and the recombinants will decrease.