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Genetics Problems- some help

  1. Apr 9, 2008 #1
    1. How could you find out whether blue eyes or brown eyes is dominant? 1. My thoughts: Would you just cross a blue-eyed individual and a brown-eyed indivdual and observe what's prevalent in their offspring?

    2. Gary has 6 fingers on each hand & 6 toes on each foot. His wife Sandy and daughter Stephanie are normal. Extra digits is a dominant trait. What fraction of their children would be expected to have extra digits? My thougts: Let's say R= dominant allele, r= recessive allele. Since Gary has extra digits, I know his genotype is rr, but how do I do if the mother is RR or Rr?

    3. Phenylketonuria (PKU) is a genetic defect harming the brain. A normal couple have a baby with PKU. What is the probability that their next child will be affected? What is the probability that if they have 2 more children, neither will be affected?
     
  2. jcsd
  3. Apr 9, 2008 #2
    1. You could do that, but I doubt you'd get many volunteers. You could also look at the offspring of a blue-eyed and brown-eyed couple.

    2. You're wrong here, rr would mean that Gary has recessive alleles for the gene and would be normal, but having extra digits means that Gary has at least one dominant allele for the trait since having extra digits is a dominant trait.

    3. Again, do a Punnett square. See how the offspring differ if one parent is heterozygous and one parent is homozygous dominant, or if both parents are heterozygous. Only one square will allow you to have a child who has PKU where neither parent has it.
     
  4. Apr 9, 2008 #3
    Thank you for the help. Below is my work. Could someone please double check?

    2. Punnett square is Rr X rr. 50% of their children would be expected to have extra digits.
    3. Let alleles A=normal, a= has PKU. Punnett square is Aa X Aa.

    Probability that their next child will be affected= 25%. Probability that if they have 2 more children neither will be affected= 75%
     
  5. Apr 9, 2008 #4
    The chance that one child will not be affected is 75%. The chance that another child will not be affected is 75%. (These are independent events).

    The chance that BOTH children will not be affected is...
     
  6. Apr 10, 2008 #5
    0.75 X 0.75 = 0.5625 ???
     
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