Stat 1040
Chapter 25 Solutions

  1. We probably have one gene-pair, with two variants, s and w. Smooth (s) is dominant, wrinkled (w) is recessive. If we cross two mixed hybrids, say s/w with s/w, we should get 25% s/s (smooth), 50% s/w or w/s (also smooth, so 75% smooth overall), and 25% w/w (wrinkled).

    If we count smooth second-generation hybrids, it is like drawing 7,324 times from a box containing 3 ones and 1 zero. The average of the box is 3/4, and the SD is 0.433, so the EV for the count is 7,324*.75=5,493 and the SE is sqrt{7,324}*.433=37. Mendel was off by only 5,493-5,474=19. The normal curve tells us that with an SE of 37, the chance of total error between -19 and 19 is about 38%.

    1. No, the child must have at least one A from that parent.
    2. Yes, if both parents are A/a or a/A, the child could get the a from each parent.
    3. No, both parents must contribute an a gene (they are both a/a), so the child must also be a/a.