Question

In a diploid plant species, and F1 with the genotype Gg Ll Tt is test-crossed to a pure-breeding recessive plant with the genotype gg ll tt. The offspring genotypes are listed in the table below:
Genotype Number



Gg Ll Tt 653

Gg Ll tt 5

Gg ll Tt 78

Gg ll tt 116

gg Ll tt 121

gg Ll tt 75

gg ll Tt 4

gg ll tt 548



Total= 1600



1) What is the order of these three linked genes?

A) L-G-T

B) G-T-L

C) G-L-T



2) Calculate the recombination fraction between G and T pair of a gene. Enter answer to three decimal places (example 0.235)



3) Calculate the recombination fraction between T and L pair of a gene.



4) Calculate the recombination fraction between G and L pair of a gene



5) What is the interference value for this data set?

Answer 1

Answer: 1) G-T-L; 2) RF = 15.375mu; 3) RF = 10.125 mu; 4)RF = 25.5 mu; 5) Interference = 63.873%

Explanation: 1) To define the order of genes, we have to identify no-crossovers, single crossovers and double crossovers among the genotype numbers. Because there is a higher probability of not having crossover, the genotype with higher number will be no crossover. On the other hand, the offsprings with the lowest numbers will be double crossover. So,

• no crossover: GgLlTt and gglltt;
• double crossover: GgLltt and ggllTt;

In order for this last two to be a double crossover, comparing to the no crossover, the allele T must be in the center. Thus, the order is G-T-L.

The others are single crossovers.

2) Recombination frequency (or fraction) is a percentage of recombination an allele can have. It can also determine the distance between genes.

The recombination fraction between G and T is: RF = ($\frac{P+R}{T}$) . 100

where P is the double crossover between G and T and R is the single crossover between the same genes;

Single crossover G-T = 121+116

RF₁ = $\frac{9+237}{1600}$ . 100 = 15.375 mu (map unit)

3) RF for T-L:

In this case, P is the double crossover between T and L and R is the single crossover between them:

Single crossover T-L = 78 + 75 = 153

RF₂ = $\frac{153+9}{1600}$ .100 = 10.125 mu

Note: T represents the total of individuals from the experiment, in this case T = 1600.

4) As the order of the genes is G-T-L and RF also represents the distance between genes, the recombination fraction between G and L is

RF₃ = 15.375 + 10.125 = 25.5 mu.

5) Interference value = 1 - coefficient of coincidence

Coefficient of coincidence = $\frac{observed double crossover}{expecteddoublecrossovers}$ . 100

• Observed double crossover = $\frac{9}{1600}$·100 = 0.5625
• Expected double crossover = product of 2 single recombination value

expected double crossover = $\frac{15.375.10.125}{100}$ = 1.557

coefficient of coincidence = $\frac{0.5625}{1.557}$ . 100 = 36.127%

Interference value = 1 - 0.36127 = 0.63873

Interference value = 63.873%