1. Name ______________________

2. One flower is heterozygous red (Rr) and it is crossed with a homozygous white (rr) plant. Use a Punnett square to determine the probability of one of their offspring having a red color. R r r r Rrrr Rr rr 50% probability of being red

3. F.O.I.L How to list all possible combinations of two alleles. First-Outside-Inside-Last RrYy First: ______ Outside: ______ Inside: ______ Last: ______ RY Ry rY ry

4. 1. In mice, black fur (B) is dominant over brown (b) and short tails (S) are dominant over long (s). Write the genotype for a heterozygous black, short-tailed mouse: ______________ Cross two of these individuals. From the Punnett Square, describe the phenotype ratios of the offspring. 9 Black Short; 3 Black Long; 3 Brown Short; 1 Brown Long BS Bs bS bs BS Bs bS bs BBSS BBSs BbSS BbSs BBSsBBssBbSsBbss BbSSBbSsbbSSbbSs BbSs Bbss bbSs bbss BbSs

5. 2. What proportion of the mice offspring of the cross BbSs x BBss will have black fur and long tails? BS Bs bS bs Bs BBSs BBss BbSs Bbss 1/2 will have black fur, long tails (50%)

6. 3. A couple has three children, all of whom have brown eyes and blond hair. Both parents are homozygous for brown eyes (BB), and one is blond (rr) while the other is a redhead (Rr). What is the probability that the next child will be a brown-eyed redhead? Br BR Br BR Br BBRr BBrr BB rr BB Rr Brown-eyed redhead child = BBRr = 50% chance

7. homozygous = __________ combination of genes, e.g. BB, bb heterozygous = __________ combination of genes, e.g. Bb genotype = What combination of __________ it has, e.g. Bb, BB phenotype = what it ________ like, e.g. brown eyes Review genes looks Same Different

8. 4. In guinea pigs, black fur color (B) is dominant over white (b), and short hair (S) is dominant over long (s).. Show the Punnett Square for a cross between a homozygous black, short-haired guinea pig and a homozygous white, long-haired guinea pig. What do the offspring look like? Homozygous black Short hair = BBSS Homozygous white, long haired = bbss BS bs BbSs All BbSs, black fur, short hair

9. 5. In humans, the presence of freckles is due to a dominant gene (F) and the non-freckled condition is due to its recessive allele (f). Dimpled cheeks (D) is dominant to non-dimpled cheeks (d). Two persons with freckles and dimpled cheeks have two children. One has freckles but no dimples and one has dimples but no freckles. a. What are the genotypes of the parents? Parents: Children F__D__ x F__D__ F__ dd, Parents must be heterozygous for each gene: FfDd x FfDd ffD___

10. b. What are the ratios of the possible phenotypes and genotypes of the children they could produce? 9 Freckled Dimpled, 3 Freckled Non Dimpled, 3 Non-freckle Dimple; 1 Non Freckle Non Dimple FfDd x FfDd FD Fd fD fd FD Fd fD fd FFDD FFDd FfDD FfDd FFDd FFdd FfDd Ffdd FfDDFfDdffDDffDd FfDd Ffdd ffDd ffdd F.O.I.L First-Outside-Inside-Last