GS Activity: HH Thumb (Teacher Version)

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.

TT Tt tt T- t- The student should answer tt.

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.

2. What must Erica's and Bob's genotypes be? Circle the right answer. TT Tt tt T- t-

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?

TT Tt tt T- t-

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.

4. What is Jill's genotype for hitchhiker's thumb likely to be? (Hint: none of their children has the trait.)
TT Tt tt T- t-

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.

5. If Jack and Jill have 12 more children, how many do you think will have hitchhiker's thumb? (circle one) None
3 or 4
About half
All of them The answer is none of them.

Since Jack is homozygous recessive and Jill is homozygous dominant all of their children will be normal for the trait.

6. If Erica and Bob have 12 more children, how many do you think will have hitchhiker's thumb? (circle one) None
3 or 4
About half
All of them The answer is three or four.

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.

7. OK - now try it on GenScopeô. Open the file HHThumb.gs and check out the chromosomes. Also, give Jack and Jill and Erica and Bob some more kids and see if you made the right predictions.

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