1. This pedigree is for a simple Mendelian trait
This pedigree is for a simple Mendelian trait. Determine whether the shaded trait is dominant or recessive, and determine the genotype for each person

2. The first step is to determine which form of the trait is recessive
The first step is to determine which form of the trait is recessive. The key to this step is to find some people on the pedigree that passed an allele to their children that they did not express. The only allele that you can have, but not express is a recessive allele.

3. Person #6 is different from both of her parents, so her parents gave her an allele that they did not express. She (#6) must be the recessive phenotype. So on this pedigree, all shaded people are recessive for the trait, and all unshaded people are dominant.

4. If we use the letter “A” for the dominant allele, and “a” for the recessive allele, #6 and #3 must both be “aa” aa aa

5. Both #1 and #2 express the dominant form of the trait, so each must have at least one “A” allele, but their daughter (#6) is “aa”, so they must each have an “a” allele. Both #1 and #2 must be “Aa” Aa Aa aa aa

6. #7 and #8 both express the dominant form of the trait
#7 and #8 both express the dominant form of the trait. They must each have at least one “A” allele. But their mother (#3) is “aa”, so they must have inherited an “a” allele from her. Both #7 and #8 are heterozygous – “Aa” Aa Aa aa Aa Aa aa

7. Both #7 and #8 had to have inherited the dominant “A” allele from their father (#4). Since all of his children inherited an “A” allele from him, it seems most likely that he would be “AA” Aa Aa
AA? aa Aa aa Aa

8. But, it is still possible for a person that is “Aa” to pass on the “A” allele twice in a row. Remember that these are PROBABILITIES. The chances of a heterozygous person passing on the same allele twice is no more uncommon than flipping heads twice. #4’s genotype could be “AA” or “Aa”, you can’t be 100% sure. Aa Aa aa
A__ aa Aa Aa

9. #5 came from a family with both parents heterozygous – “Aa”
#5 came from a family with both parents heterozygous – “Aa”. A cross between two “Aa” people should produce offspring in a ratio of 1 AA:2 Aa:1 aa. Both “AA” and “Aa” are possible for #5 aa
A__ Aa Aa
A__ aa Aa Aa

10. We already determined that #6 was “aa” and #7 was “Aa”
We already determined that #6 was “aa” and #7 was “Aa”. A punnett square would show that we can predict that ½ of their children should be “Aa” (expressing the dominant phenotype) and ½ should be “aa” (expressing the recessive phenotype) aa
A__ Aa Aa aa Aa
A__ Aa

11. Aa Aa aa A__ aa Aa Aa A__ Aa Aa
#9 and #10 both express the dominant phenotype, and inherited a recessive allele from their mother. Both #9 and #10 are heterozygous – “Aa” Aa Aa aa
A__ aa Aa Aa
A__ Aa Aa

12. For this pedigree of ABO blood type: Determine the genotype for each person. Determine the probability of type A, B, AB and O for #11 and #12

13. This pedigree is for a sex-linked (X) recessive trait
This pedigree is for a sex-linked (X) recessive trait. Determine the sex chromosome combination (XX or XY) and the genotype for each person.