Review Questions for Mendelian Genetics

Match the description with the genetics terms. Each question can have more than one answer. 1.The mixture of alleles in an individual 2.The specific form of a gene. 3.For example, tall, short, white, purple 4.Where a gene is located on a chromosome 5.For example, BB or Tt 6.Homozygous dominant 7.heterozygous 8.hybrid 9.For example, B or b 10.For example, flower color 11.For example, green seeds or yellow seeds 12.Can contain multiple genes 13.The specific genetic makeup of an individual A.genotype B.phenotype C.allele D.gene E.Chromosome F.Locus A C B, C F A A A A C D B E A

Match the description with the genetics terms. Each question can have more than one answer. 1.Genotype of an individual with a dominant phenotype 2.Requires mating with a recessive individual 3.Has the alleles G and t from an individual that was GgTt 4.Used to determine the possible offspring from a cross 5.Genotype of an individual with the recessive trait 6.A way to determine the genotype of an individual with the dominant trait 7.Contains two different alleles 8.For example, hh 9.hybrid D, E A B C F A F D A.Test cross B.Gamete C.Punnett Square D.Heterozygous E.Homozygous dominant F.Homozygous recessive D

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.

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 1/8 or 12.5% chance of having aaBb child ab AABb AbaBab AaBb AAbb Aabb AaBb aaBb Aabb aabb

If parents are aaBB and AaBb, assuming independent assortment and random recombination, what is the chance that they have a child who is AaBb? a A a Aa aa BB Bb ½ chance of getting Aa B B b ½ chance of getting Bb (1/2)(1/2) = ¼ or 25% of having AaBb child 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: Possible gametes:

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: A A a AA Aa BB Bb 1/4 chance of getting aa B B b ½ chance of getting Bb (1/4)(1/2) = 1/8 or 12.5% chance of having aaBb child a Aa aa

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: A A a AA Aa Bb bb 1/2 chance of getting Aa b B b ½ chance of getting Bb (1/2)(1/2) = 1/4 or 25% chance of having AaBb child a Aa aa

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: A A a AA Aa Bb bb 1/4 chance of getting aa b B b ½ chance of getting Bb (1/4)(1/2)(1/4) = 1/32 chance of having aaBbDD child a Aa aa Dd D d Dd 1/4 chance of getting DD dd DDDd

If one parent has the genotype AabbDd assuming independent assortment and random recombination, which of the following are possible gametes made by that individual? a.Aa b.AbD c.ABd d.Aabbdd e.D f.d g.B h.abd