Review Questions for Mendelian Genetics

1. Review Questions for Mendelian Genetics

2. Match the description with the genetics terms. Each question can have more than one answer. • The mixture of alleles in an individual • The specific form of a gene. • For example, tall, short, white, purple • Where a gene is located on a chromosome • For example, BB or Tt • Homozygous dominant • heterozygous • hybrid • For example, B or b • For example, flower color • For example, green seeds or yellow seeds • Can contain multiple genes • The specific genetic makeup of an individual A • genotype • phenotype • allele • gene • Chromosome • Locus C B, C F A A A A C D B E A

3. Match the description with the genetics terms. Each question can have more than one answer. • Genotype of an individual with a dominant phenotype • Requires mating with a recessive individual • Has the alleles G and t from an individual that was GgTt • Used to determine the possible offspring from a cross • Genotype of an individual with the recessive trait • A way to determine the genotype of an individual with the dominant trait • Contains two different alleles • For example, hh • hybrid D, E A • Test cross • Gamete • Punnett Square • Heterozygous • Homozygous dominant • Homozygous recessive B C F A D F D

4. 7 Steps to doing genetics problems • Define alleles • Parental genotypes • Gametes that parents can produce • Punnett Square using gametes • Fill in Punnett Square to determine offspring genotypes • Write genotype ratios & phenotype ratios of offspring • Answer the question and circle your answer.

5. If parents are Aabb and AaBb, assuming independent assortment and random recombination, what is the chance that they have a child who is aaBb? • Determine the gametes that each parent can produce. • Create your Punnett Square. • Answer the question Ab ab aB AB Ab ab AABb Aabb AAbb AaBb AaBb Aabb aaBb aabb 1/8 or 12.5% chance of having aaBb child

6. If parents are aaBB and AaBb, assuming independent assortment and random recombination, what is the chance that they have a child who is AaBb? • Use the product law: • Chance of aa X Aa making Aa: • Chance of BB X Bb making Bb: • Multiply the two probabilities to get your answer: Aa ½ chance of getting Aa aa a b Possible gametes: a A B ½ chance of getting Bb BB Possible gametes: B Bb (1/2)(1/2) = ¼ or 25% of having AaBb child

7. If parents are AaBB and AaBb, assuming independent assortment and random recombination, what is the chance that they have a child who is aaBb? • Use the product law: • Chance of Aa X Aa making aa: • Chance of BB X Bb making Bb: • Multiply the two probabilities to get your answer: 1/4 chance of getting aa AA Aa A aa a Aa a b A B ½ chance of getting Bb BB B Bb (1/4)(1/2) = 1/8 or 12.5% chance of having aaBb child

8. If parents are Aabb and AaBb, assuming independent assortment and random recombination, what is the chance that they have a child who is AaBb? • Use the product law: • Chance of Aa X Aa making Aa: • Chance of BB X Bb making Bb: • Multiply the two probabilities to get your answer: 1/2 chance of getting Aa AA Aa A aa a Aa a b A B ½ chance of getting Bb Bb b bb (1/2)(1/2) = 1/4 or 25% chance of having AaBb child

9. If parents are AabbDd and AaBbDd, assuming independent assortment and random recombination, what is the chance that they have a child who is aaBbDD? • Use the product law: • Chance of Aa X Aa making aa: • Chance of bb X Bb making Bb: • Chance of Dd X Dd making DD: • Multiply the two probabilities to get your answer: 1/4 chance of getting aa AA Aa A a Aa aa a d b B D A ½ chance of getting Bb Bb b bb D d 1/4 chance of getting DD DD Dd Dd dd (1/4)(1/2)(1/4) = 1/32 chance of having aaBbDD child

10. If one parent has the genotype AabbDd assuming independent assortment and random recombination, which of the following are possible gametes made by that individual? • Aa • AbD • ABd • Aabbdd • D • d • B • abd