Homework Assignment #7

Answers

1:a) p = 1/2(1 - e^-2d) = 1/2(1- e^-2) = 1/2(1) - 1/2(1/e²)
1:a) p = 0.5 - 0.068
1:a) p = 0.432

1:b) p = distance in cM only when loci are located less
1:b) than 10cM apart. Otherwise, distance is
1:b) underestimated by observed percent recombination
1:b) due to undetected double crossovers.

2:a) x = 9/7, y = 13/3, X = 3.84

2:a)    

                                      n > 49

2:b)

If 35 or more plants out of 49 are of the dominant phenotype, then the most likely segregation ratio is 13:3.

X² = (35 - 39.81)² + (14 - 9.19)² = 0.581 + 2.518 = 3.09
           39.81              9.19

                                            (accept 13:3 ratio)

X² = (34 - 27.56)² + (15 - 21.44)² = 1.505 + 1.934 = 3.44
            27.56             21.44

                                           (accept 9:7 ratio)

X = (34 - 39.81)² + (15 - 9.19)² = 0.848 + 3.673 = 4.52
           39.81              9.19

                                           (reject 13:3 ratio)

3. P = 0.01, q =15/16, find n

3. n =  log(p)  =      -2     = 71.35
3.        log(q)    -0.02803

At least 72 plants must be grown.

4. For repulsion phase linkage with F₂ data

i =    2(1 + 2p²)  
    (1 - p²) (2 + p²)

For coupling data we can use the same formula, with 1-p substituted for p.