Name ________________________ Date ___________________
Hitchhiker's Thumb Activity: Understanding how a trait can skip generations
GenScope File: HHThumb.gs
If you need to reconstruct this file, Jill is TT, homozygous dominant. Jack is tt, homozygous recessive. Bob is heterozygous Tt.
The file opens with the pedigree window
showing. The pedigree shows hitchhiker's thumb for Jack, Jill, their children
and their grandchildren. Hitchhiker's thumb is an autosomal, recessive
condition (tt = hitchhiker's thumb, T = non-hitchhiker's thumb).
How come none of Jack and Jill's kids
have hitchhiker's thumb but one of their grandchildren does???
Here are some questions that will help you get started:
1. What is Jim's genotype for hitchhiker's thumb? Circle the right answer.
All of these questions should be answered off the computer. The first question may be answered by inspection of the pedigree itself. It presupposes that the student understands that in order for a recessive trait to "show", two recessive alleles must be present.
Both of them are Tt.
By reasoning backwards, the student knows that each of these individuals must carry the allele t, since one of their kids, Jim, has the recessive trait (tt). But, since neither one of them has hitchhiker's thumb, they must each also have the dominant allele T.
3. What is Jackís genotype for hitchhiker's thumb?
Jack is also tt and has the recessive trait.
Again, by inspection of the pedigree, the student can see that Jack has the trait, and therefore is tt.
The student should answer TT. Because Jack has the trait and none of the children do, the student will probably guess that Mom is homozygous dominant for normal thumbs.
Since none of the children has the trait, but Jack does, the student will probably have figured out that Jill must be homozygous dominant, TT. Any alleles that she contributes to the children will overshadow the recessive alleles from Jack. They should also be aware of the fact that, despite the appearance of normality, each of the children of this marriage is a carrier of the allele. The next generation will have some children with the trait, if the new spouses also carry the trait.
Since Jack is homozygous recessive and Jill is homozygous dominant all of their children will be normal for the trait.
Since both Erica and Bob are heterozygous for the trait (Tt), one fourth of their children will have hitchhiker's thumb. By doing a Punnett square, the student will determine that 1/4 will be homozygous dominant (TT), 1/2 will be heterozygous (Tt) and 1/4 recessive (tt). Thus, 3/4 of the children will look normal, that is, their phenotype will be straight thumbs and 1/4 will have hitchhiker's thumb. This problem is another attempt to get the students to think about probability. So much of genetics is based on statistical analysis of data that the students have to get used to thinking about general trends, not absolute numbers.
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