Analyzing Human Genetics Using Pedigree Charts - Free Printable

This is a genetics problem involving a pedigree chart for the trait of polydactyly, which is stated to be dominant. The task is to fill out the pedigree, meaning we need to assign genotypes (DD, Dd, or dd) to each individual based on their phenotype (shaded = affected, unshaded = unaffected) and the rules of Mendelian inheritance.

Let's break this down step by step.

Step 1: Understand the Trait and Notation
- Polydactyly is dominant. Let's use 'D' for the dominant allele (causes polydactyly) and 'd' for the recessive allele (normal number of digits).
- An individual with genotype DD or Dd will have polydactyly (shaded symbol).
- An individual with genotype dd will have the normal phenotype (unshaded symbol).

Step 2: Analyze the Pedigree Generation by Generation

We'll start from the top (Generation I) and work our way down.

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Generation I:
- Individual I-1 (male, unshaded): Must be dd (since he is unaffected, he must have two recessive alleles).
- Individual I-2 (female, shaded): Has polydactyly, so her genotype is either DD or Dd.

They have three children in Generation II:
- II-1 (female, unshaded): Unaffected, so she must be dd.
- II-2 (male, shaded): Affected, so he is D_ (either DD or Dd).
- II-3 (male, shaded): Affected, so he is D_ (either DD or Dd).

Since I-1 is dd, he can only pass on a 'd' allele to his children. Therefore, any child who is affected (has polydactyly) must have received a 'D' allele from I-2.

So:
- II-2 (affected) must be Dd (received 'd' from father, 'D' from mother).
- II-3 (affected) must be Dd (received 'd' from father, 'D' from mother).
- II-1 (unaffected) is dd (received 'd' from both parents).

This also tells us that I-2 (the mother) must be Dd, because if she were DD, all her children would be Dd and thus affected. But she has an unaffected child (II-1), so she must carry a 'd' allele.

Conclusion for Generation I:
- I-1: dd
- I-2: Dd

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Generation II:
We already determined:
- II-1: dd
- II-2: Dd
- II-3: Dd

Now let's look at their spouses and children.

II-2 (Dd) marries II-4 (unshaded female):
- II-4 is unshaded, so she is dd.
- Their child is III-1 (unshaded female). Since II-2 is Dd and II-4 is dd, the possible genotypes for their child are Dd or dd. III-1 is unshaded, so she must be dd.

II-3 (Dd) marries II-5 (unshaded female):
- II-5 is unshaded, so she is dd.
- They have three children: III-2 (shaded female), III-3 (shaded male), and III-4 (shaded male).
- Since II-3 is Dd and II-5 is dd, each child has a 50% chance of being Dd (affected) or dd (unaffected).
- All three children are affected, so they must all be Dd (they received 'd' from mother and 'D' from father).

Conclusion for Generation II:
- II-1: dd
- II-2: Dd
- II-3: Dd
- II-4: dd
- II-5: dd

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Generation III:
- III-1 (unshaded): dd (as determined above).
- III-2 (shaded): Dd (as determined above).
- III-3 (shaded): Dd (as determined above).
- III-4 (shaded): Dd (as determined above).

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Final Answer:

Here are the genotypes for each individual in the pedigree:

Generation I:
- I-1: dd
- I-2: Dd

Generation II:
- II-1: dd
- II-2: Dd
- II-3: Dd
- II-4: dd
- II-5: dd

Generation III:
- III-1: dd
- III-2: Dd
- III-3: Dd
- III-4: Dd

This assignment is consistent with the dominant inheritance pattern of polydactyly and the observed phenotypes in the pedigree.