4. Three genes in the lab bug, an insect specially bred for genetics problems, a

Question

4. Three genes in the lab bug, an insect specially bred for genetics problems, are linked as indicated in the following gene map:
-----------------A/a------10 cM-----B/b------------------25 cM-------------------C/c------ Assume a Coincidence Coefficient of 0.80. Parental AAbbCC and aaBBcc lab bugs are mated to yield an AaBbCc F1 generation. The F1 individuals are then test-crossed with aabbcc lab bugs.

Out of 1,000 test-cross offspring (the lab bug is very prolific),
(a) predict the number of test-cross offspring exhibiting noncrossover, single crossover, and double crossover genotypes / phenotypes, and
(b) give the specific allelic arrangement in those offspring. (HINT: Look at the parental generation to see which alleles are linked together.)

Explanation / Answer

A -Given that Coefficient of coincidence is 0.80 so

The observed frequency of double recombinant = 0.80 *expected frequency of double recombinant

Map distance between a and b is 10cM

Map distance between b and c is 25cM

expected frequency of double recombinant is given by

Map distance between color and body type /100 * map distance body type and disposition/100

10/100 * 25/100 = 0.025

Observed frequency of double recombinant we need to find out

Coefficient of coincidence = the observed frequency of double recombinant/ expected frequency of double recombinant

0.8 = X /0.025

X = 0.025*0.80 =0.02

The observed frequency of double recombinant = sum of the frequency of both the double recombination

Number of progeny = 1000

so we will take the double recombinant frequency as half of the total double recombinants that we got.

Number of progeny of A, C and C = 0.02*1000/2 =10

Number of progeny of a, b and c = 0.02*1000/2 =10

So the double crossover progenies are ABC and abc

The distance between a+ and b+ is 10 cM.

A total number of recombinant/total number of progeny = frequency of recombinant.

The frequency of recombination between A and B will be a sum of frequencies of A, B and c, a, b, and C, a, b c and A B and C

Linkage distance = total number of recombinant * 100 / total number of progeny

10 = total number of recombinant * 100 /1000

10000/100 = total number of recombinants

a, b c and A B and C = 10+ 10 = 20

So the number of recombinants of  A, B and c, a, b, and C, will be 100-20= 80

So the number of progeny having A, B and c, phenotype will be 54/2 = 40

and a, b, and C phenotype will be 40

The distance between b+ and c+ is 25 cM.

The frequency of recombination between a+ and b+ will be a sum of frequencies of A,b and c, a, b, and C, a, b c and A B and C

Linkage distance = total number of recombinant * 100 / total number of progeny

25= total number of recombinant * 100 *1000

25000/100 = total number of recombinants

a, b c and A B and C = 10+ 10 = 20

A,b and c, a, b, and C = 250-20 = 230

the number of progeny having Abc phenotype will be 230/2 = 115

and a, b, and C genotype will be 115

Given that total number of progeny = 1000

number of progeny having a+b+c+ and a, b, c phenotype will be

1000-(115 +115+40+40+10+10)

1000-154 =

670

number of progney having AbC phenotype = 670/2 = 335

number of progney having aBc phenotype = 670/2 = 335

B- Allelic arrangement -

AbC 335

aBc 335

ABC 10

abc 10

ABc 40

abC 40

aBC 115

Abc 115

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.