Question

Help with both please!

1. Three genes of corn, R, D, and Y, lie on chromosome 9. The map of the three genes, using map units is as follows: R 10 20 In a testcross of a trihybrid plant (RD Y/r d y) arising from true-breeding RD YRD Y and r d y/r d y parents, how many Rd Y/r dy plants would be expected in a progeny of 1000 if there were no interference? If the coefficient of coincidence were 0.3, how would this change your answer? 2. In loppins (small, diploid invertebrates), chocolate-dependent problem-solving ability is an autosomal recessive trait. You have been hired to map the position of the gene controlling this phenotype (called "lu”) using two SSLP markers (called “A” and “B”). You first cross two pure-breeding loppins: lu+A repeats) B5 repeats) /lu+A/3 repeats) B15 repeats) x lu A6 repeats) B18 repeats) lu A6 repeats) B18 repeats) The Fı is then crossed back to the lu lu parent. Which of the following results would indicate that the lu gene is much closer to SSLP "A" than to SSLP "B"? A) lower proportion of A13 repeats) B/5 repeats) / A3 repeats) B/repeats) than Al3 repeats) B/8 repeats) / 413 repeats) B/5 repeats) individuals B) lower proportion of lu+A[3 repeats) ſu A6 repeats) than lu+A[3 repeats) ſu Al repeats) individuals C) lower proportion of lu+A13 repeats [lu A6 repeats) than lu+Alo repeats) ſu Alb repeats) individuals D) lower proportion of lu+A6 repeats) Tlu A3 repeats) than lu Bl8 repeatsu Bls repeats) individuals E) lower proportion of lu+A6 repeats llu A6 repeats than lu+B/8 repeats lu B/8 repeats individuals

Answer 1

Answer-

According to the given question-

Three genes lies on the chromosome number 9 with the distance between gene R and D is 10 m.u. and the distance between D and Y is 20 m.u.

R 10 D 20 Y

following informations and instruction given in the question-

number of progeny = 1000 and no interference is present then-

To get  a corn plant that is r d y/ r d y from selfing of R d y / r D Y , then both gametes must be derived from a crossover between R  and D.

so first we have to calculate the frequency of the r d y gamete is
frequency of the r d y gamete = 1/2 probability (Crossing Over R–D)  $\times$  probability( no Crossing Over D–Y)

= 1/2(0.10)(0.80)

frequency of the r d y gamete = 0.04

So the frequency of the homozygous corn = (0.04)² = 0. 0016.

now we have to cross between

R d y / r D Y  $\times$ r d y / r d y   

here the parental corn are those in which the crossing over does not occur.

parental corn  = probability(no Crossing Over R–D) $\times$ probability(no Crossing over D–Y)
= (0.90)(0.80)

parental corn= 0.72

each parents should be represented in equal proportions so

R d y = 0.36

r D Y = 0.36

now calculate the the frequency of the r d y  gamete -

frequency of the r d y  gamete = 1/2 probability (Crossing Over R–D) $\times$ probability ( no Crossing Over D–Y)

= 1/2(0.10)(0.80)

frequency of the r d y  gamete = 0.04

frequency of the r d y  gamete = frequency of the R D Y  gamete = 0.04.

now calculate the the frequency of the R d Y gamete-

= 1/2 probability (Crossing Over D–Y) $\times$ probability ( no Crossing Over R–D)

frequency of the R d Y gamete = 1/2(0.20)(0.90) = 0.09

frequency of the R d Y gamete = frequency of the r D y   gamete  = 0.09

now calculate the the frequency of the R D y gamete-  

1/2 probability (Crossing Over R–D) $\times$ probability ( Crossing Over D–Y)

frequency of the R D y gamete = 1/2(0.10)(0.20) = 0.01

frequency of the R D y gamete = frequency of the r d Y   gamete = 0.01

here the number of progeny is 1000

so number of

R d y = 1000  $\times$ 0.36 = 360

r D Y= 1000  $\times$ 0.36 = 360

r d y = 1000  $\times$ 0.04= 40

R D Y= 1000  $\times$ 0.04= 40

R d Y= 1000  $\times$ 0.09 = 90

r D y = 1000  $\times$ 0.09 = 90

R D y = 1000  $\times$ 0.01 = 10

r d Y= 1000  $\times$ 0.01 = 10

here the valve of coefficient of coincidence= 0.3

and we know that the

Interference (I) = 1 – coefficient of coincidence

so Interference (I) = 1- 0.3 = 0.7

Interference (I)= 0.7

now we have to calculate the value based on Interference 0.7 or 70 % interference.

coefficient of coincidence = observed Double Crossing Over /expected Double Crossing Over

Interference (I) = 1 – observed Double Crossing Over / expected Double Crossing Over

0.7 = 1 –  observed Double Crossing Over / ( 0.10 ) (0.20)

0.7 (0.10) (0.20)   = (0.10) (0.20)  –  observed Double Crossing Over

observed Double Crossing Over= 0.02 – 0.014= 0.006

The distance between R –D = 10% = 100% [probability (Crossing over R–D) + probability (Double Crossing Over)]

so the probability (Crossing over R–D) = 0.10 – 0.006= 0.094

The distance between the  D–Y = 20% = 100% [probability (Crossing over D–Y) + probability (Double Crossing Over)]

probability (Crossing over D–Y) = 0.20 – 0.006 = 0.194

Therefore the  probability of (parental corn ) = 1 – probability (Crossing over  R–D) – probability (Crossing over D–Y) – probability (observed Double Crossing Over)

= 1 – 0.094 – 0.194 – 0.006= 1 – 0.294 = 0.706
probability of (parental corn ) = 0.706

we know that

each parents should be represented in equal proportions so

R d y = 0.706 / 2 = 0. 353

r D Y = 0.706 / 2 = 0.353

so the number of progeny from the 1000

R d y = 1000  $\times$ 0.353 = 353

r D Y= 1000  $\times$ 0.353 = 353

r d y = 1000  $\times$ 0.094 / 2 = 47

R D Y= 1000  $\times$ 0.094 / 2 = 47

R d Y= 1000  $\times$ 0.194 / 2 = 97

r D y = 1000  $\times$ 0.194 / 2 = 97

R D y = 03

r d Y= 03

According to the Chegg guidelines i am only suppose to answer only first question or first four subpart of first question. please post question separately if you want the answer. thanks.