Question

23. R. A. Emerson crossed two different pure breeding lines type Fl that was heterozygous for three alleles that determine recessive phenotypes. Terent pure-breeding lines of corn and obtained a phenotypically wild- anther, br, brachytic; and f, fine. He testcrossed the F1 with a tester that was homoly mine recessive phenotypes: an determines the three genes and obtained these progeny phenoty with a tester that was homozygous recessive for ained these progeny phenotypes: 355 anther: 339 brachytic, fine: 88 completely W wild type: 55 anther y pe, Ssanther, brachytic fine: 21 fine. 17 anther brachytic: 2 brachytic: 2 anther, fine. a. Who were the genotypes of the parental lines? b. Draw a linkage map for the three genes (include map distances). c. Calculate the interference value of coincidence?

Answer 1

Answer:

a): Genotype of original parents

B):

an 1 + + + + + + an f f br br +'

• an + +355
• br f + 339   
• an br f 88
• ++ + 55
• br + + 2
• an f + 2
• + + f 21
• an br + 17

$\rightarrow$Total population = 879

• In any double cross, the maximum number of progeny is the progenies which have the same genotype as that of the parents and the progeny which have the least number is the progeny which are the result of a double crossover.
• So looking at the progenies we can easily conclude that 'an + +' and 'br f +' are the original parent type and br + + and an f + are the double recombinant.
• This also gives us the idea about the gene which is present in the middle as only the middle gene is transferred to the other chromosome during the double crossover. So the middle gene is f.

$\rightarrow$ The distance between f and an.

• If we want to find the distance between the an and f then we need to find out the crossover frequency and divide it by the total number of progeny
• can br f 88
• ++ + 55
• br + + 2
• an f + 2

$\rightarrow$Linkage distance = total number of recombinant * 100 / total number of progeny

• 147 * 100/ 879
• = 16.79

$\rightarrow$So, map distance between f and an is 16.79 map units or 16.79cM

• The distance between f and br
• If we want to find the distance between the f and br then we need to find out the crossover frequency and divide it by the total number of progeny
• br + + 2
• an f + 2
• + + f 21
• an br + 17

$\rightarrow$Linkage distance = total number of recombinant * 100 / total number of progeny

• 42*100/ 879
• = 4.77

$\rightarrow$So, map distance between f and br is 4.77 map units or 4.77cM

16.79 f 4.77 bn

C):

$\rightarrow$Since the flies have undergone the double recombination so we can expect the recombination interference

• In order to find out the recombination interference we need to find out the coefficient of coincidence (c.o.c.)

$\rightarrow$Coefficient of coincidence is given by

• the observed frequency of double recombinant/ expected frequency of double recombinant

$\rightarrow$expected frequency of double recombinant is given by

• Map distance between an and f /100 * map distance between f and br /100 = 16.79/100*4.77/100 = 0.008
• Observed frequency of double recombinant = 2+2 /879
• 0.0045
• Coefficient of coincidence = 0.0045/0.008= 0.56
• Recombination interference is given by
• 1- C.o.C
• 1-0.56 =0.44

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