First determine the genotypes of the parents. We know that the woman has normal pigment which means she must have at least one A. Her father is albino and because the albino allele is recessive, his genotype is aa. What does this make her skin genotype?



We also know that the woman cannot taste PTC. Because the ability to taste PTC is dominant, what does this make her genotype for the tasting trait?



Putting both traits together, we see that the woman's overall genotype is Aatt.





Now, what about her husband? You have been told that he is homozygous for normal pigment. What is his genotype for skin color?



He is a taster and so must have at least one T. However, we also know that his mother can not taste PTC so she must be homozygous recessive. With this information, what is his genotype for taste?



Putting both traits together, you see that his overall genotype is AATt.





Now that you have the genotypes of the parents, perform a punnett square to determine what the F1 offspring would be (remember to place the gametes of one parent across the top and the gametes of the other parent down the left side. It does not matter which parent's gametes are placed across the top and which are placed down the side.):





Your punnett square should look similar to the one below. If it does not, review the material on gamete formation and how to set up the punnett square and try again.

At At at at
AT AATt AATt AaTt AaTt
AT AATt AATt AaTt AaTt
At AAtt AAtt Aatt Aatt
At AAtt AAtt Aatt Aatt





Now return to the question page and select the correct answers.






Below is the punnett square produced from this cross. Simply count the number of offspring with the aa genotype necessary for the albino trait and the T allele necessary to be a taster.

At At at at
AT AATt AATt AaTt AaTt
AT AATt AATt AaTt AaTt
At AAtt AAtt Aatt Aatt
At AAtt AAtt Aatt Aatt





Now return to the question page and select the correct answers.