Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Three point linkage analysis Genetic analysis has shown that the recessive genes

ID: 218171 • Letter: T

Question

Three point linkage analysis

Genetic analysis has shown that the recessive genes an (Anther ear), br (brachytic) and f (fine stripe) are all found on chromosome #1 of maize. When a plant that is heterozygous for each of these markers is testcrossed with a homozygous recessive plant, the following results are obtained:

Testcross Progeny Numbers:

wild type- 3 (+++)

fine- 48 (++f)

brachytic- 400 (+br+)

brachytic fine – 42 (+brf)                          Total Offsprings = 1000

anther -45 (an++)

anther fine -402 (an+f)

anther brachytic -56 (anbr+)

anther brachytic fine- 4 (anbrf)

Calculate recombination frequencies between each of these three pairs of genes.

Draw a genetic map for the location of these 3 genes on chromosome #1 of maize. Be sure to show the map distances between each loci.

Calculate the interference.

Explanation / Answer

Answer:

Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.

Hence, the parental (non-recombinant) genotypes is +br+ / an+f.

1).

If single crossover occurs between + & br.

Normal combination: +br/an+

After crossover: ++/an br

++ progeny= 3+48=51

an br= 56+4=60

Total this progeny = 111

The recombination frequency between +&br = (number of recombinants/Total progeny) 100

RF = (111/1000)100 = 11.1%

2).

If single crossover occurs between br&+.

Normal combination: br+/+f

After crossover: brf/++

br f progeny= 42+4=46

++ progeny = 3+45=48

Total this progeny = 94

The recombination frequency between br&+ = (number of recombinants/Total progeny) 100

RF = (94/1000)100 = 9.4%

3).

If single crossover occurs between +&+.

Normal combination: ++/an f

After crossover: +f/an +

+f progeny= 48+42=90

an+ progeny = 45+56=101

Total this progeny = 191

The recombination frequency between +&+ = (number of recombinants/Total progeny) 100

RF = (191/1000)100 = 19.1%

Recombination frequency (%) = Distance between the genes (cM)

an----------11.1cM-------br-----9.4cM------f

Expected double crossover frequency = (RF between w & m) * (RF between m & f)

= 0.111 * 0.094 = 0.01

The observed double crossover frequency = 3+4 / 1000 = 0.007

Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency

= 0.007 / 0.01

= 0.7                    

Interference = 1-COC

= 1-0.7 = 0.3

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote