Assignment #5 Answers

1-a).

  Prob.(c.o. event) = 2p
  Prob.(Ab gamete given a c.o. event has occurred)= 1/4

  P(Ab gamete) = 2p(1/4) = p/2.

1-b).

  Prob(Ab gamete) = p/2. When p = 1/4, then:
  P(Ab) = p/2 = (1/4)/2 = 1/8.

  P(Aabb) = 1/8(1) = P(Ab)P(ab) = 1/8.

1-c).

  Prob. aaBb gamete = p/2. When p = 1/2, then:
  P(aB) = p/2 = (1/2)/2 = 1/4.

  P(aaBb) = P(aB)P(ab) = 1/4(1) = 1/4.
  Same probability as for independent assortment.

2-a). 1-p 2 1-p 2 p 2 p 2 AB ab Ab aB 1-p 4 AB 1- p 2 ab p 2 Ab p 2 aB               X           X   X P(aaB_) = P(aaBB) + P(aaBb) + P(aaBb) P(aaB_) = (p/2)2 + ([1-p]/2)(p/2) + ([1-p]/2)(p/2) P(aaB_) = p2/4 + (p-p2)/4 + (p-p2)/4 P(aaB_) = 1/4[p2 + p - p2 + p - p2] P(aaB_) = 1/4[2p - p2] = p(2-p)/4

2-b).

  P(aaB_) progeny when p = 0.2 is:
  p(2-p)/4 = 0.2(1.8)/4 = 0.09

2-c).

  P(aaB_ progeny when p = 0.5 is:
  p(2-p)/4 = 0.5(1.5)/4 = 0.1875 = 3/16

  We expect 3/16 aaB_ progeny when p = 0.5 or no linkage

2-d).

  P(aaBb) = p/2[(1-p)/2)] + p/2[(1-p)/2]
  P(aaBb) = (p-p²)/4 + (p-p²)/4
  P(aaBb) = 1/4(2p-2p²) = 1/2(p-p²)

  When p = 0.2, 1/2(p-p²) = 1/2(0.2 - 0.04) = 0.08

3-a).

  Coupling data due to an excess of aabb progeny.

3-b).

  st/ru = [90(90)]/[660(160)] = 8100/105600 = 0.0767

  From the table, when the ratio is 0.0767, p =0.2.