The information in the table shows the possible genotypes resulting from the mating of two heterozygotes aadd. the allele a codes for normal pigmentation, a for albinism, d for dimpled chin, and d for non-dimpled chin. what fraction of offspring would be expected to be albino with a non-dimpled chin? a. 1/2 b. 1/4 c. 3/4 d. 2/8 e. 1/16

Admitting that the "a" is a capital A for normal pigmentation and "d" is a capital D for dimpled chin, meaning that these are the dominant traits, the fraction expected to be albino with a non-dimpled chin is of 1/16.
When two heterozygous are crossed and two characteristics are being analysed, the offspring quantity that will possess the two recessive traits can be represented by 1/16. This is easily confirmed when a Punnett square is made. Considering that both parents were heterozygous, on both sides of the crossing in the Punnett square, you would have the following alleles' combination: AD, Ad, aD, and ad. The offspring that would be homozygous recessive (aadd) would correspond to only 1/16. 

  a. chromosomes are duplicated before cell division so that each new daughter cell has a complete set

Need the answers if its a multiple choice question