Question

In corn the glossy trait (AA) gives glossy leaves and the ramosa trait (BB) determines branching of ears, a test cross produced the following results:

Normal leaved and normal branches                   395

Glossy leaved and ramosa branching 382

Normal leaved with ramosa branching 223

Glossy leaved with normal branching 247

Total of 1247 offspring.

Are the glossy and ramosa genes linked ? Test your hypothesis. If they are linked calculate distance between them.

Answer 1

Solution

Are the glossy and ramosa genes linked?

Null hypothesis: genes are not linked.

Testing of null hypothesis: If genes are not linked they must follow the law of independent assortment according to which test cross (F1 generation offspring is crossed with a homozygous recessive individual) produces a ratio of 1:1:1:1 in F2 generation.

Glossy leaves with ramosa branching (AaBb) = 382

glossy leaves with normal branching (Aabb) = 247

normal leaves with ramosa branching (aaBb) = 223

normal leaves with normal branching (aabb) = 395

According to null hypothesis if genes are not linked they should be 1:1:1:1

therefore 1x +1x+ 1x+ 1x= 1247

4x=1247

x= 311.75 (expected value)

chi square= (observed- expected)²/ expected

= (382-311.75)²/311.75 + (247-311.75)²/311.75+ (223-311.75)²/ 311.75+ (395-311.75)²/ 311.75

= 75.6

degree of freedom (df) = number of genes minus 1 so it is 2-1= 1

So when we look at chi square distribution table (https://people.smp.uq.edu.au/YoniNazarathy/stat_models_B_course_spring_07/distributions/chisqtab.pdf) for degree of freedom 1, 75.6 is a very high value so null hypothesis is rejected. therefore It is certain that two genes are linked together.

Genetic distance between two genes

To find the distance we need to find the recombination frequency which is calculated by total number of recombinants/ total number of offsprings*100

parental are always the ones which are in greater number therefore

Glossy leaves with ramosa branching (AaBb) = 382 (parental)

glossy leaves with normal branching (Aabb) = 247 (recombinant)

normal leaves with ramosa branching (aaBb) = 223 (recombinant)

normal leaves with normal branching (aabb) = 395 (parental)

Therefore according to formula= (247+223)/1247*100= 37.69

therefore it can be said that glossy leaves (AA) and ramosa branching (BB) are linked and 37.69 cM (centimorgan) apart.