GENETICS PRACTICE 1: BASIC MENDELIAN GENETICS

Period Date GENETICS PRACTICE 1: BASIC MENDELIAN GENETICS Solve these genetics problems. Be sure to complete the Punnett square to show how you derived your solution. 1. In humans the allele for albinism is recessive to the allele for normal skin pigmentation. If two heterozygotes have children, what is the chance that a child will have normal skin pigment? What is the chance that a child will be albino? A a A AA normal pigment: 75% chance of AA & albino: a aa 25% chance of aa a. If the child is normal, what is the chance that it is a carrier (heterozygous) for the albino allele? (CAREFUL!) 2/3 or 67% 2. In purple people eaters, one-horn is dominant and no horns is recessive. Show the cross of a purple people eater that is heterozygous for horns with a purple people eater that does not have horns. Summarize the genotypes & phenotypes of the possible offspring? H h h Hh hh h Hh hh 50% chance of Hh (horned) 50% chance of hh (no horn) 3. In humans, the brown-eye (B) allele is dominant to the blue-eye allele (b). If two heterozygotes mate, what will be the likely genotype and phenotype ratios of the offspring. B b B BB Bb b Bb bb 75% chance of brown-eyed 25% chance of blue-eyed 25% chance of BB 50% chance of Bb 25% chance of bb 1 of 2 Developed by Kim B. Foglia www.explorebiology.com 2008

4. In seals, the gene for the length of the whiskers has two alleles. The dominant allele (W) codes long whiskers & the recessive allele (w) codes for short whiskers. What percentage of offspring would be expected to have short whiskers from the cross of two long-whiskered seals, one that is homozygous dominant and one that is heterozygous? W w W WW Ww W WW Ww 0% chance of short whiskers 5. In pea plants, the green color allele (G) is dominant over yellow color allele (g) for seed color and tall (T) is the dominant allele in plant height. Parents heterozygous for both traits are cross-pollinated. Determine the frequency for the four different phenotypes of the offspring. GT Gt gt gt GT GGTT GGTt GgTT GgTt Gt GGTt GGtt GgTt Ggtt gt GgTT GgTt ggtt ggtt gt GgTt Ggtt ggtt ggtt Tall plant, green seeds: 9/16 Tall plant, yellow seeds: 3/16 Short plant, green seeds: 3/16 Short plant, yellow seeds: 1/16 6. Now let s try a shortcut way of solving that same dihybrid cross. Because of Mendel s (2 nd ) Law of Independent Assortment, you can work with the color gene and the height gene separately so set up two separate monohybrid crosses from those same parents: G g G GG Gg g Gg gg T t T TT Tt t Tt tt Now use the laws of probability to calculate your frequencies of each trait alone and combined: height color = P height color = P Tall plant, green seeds ¾ ¾ 9/16 Short plant, green seeds ¼ ¾ 3/16 Tall plant, yellow seeds ¾ ¼ 3/16 Short plant, yellow seeds ¼ ¼ 1/16 2 of 2

Period Date GENETICS PRACTICE 2: BEYOND THE BASICS Solve these genetics problems. Be sure to complete the Punnett square to show how you derived your solution. INCOMPLETE DOMINANCE 1. In radishes, the gene that controls color exhibits incomplete dominance. Pure-breeding red radishes crossed with pure-breeding white radishes make purple radishes. What are the genotypic and phenotypic ratios when you cross a purple radish with a white radish? R W W RW WW W RW WW 50% purple (RW) 50% white (WW) 2. Certain breeds of cattle show incomplete dominance in coat color. When pure breeding red cows are bred with pure breeding white cows, the offspring are roan (a pinkish coat color). Summarize the genotypes & phenotypes of the possible offspring when a roan cow is mated with a roan bull R W R RR RW 25% red (RR) 50% roan (RW) W RW WW 25% white (WW) CO-DOMINANCE 3. A man with type AB blood marries a woman with type B blood. Her mother has type O blood. List the expected phenotype & genotype frequencies of their children. I A I B I B I A I B I B I B 25% type AB blood (I A I B ) 50% type B blood (I B I B & I B i) i I A i I B i 25% type A blood (I A i) 1 of 3 Developed by Kim B. Foglia www.explorebiology.com 2008

4. The father of a child has type AB blood. The mother has type A. Which blood types can their children NOT have? I A _ x I A I B = no O 5. A woman with type A blood and a man with type B blood could potentially have offspring with what blood types? I A _ x I B _ = AB, A, B, O 6. The mother has type A blood. Her husband has type B blood. Their child has type O blood. The father claims the child can t be his. Is he right? No! I A _ x I B _ = AB, A, B, O 7. The mother has type B blood. Her husband has type AB blood. Their child has type O blood. The father claims the child can t be his. Is he right? Yes! I B _ x I A I B = AB, A, B 8. The mother has type AB blood. The father has type B blood. His mother has type O blood. What are all the possibilities of blood type for their children? I A I B x I B i = AB, A, B LETHAL DOMINANT 9. Achondroplasia (dwarfism) is caused by a dominant gene. A woman and a man both with dwarfism marry. If homozygous achondroplasia results in death of embryos, list the genotypes and phenotypes of all potential live-birth offspring A A AA a 50% dwarfism () 25% normal (aa) What is the expected ratio of dwarfism to normal offspring? a aa 67% dwarfism : 33% normal SEX-LINKED 10. The genes for hemophilia are located on the X chromosome. It is a recessive disorder. List the possible genotypes and phenotypes of the children from a man normal for blood clotting and a woman who is a carrier. (HINT: You have to keep track of what sex the children are!) X H Y X H X H X H X H Y 50% females normal 50% females carrier 50% males normal X h X H X h X h Y 50% males hemophilia A CHALLENGE: Remember those roan cows from question 2? They also have a second gene for horn vs. hornless cattle. The allele for horns dominates the allele for hornless. If a bull and cow are heterozygous for both genes, what are the probabilities for each possible phenotype? 2 of 3

Solving it the short (probability) way: R r R RR Rr r Rr rr H h H HH Hh h Hh hh Red, horn RR x H_ = ¼ x ¾ = 3/16 Roan, horn Rr x H_ = 2/4 x ¾ = 6/16 Red, no horn RR x hh = ¼ x ¼ = 1/16 Roan, no horn Rr x hh = 2/4 x ¼ = 2/16 White, horn rr x H_ = ¼ x ¾ = 3/16 White, no horn rr x hh = ¼ x ¼ = 1/16 Usually your 9/16 if red showed simple dominance to white Usually your 3/16 if red showed simple dominance to white Solving it the long way: RH Rh rh rh RH Rh rh rh RRHH red, horn RRHh red, horn RrHH RrHh RRHh red, horn RRhh red, no horn RrHh Rrhh roan, no horn RrHh RrHh rrhh white, horn rrhh white, horn RrHh Rrhh roan, no horn rrhh white, horn rrhh white, no horn RRH_ red, horn: 3/16 RrH_ : 6/16 RRhh red, no horn: 1/16 Rrhh roan, no horn: 2/16 rrh_ white, horn: 3/16 rrhh white, no horn: 1/16 3 of 3

Period Date GENETICS PRACTICE 3: PROBABILITY PRACTICE 1. In humans, curly hair is dominant over straight hair. A woman heterozygous for hair curl marries a man with straight hair and they have children. a. What is the genotype of the mother? b. What gametes can she produce? c. What is the genotype of the father? d. What gametes can he produce? e. What is the probability that the 1st child will have curly hair? f. What is the probability that the 2nd child will have curly hair? 2. List all the gametes that are possible with each of the following genotypes. Ab, ab a. bb d. AABb AB, ab b. BB e. AAbb AB, Ab, ab, ab c. Bb f. aabb 3. What is the probability of getting the gamete (ab) from each of the following parents? a. bb b. aabb c. Bb d. AABb e. AAbb 4. In a certain strain of mice, black coat (B) is dominant over white coat (b). Describe what you would do to determine the genotype of a male with a black coat and how this would enable you to choose between the genotypes BB or Bb. Hh H or h hh h AB, Ab Ab ab ½ = 50% ½ = 50% a 1/2 ; bb b 2/2, so ½ x 1 p= ½ =.5 aa a 2/2 ; Bb b 1/2, so 1 x ½ p= ½ =.5 a 1/2 ; Bb b 1/2, so ½ x ½ p= ¼ =.25 AA a 0/2 ; Bb b 1/2, so p=0 AA a 0/2 ; bb b 2/2, so p=0 Test cross: cross with bb (homozygous recessive) BB x bb 100% black vs. Bb x bb 1 black : 1 white 1 of 3 Developed by Kim B. Foglia www.explorebiology.com 2008

5. What is the probability of each of the following sets of parents producing the given genotypes in their offspring? Parents Genotype Offspring Genotype Probability x 2/4 =.5 x aa 2/4 =.5 Bb x BB Bb x AABb Bb x Bb AABB aabb Bb x ¼ AA; Bb x BB ½ BB ¼ x ½ = 1/8 x AA 0 aa; Bb x Bb ¼ bb 0 x ¼ = 0 x ½ ; Bb x Bb ½ Bb ½ x ½ = ¼ 6. If an offspring has the genotype, what possible combinations of parental genotypes could have produced this offspring? AA x aa, AA x, x aa, x 7. In corn, the trait for tall plants (T) is dominant to the trait for dwarf plants (t) and the trait for colored kernels (C) is dominant to the trait for white kernels (c). In a particular cross of corn plants, the probability of an offspring being tall is 1/2 and the probability of a kernel being colored is 3/4. Which of the following most probably represents the parental genotype? Include your work to show how you derived your solution. a. TtCc x ttcc b. TtCc x TtCc c. TtCc x ttcc d. TTCc x ttcc e. TTCc x TtCC Tt x tt ½ Tt ½ tt Cc x Cc ¾ CC, Cc ¼ cc 8. In humans, the allele for albinism (lack of pigment) is recessive to the allele for normal skin pigmentation. a. If two heterozygous parents have children what is the chance that a child will be albino? b. If the child is normal, what is the chance that it is a carrier (heterozygous) for the albino allele? x aa: ¼ = 25% vs. AA 2/3 = 67% c. If normal parents have an albino child, what is the probability that their next child will be normal for pigment? or AA 3/4 = 75% 2 of 3

A CHALLENGE In a cross between a female BbccDdee and a male bbccddee, what proportion of the progeny will be the same phenotype as the female parent? (Assume independent assortment of all genes and complete dominance). x AA, ¾ Bb x bb Bb ½ cc x Cc cc ½ Dd x Dd DD, Dd ¾ ee x ee ee 1 = 9/64 3 of 3

Period Date GENETICS PRACTICE 4: PEDIGREES PEDIGREE #1 Could this trait be inherited as a simple If YES, then suggested genotypes of father mother a. autosomal recessive? YES NO x b. autosomal dominant? YES NO x c. X-linked recessive? YES NO x d. X-linked dominant? YES NO x e. Y-linked trait? YES NO x X A Y X A X a PEDIGREE #2 Could this trait be inherited as a simple If YES, then suggested genotypes of father mother a. autosomal recessive? YES NO x b. autosomal dominant? YES NO x c. X-linked recessive? YES NO x d. X-linked dominant? YES NO x e. Y-linked trait? YES NO x X A Y X A X a 1 of 2 Developed by Kim B. Foglia www.explorebiology.com 2008

PEDIGREE #3 Could this trait be inherited as a simple If YES, then suggested genotypes of father mother a. autosomal recessive? YES NO x b. autosomal dominant? YES NO x c. X-linked recessive? YES NO x d. X-linked dominant? YES NO x e. Y-linked trait? YES NO x PEDIGREE #4 Could this trait be inherited as a simple If YES, then suggested genotypes of father mother a. autosomal recessive? YES NO x b. autosomal dominant? YES NO x c. X-linked recessive? YES NO x d. X-linked dominant? YES NO x e. Y-linked trait? YES NO x aa X a Y X A Y aa X A X a X a X a 2 of 2