Mendelian Genetics Review Worksheet - Answers

Mendelian Genetics Review Worksheet ANSWERS 1. A research botanist is trying to replicate Gregor Mendel’s work with peas. The scientist crosses a true breeding purple flowered plant with a true breeding white flowered plant. All of the offspring have purple flowers. The scientist then self-pollinates plants of the F1 generation, and harvests 125 seeds. P generation: PP x pp  F1 Generation: Pp x Pp P p P PP Pp Pp pp p a. Which is the dominant gene? Purple flower color b. How many of the offspring in the F2 generation would you expect to have purple flowers? 3/4 phenotypes with at least 1 dominant allele x 125 seeds = 94 purple offspring c. How many of the offspring in the F2 generation would you expect to be heterozygous? 1/2 heterozygote genotype x 125 seeds = 63 heterozygote offspring 2. The research botanist is now looking at wrinkled and smooth seeds. The scientist crosses a true breeding smooth seeded plant with a true breeding wrinkle seeded plant. All of the offspring have smooth seeds and are allowed to self-pollinate. The botanist then harvests 535 seeds. P generation: RR x rr  F1 Generation: Rr x Rr R r R RR Rr r Rr rr a. Which is the dominant gene? Smooth seed shape b. How many of the offspring in the F2 generation would you expect to have wrinkled seeds? 1/4 homozygous recessive genotype x 535 seeds = 134 wrinkled offspring c. How many of the offspring in the F2 generation would you expect to be homozygous dominant? 1/4 homozygous dominant genotype x 535 seeds = 134 offspring 3. The botanist then breeds a plant that has white flowers and is heterozygous for seed type with a wrinkled seed plant that is heterozygous for flower color. ppRr x Pprr a. Draw the Punnett Square for this cross. pR pr Pr PpRr Pprr Pr PpRr Pprr pr ppRr pprr pr ppRr pprr pR PpRr PpRr ppRr ppRr pr Pprr Pprr pprr pprr b. What do you expect to be the phenotypic ratio in the offspring? Purple, smooth = 4/16 = 1/4 or 25% Purple, wrinkled = 4/16 = 1/4 or 25% White, smooth = 4/16 = 1/4 or 25% White, wrinkled = 4/16 = 1/4 or 25% 4. In corn, red kernel color is dominant to yellow kernel color and a tall plant is dominant to a short one. A geneticist breeds a plant that is heterozygous for both traits with a plant that is homozygous for the red kernel allele and heterozygous for height. What is the expected ratio of phenotypes for this cross? RrTt x RRTt RT Rt rT rt RT RRTT RRTt RrTT RrTt RRTt RRtt RrTt Rrtt Rt RT RRTT RRTt RrTT RrTt RRTt RRtt RrTt Rrtt Rt Red, tall = 12/16 = 3/4 Red, short = 4/16 = 1/4 5. In humans, a hairline above the forehead that comes to a point is referred to as a widow’s peak. The trait is controlled by two genes at a single locus, with a peak being dominant over a straight hairline. Given this, which is not possible: two parents with widow’s peaks having a child without one, or two parents without a peak having a child that has one? Justify your answer. Widow’s peak is dominant, so if the parents do not show the trait they must be homozygous recessive (aa). Two homozygous recessive parents (aa x aa) cannot have a child with a dominant allele (AA or Aa) as they have no dominant allele to give. If both parents have widow’s peak, they each have at least one dominant allele (AA or Aa). If they are both heterozygous carriers (Aa x Aa), then there is a 25% chance that the child will be homozygous recessive (aa) and not show a widow’s peak. 6. In mice, the allele for a brown coat is dominant to the allele for a white coat, and the allele for a long tail is dominant to the allele for a short tail. A geneticist breeds a white mouse that is heterozygous for tail length with a short-tailed mouse that is heterozygous for coat color. What are the phenotypes of the parents? What is the expected ratio of phenotypes from this cross? bbTt x Bbtt Parental phenotypes: white with a long tail and brown with a short tail bT bt bT bt BbTt Bbtt BbTt Bbtt Bt BbTt Bbtt BbTt Bbtt Bt bbTt bbtt bbTt bbtt bt bbTt bbtt bbTt bbtt bt Brown, long = 4/16 = 1/4 or 25% Brown, short = 4/16 = 1/4 or 25% White, long = 4/16 = 1/4 or 25% White, short = 4/16 = 1/4 or 25% 7. In humans, blood group alleles IA and IB have codominance with each other, while both genes are dominant over the allele i. A woman with type A blood gives birth to an infant with type O blood. Mom is Type A, so her genotype is either IAIA or IAi. Baby is Type O, so its genotype must be ii – one i came from mom and one came from dad. This means mom must be IAi. Dad could be any blood type that includes at least one i allele. a. Which of the following blood types could possibly be the father? O A B AB b. What is the mother’s genotype? IA i I A IA IA I B 8. A woman with type O blood gives birth to an infant with type B blood. Mom is Type O, so her genotype must be ii. Baby is Type B, so its genotype is either IBIB or IBi. Since mom can only give i alleles, baby must be IBi. Dad could be any blood type that includes at least one IB allele. a. Which of the following blood types could possibly be the father? O A B AB b. What is the child’s genotype? ii I A IA IA i IB IB IB i I A IB 9. A woman with type A blood has three children, one with type O blood, one with AB blood and one with type A blood. What are the mother and father’s genotypes? Mom is Type A, so her genotype is either IAIA or IAi. Baby 1 is Type O and must be ii – one i came from mom and one i came from dad. This means mom must be IAi. Baby 2 is Type AB and must be IAIB – the IA came from mom, as she cannot give a IB, and the IB must have come from dad. Since baby 1 also needs an i from dad, dad’s genotype must be IBi. Baby 3 is Type A, and must be IAi – the IA allele came from mom and the i allele came from dad. 10. In humans, the gene for color blindness is carried on the X chromosome. Human gender is determined by the X and Y chromosome, where XX is female and XY is male. What is the probability that a female is carrying the color blindness gene if both her parents had normal vision but her maternal grandfather was colorblind? Maternal Grandfather (colorblind) = X aY Parents: Normal vision mother whose father was color blind = X AXa Normal vision father = XAY XA Y XA XA XA XA Y Xa XA Xa Xa Y Probability a female is carrying the gene = 50% (of the females, half are carriers) 11. In Labrador retrievers, coat color is controlled by alleles at two loci. For one gene pair, black fur color is dominant to brown, but another gene determines if any melanin pigment is deposited in the hair shaft. Two recessive alleles of this gene block deposition, and a yellow fur results. Thus, a retriever with at least one dominant allele at both loci is black, a retriever with two recessive alleles at the second loci is yellow, and a retriever with two recessive alleles at the first loci and at least one dominant allele at the second is chocolate. A dog breeder crosses a chocolate lab that is heterozygous for the deposition allele with a black lab that is heterozygous at both loci. What is the expected phenotypic ratio from this cross? bbEe x BbEe bE be bE be BE BbEE BbEe BbEE BbEe Be BbEe Bbee BbEe Bbee bE bbEE bbEe bbEE bbEe be bbEe bbee bbEe bbee Black (at least 1 B and at least one E) = 6/16 = 3/8 or 37.5% Brown (bb and at least one E) = 6/16 = 3/8 or 37.5% Yellow (ee, regardless of B or b) = 4/16 = 2/8 = 1/4 or 25% 12. In cats, the tortoiseshell gene is on the X chromosome, and the tortoiseshell phenotype requires the presence of two different alleles, one for orange fur and one for black fur. Cat gender is determined by the X and Y chromosome, where XX is female and XY is male. A cat breeder crosses a female tortoiseshell cat with a male black cat. XB XO x XB Y XB Y XB XO XB XB XB X O XB Y XO Y a. What percentage of the kittens will be black and male? 25% (out of all the kittens, 1/4 have the genotype X B Y) b. What percentage of the kittens will be tortoiseshell and female? 25% (out of all the kittens, 1/4 have the genotype X B XO) 13. A cat breeder crosses a female black cat with a male orange cat. What percentage of the kittens will be tortoiseshell and female? What color(s) could the male kittens be? XB X B x X O Y XO Y XB XB XO XB Y XB XB XO XB Y 50% (out of all the kittens, 1/2 have the genotype X B XO) The male kittens can only be black Mendelian Genetics Review Worksheet 1. A research botanist is trying to replicate Gregor Mendel’s work with peas. The scientist crosses a true breeding purple flowered plant with a true breeding white flowered plant. All of the offspring have purple flowers. The scientist then self-pollinates plants of the F1 generation, and harvests 125 seeds. a. Which is the dominant gene? b. How many of the offspring in the F2 generation would you expect to have purple flowers? c. How many of the offspring in the F2 generation would you expect to be heterozygous? 2. The research botanist is now looking at wrinkled and smooth seeds. The scientist crosses a true breeding smooth seeded plant with a true breeding wrinkle seeded plant. All of the offspring have smooth seeds and are allowed to self-pollinate. The botanist then harvests 535 seeds. a. Which is the dominant gene? b. How many of the offspring in the F2 generation would you expect to have wrinkled seeds? c. How many of the offspring in the F2 generation would you expect to be homozygous dominant? 3. The botanist then breeds a plant that has white flowers and is heterozygous for seed type with a wrinkled seed plant that is heterozygous for flower color. a. Draw the Punnett Square for this cross. b. What do you expect to be the phenotypic ratio in the offspring? 4. In corn, red kernel color is dominant to yellow kernel color and a tall plant is dominant to a short one. A geneticist breeds a plant that is heterozygous for both traits with a plant that is homozygous for the red kernel allele and heterozygous for height. What is the expected ratio of phenotypes for this cross? 5. In humans, a hairline above the forehead that comes to a point is referred to as a widow’s peak. The trait is controlled by two alleles at a single locus, with a peak being dominant over a straight hairline. Given this, which is not possible: two parents with widow’s peaks having a child without one, or two parents without a peak having a child that has one? Justify your answer. 6. In mice, the allele for a brown coat is dominant to the allele for a white coat, and the allele for a long tail is dominant to the allele for a short tail. A geneticist breeds a white mouse that is heterozygous for tail length with a short-tailed mouse that is heterozygous for coat color. What are the phenotypes of these parent mice? What is the expected ratio of phenotypes from this cross? 7. In humans, blood group alleles IA and IB have codominance with each other, while both genes are dominant over the allele i. A woman with type A blood gives birth to an infant with type O blood. a. Which of the following blood phenotypes could possibly be the father? O A B AB b. What is the mother’s genotype? IA i I A IA IA I B 8. A woman with type O blood gives birth to an infant with type B blood. a. Which of the following blood phenotypes could possibly be the father? O A B AB b. What is the child’s genotype? ii I A IA IA i IB IB IB i IA IB 9. A woman with type A blood has three children, one with type O blood, one with AB blood and one with type A blood. What are the mother and father’s genotypes? 10. In humans, the gene for color blindness is carried on the X chromosome. Human gender is determined by the X and Y chromosome, where XX is female and XY is male. What is the probability that a female is carrying the color blindness gene if both her parents had normal vision but her maternal grandfather was colorblind? 11. In Labrador retrievers, coat color is controlled by alleles at two loci. For one gene pair, black fur color is dominant to brown, but another gene determines if any melanin pigment is deposited in the hair shaft. Two recessive alleles of this gene block deposition, and a yellow fur results. Thus, a retriever with at least one dominant allele at both loci is black, a retriever with two recessive alleles at the second loci is yellow, and a retriever with two recessive alleles at the first loci and at least one dominant allele at the second is brown. A dog breeder crosses a brown lab that is heterozygous for the deposition allele with a black lab that is heterozygous at both loci. What is the expected phenotypic ratio from this cross? 12. In cats, the tortoiseshell gene is on the X chromosome, and the tortoiseshell phenotype requires the presence of two different alleles, one for orange fur and one for black fur. Cat gender is determined by the X and Y chromosome, where XX is female and XY is male. A cat breeder crosses a female tortoiseshell cat with a male black cat. a. What percentage of the kittens will be black and male? b. What percentage of the kittens will be tortoiseshell and female? 13. A cat breeder crosses a female black cat with a male orange cat. What percentage of the kittens will be tortoiseshell and female? What color(s) could the male kittens be?