CARNES AP BIO | "Nothing in biology makes sense except in ...



Directions: Show your work for each problem below. Remember that for any gene or trait, 'homozygous dominant' means AA, 'heterozygous' (also called a carrier) means Aa and 'homozygous recessive' means aa. The following website is very useful for Punnett square practice:



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 phenotypically have normal skin pigment? What is the chance that the child will be phenotypically albino?

• What is the genotype of parent #1? ________________

• What is the genotype of parent #2? ________________

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2. If the child is normal (from question #1 above), what is the chance that it is a carrier (heterozygous) for the albino allele…CAREFUL? _______________________

3. In seals, the gene for the length of the whiskers has two alleles. The dominant allele (W) codes for long whiskers, and the recessive allele (w) codes for short whiskers. What percentage of the offspring would be expected to have short whiskers from the cross of two long-whiskered seals, one that is homozygous and one that is heterozygous?

• What is the genotype of parent #1? _________________

• What is the genotype of parent #2? _________________

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4. A woman is heterozygous for Tay-Sachs. What are the chances of her giving birth to a child with Tay-Sachs (a homozygous recessive disorder) if the father is normal (homozygous dominant)? What are the chances of giving birth to a carrier like herself?

• What is the genotype of the female parent? ___________________

• What is the genotype of the male parent? ____________________

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5. In pea plants, the green color allele (G) is dominant over the yellow color allele (g) for seed color and tall (T) is dominant over short (t) in plant height. Parents who are heterozygous for both traits are cross-pollinated.

• Determine the genotypes of the female parent: _______________ and male parent: __________________.

• Determine the gamete possibilities for sperm: ____________________ and egg: ____________________.

• Work out the dihybrid Punnett square.

• Determine the phenotype frequency for the offspring.

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6. Now let’s try a short-cut method for solving the same dihybrid cross. Because of Mendel’s 2nd Law of Independent Assortment, you can work with the color gene and the height gene separately…so set up two different monohybrid crosses from the same parents.

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7. Imagine that in squirrels, gray color (G) is dominant over black color (g). A black squirrel has the genotype gg, but a gray squirrel can have the genotype GG or Gg. Crossing a gray squirrel with what genotype would let you know with the most certainty the genotype of the gray squirrel? What is this method of determining the genotype of an unknown called?

_____________________________________.

8. When and how is this method (from #8 above) useful in genetics? ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

9. What is the number of possible gametes produced from a genotype of RrBBCcDDEe? _________________________. What is an EASY way to determine this? ________________________________________________________________________________________________________________________________________________________________________________________________________________

10. Both Dick and Jane have freckles (dominant) and attached earlobes (recessive). Jane’s mother did not have freckles, but her father did. Dick’s mother had freckles but his father did not. Six of Dick and Jane’s children do not have freckles.

• What are the chances that their next child will have freckles and attached earlobes? _____________

• Determine Dick and Jane’s genotypes. Dick: ________________, Jane: __________________

• Work a dihybrid Punnett. List the expected phenotype frequencies of the possible offspring.

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11. Yellow hair (Y) can be crossed with red hair (R) to create strawberry-blond. A yellow haired person is crossed with a strawberry-blonde person. Complete a Punnett square, using Incomplete Dominance. What is the genotypic & phenotypic ratio of the children?

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12. 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 color (pinkish coat). Summarize the genotypes and phenotypes of the possible offspring when a roan cow is mated with a roan bull.

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13. A man with type AB blood marries a woman with type B blood. Her mother has type O blood. List the expected genotype and phenotype frequencies of their children.

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14. Achondroplasia (dwarfism) is caused by a dominant gene. A woman and a man both with dwarfism marry. If homozygous achondroplasia results in the DEATH of embryos, list the genotypes and phenotypes of all potential live-birth offspring.

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15. Pam and Roy have a son, Dwight, who has a hitchhiker’s thumb (recessive phenotype) and can tongue roll (dominant phenotype). Pam too has a normal thumb but she can tongue roll. Roy believes his son was switched at birth because he has a normal thumb and cannot tongue roll, though both he and Pam’s fathers had hitchhiker’s thumbs. The hospital’s lawyer has contacted you to help show Pam and Roy that Dwight is their son. In the space provided below, work a dihybrid Punnett square for the cross showing that Dwight’s phenotypes could be produced by Pam and Roy. Give a brief explanation of your logic.

16. 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 the sex of the children!!!

• What is the genotype of the father? __________________

• What is the genotype of the mother? _________________

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17. Create a linkage map of the relative location of the following four genes on a chromosome based on their crossover frequencies:

• X and Z = 5%

• Y and W = 15%

• W and X = 30%

• Y and X = 45%

• Y and Z = 50%

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