Temple University Gray Male Is Homozygous Recessive Discussion

Temple University Gray Male Is Homozygous Recessive Discussion Temple University Gray Male Is Homozygous Recessive Discussion Help me with my homework??????????????????????????????????????????????????????????????? fall_2020_genetics_assignment_for_dna_class__1_.docx punnett_sq_example.pptx Name ______________________________ GENETICS ASSIGNMENT-FALL 2020 1. In cats, black coat color is dominant over gray. A female black cat whose mother is gray mates with a gray male. If this female has a litter of kittens, what is the probability that the mating will produce gray kittens? Show your work in a Punnett square and give the probability. 2. In foxes, silver-black coat color is governed by a recessive allele “b” and red color by the dominant allele “B”. Determine the genotypic and phenotypic ratios expected from the following matings. Be sure to show your work and give the ratios for each mating. A. Pure red x carrier B. Silver-black x Pure red 3. Any measurable or distinctive characteristic or trait possessed by an individual is referred to as the individual’s _________________________________. 4. If an individual is the result of the union of gametes carrying different alleles one would refer to that individual’s genotype as being ___________________________. 5. The normal cloven-footed (a hoof which is split into two toes) condition in swine is produced by the homozygous recessive genotype “mm”. A mule-footed condition is produced by the dominant gene “M”. White coat color is governed by the dominant allele “B” while the black coat color is governed by its recessive allele “b”. Suppose a white, mule-footed sow (female) is mated to a black, cloven-footed boar (male) and produces several litters. Of the 26 offspring produced from this mating, all were found to be white with mule feet. A. What is the most probable genotype of the sow?______________________ B. ORDER NOW FOR CUSTOMIZED AND ORIGINAL NURSING PAPERS The next litter produced 8 white, mule-footed offspring and 1 white, cloven-footed pig. Now, what is the most probable genotype of the sow?____________________ 6. Hemophilia is a sex linked recessive trait. A. Name ______________________________ GENETICS ASSIGNMENT-FALL 2020 1. In cats, black coat color is dominant over gray. A female black cat whose mother is gray mates with a gray male. If this female has a litter of kittens, what is the probability that the mating will produce gray kittens? Show your work in a Punnett square and give the probability. 2. In foxes, silver-black coat color is governed by a recessive allele “b” and red color by the dominant allele “B”. Determine the genotypic and phenotypic ratios expected from the following matings. Be sure to show your work and give the ratios for each mating. A. Pure red x carrier B. Silver-black x Pure red 3. Any measurable or distinctive characteristic or trait possessed by an individual is referred to as the individual’s _________________________________. 4. If an individual is the result of the union of gametes carrying different alleles one would refer to that individual’s genotype as being ___________________________. 5. The normal cloven-footed (a hoof which is split into two toes) condition in swine is produced by the homozygous recessive genotype “mm”. A mule-footed condition is produced by the dominant gene “M”. White coat color is governed by the dominant allele “B” while the black coat color is governed by its recessive allele “b”. Suppose a white, mule-footed sow (female) is mated to a black, cloven-footed boar (male) and produces several litters. Of the 26 offspring produced from this mating, all were found to be white with mule feet. A. What is the most probable genotype of the sow?______________________ B. The next litter produced 8 white, mule-footed offspring and 1 white, cloven-footed pig. Now, what is the most probable genotype of the sow?____________________ 6. Hemophilia is a sex linked recessive trait. A. Is it possible that two normal parents could produce a girl with hemophilia?_________ B.Temple University Gray Male Is Homozygous Recessive Discussion A boy with hemophilia? ___________________ Explain each answer and support your reasoning with a Punnet Square 7. Piebald spotting is a condition found in humans in which there are patches of skin that lack pigmentation. The condition results from the inability of pigment producing cells to migrate properly during development. Two adults with piebald spotting have one child who has the trait and a second child with normal skin pigmentation. A. Is the piebald trait dominant or recessive? _____________________________________ B. Explain how you came to this conclusion? In other words, prove this by showing what the situation would have to be in order to be the opposite of what you chose. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ C. What are the genotypes of the parents? ________________________________________ 8. Shown below is one pair of chromosomes from a female and male cat. Each carries 4 genes as shown. E s l c chromosome from egg of female cat e S l C chromosome from sperm of male cat A. Given the information below, describe what this kitten will look like ___________________ __________________________________________________________________________ B. (E) is dominant for electric green eyes and (e) is recessive for brown eyes (S) is dominant for silky hair and (s) is recessive for hairless (L) is dominant for long hair and (l) is recessive for straight hair (C) is dominant for calico color and (c) is recessive for solid color 9. A rare recessive allele inherited in a Mendelian manner, causes the disease Cystic Fibrosis. Suppose a man who is phenotypically normal, and whose mother was normal but his father had cystic fibrosis, marries a woman who has Cystic Fibrosis. The parents of the woman are both normal. They are concerned that their children may inherit this disorder and decide to come to you, their genetic counselor. You begin by drawing a pedigree chart to show them the inheritance pattern in their family thus far. In the space below draw the pedigree chart which you would show them. Label the man, woman and each of their parents giving their possible genotypes. 10. Bronze turkeys have at least one dominant allele “R”. Red turkeys are homozygous for its recessive allele and have the genotype “rr”. Another dominant allele “H” produces normal feathers, and the recessive genotype “hh” produces feathers lacking webbing resulting in a condition known as “hairy”. Show the resulting F2 progeny from a cross between a homozygous bronze, hairy bird and one which is a homozygous red, normal-feathered bird. Be sure you read the question carefully. Show all of your work using Punnett Square(s) and be sure to give both the genotypic and phenotypic ratios showing all of the genotypes and phenotypes clearly in your answer. Name ______________________________ GENETICS ASSIGNMENT-FALL 2020 1. In cats, black coat color is dominant over gray. A female black cat whose mother is gray mates with a gray male. If this female has a litter of kittens, what is the probability that the mating will produce gray kittens? Show your work in a Punnett square and give the probability. 2. In foxes, silver-black coat color is governed by a recessive allele “b” and red color by the dominant allele “B”. Determine the genotypic and phenotypic ratios expected from the following matings. Be sure to show your work and give the ratios for each mating. A. Pure red x carrier B. Silver-black x Pure red 3. Any measurable or distinctive characteristic or trait possessed by an individual is referred to as the individual’s _________________________________. 4. If an individual is the result of the union of gametes carrying different alleles one would refer to that individual’s genotype as being ___________________________. 5. The normal cloven-footed (a hoof which is split into two toes) condition in swine is produced by the homozygous recessive genotype “mm”. A mule-footed condition is produced by the dominant gene “M”. White coat color is governed by the dominant allele “B” while the black coat color is governed by its recessive allele “b”. Suppose a white, mule-footed sow (female) is mated to a black, cloven-footed boar (male) and produces several litters. Of the 26 offspring produced from this mating, all were found to be white with mule feet. A. What is the most probable genotype of the sow?______________________ B. The next litter produced 8 white, mule-footed offspring and 1 white, cloven-footed pig. Now, what is the most probable genotype of the sow?____________________ 6. Hemophilia is a sex linked recessive trait. A. Is it possible that two normal parents could produce a girl with hemophilia?_________ B. A boy with hemophilia? ___________________ Explain each answer and support your reasoning with a Punnet Square 7. Piebald spotting is a condition found in humans in which there are patches of skin that lack pigmentation. The condition results from the inability of pigment producing cells to migrate properly during development. Two adults with piebald spotting have one child who has the trait and a second child with normal skin pigmentation. A. Is the piebald trait dominant or recessive? _____________________________________ B. Explain how you came to this conclusion? In other words, prove this by showing what the situation would have to be in order to be the opposite of what you chose. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ C. What are the genotypes of the parents? ________________________________________ 8. Shown below is one pair of chromosomes from a female and male cat. Each carries 4 genes as shown. E s l c chromosome from egg of female cat e S l C chromosome from sperm of male cat A. Given the information below, describe what this kitten will look like ___________________ __________________________________________________________________________ B. (E) is dominant for electric green eyes and (e) is recessive for brown eyes (S) is dominant for silky hair and (s) is recessive for hairless (L) is dominant for long hair and (l) is recessive for straight hair (C) is dominant for calico color and (c) is recessive for solid color 9. A rare recessive allele inherited in a Mendelian manner, causes the disease Cystic Fibrosis. Suppose a man who is phenotypically normal, and whose mother was normal but his father had cystic fibrosis, marries a woman who has Cystic Fibrosis.Temple University Gray Male Is Homozygous Recessive Discussion The parents of the woman are both normal. They are concerned that their children may inherit this disorder and decide to come to you, their genetic counselor. You begin by drawing a pedigree chart to show them the inheritance pattern in their family thus far. In the space below draw the pedigree chart which you would show them. Label the man, woman and each of their parents giving their possible genotypes. 10. Bronze turkeys have at least one dominant allele “R”. Red turkeys are homozygous for its recessive allele and have the genotype “rr”. Another dominant allele “H” produces normal feathers, and the recessive genotype “hh” produces feathers lacking webbing resulting in a condition known as “hairy”. Show the resulting F2 progeny from a cross between a homozygous bronze, hairy bird and one which is a homozygous red, normal-feathered bird. Be sure you read the question carefully. Show all of your work using Punnett Square(s) and be sure to give both the genotypic and phenotypic ratios showing all of the genotypes and phenotypes clearly in your answer. Name ______________________________ GENETICS ASSIGNMENT-FALL 2020 1. In cats, black coat color is dominant over gray. A female black cat whose mother is gray mates with a gray male. If this female has a litter of kittens, what is the probability that the mating will produce gray kittens? Show your work in a Punnett square and give the probability. 2. In foxes, silver-black coat color is governed by a recessive allele “b” and red color by the dominant allele “B”. Determine the genotypic and phenotypic ratios expected from the following matings. Be sure to show your work and give the ratios for each mating. A. Pure red x carrier B. Silver-black x Pure red 3. Any measurable or distinctive characteristic or trait possessed by an individual is referred to as the individual’s _________________________________. 4. If an individual is the result of the union of gametes carrying different alleles one would refer to that individual’s genotype as being ___________________________. 5. The normal cloven-footed (a hoof which is split into two toes) condition in swine is produced by the homozygous recessive genotype “mm”.Temple University Gray Male Is Homozygous Recessive Discussion A mule-footed condition is produced by the dominant gene “M”. White coat color is governed by the dominant allele “B” while the black coat color is governed by its recessive allele “b”. Suppose a white, mule-footed sow (female) is mated to a black, cloven-footed boar (male) and produces several litters. Of the 26 offspring produced from this mating, all were found to be white with mule feet. A. What is the most probable genotype of the sow?______________________ B. The next litter produced 8 white, mule-footed offspring and 1 white, cloven-footed pig. Now, what is the most probable genotype of the sow?____________________ 6. Hemophilia is a sex linked recessive trait. A. Is it possible that two normal parents could produce a girl with hemophilia?_________ B. A boy with hemophilia? ___________________ Explain each answer and support your reasoning with a Punnet Square 7. Piebald spotting is a condition found in humans in which there are patches of skin that lack pigmentation. The condition results from the inability of pigment producing cells to migrate properly during development. Two adults with piebald spotting have one child who has the trait and a second child with normal skin pigmentation. A. Is the piebald trait dominant or recessive? _____________________________________ B. Explain how you came to this conclusion? In other words, prove this by showing what the situation would have to be in order to be the opposite of what you chose. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ C. What are the genotypes of the parents? ________________________________________ 8. Shown below is one pair of chromosomes from a female and male cat. Each carries 4 genes as shown. E s l c chromosome from egg of female cat e S l C chromosome from sperm of male cat A. Given the information below, describe what this kitten will look like ___________________ __________________________________________________________________________ B. (E) is dominant for electric green eyes and (e) is recessive for brown eyes (S) is dominant for silky hair and (s) is recessive for hairless (L) is dominant for long hair and (l) is recessive for straight hair (C) is dominant for calico color and (c) is recessive for solid color 9. A rare recessive allele inherited in a Mendelian manner, causes the disease Cystic Fibrosis. Suppose a man who is phenotypically normal, and whose mother was normal but his father had cystic fibrosis, marries a woman who has Cystic Fibrosis. The parents of the woman are both normal. They are concerned that their children may inherit this disorder and decide to come to you, their genetic counselor. You begin by drawing a pedigree chart to show them the inheritance pattern in their family thus far. In the space below draw the pedigree chart which you would show them. Label the man, woman and each of their parents giving their possible genotypes. 10. Bronze turkeys have at least one dominant allele “R”. Red turkeys are homozygous for its recessive allele and have the genotype “rr”. Another dominant allele “H” produces normal feathers, and the recessive genotype “hh” produces feathers lacking webbing resulting in a condition known as “hairy”. Show the resulting F2 progeny from a cross between a homozygous bronze, hairy bird and one which is a homozygous red, normal-feathered bird. Be sure you read the question carefully. Show all of your work using Punnett Square(s) and be sure to give both the genotypic and phenotypic ratios showing all of the genotypes and phenotypes clearly in your answer. First generation Cross : GG (homozygous dominant) x gg (homozygous recessive) G G Gg Gg g G g Gg Gg Genotype: 100% hetero Phenotype: 100% purple Second generation Cross : Gg (heterozygous) x Gg (heterozygous) G g GG Gg G G g Gg gg Genotype: 25%. Temple University Gray Male Is Homozygous Recessive Discussion homozygous dominant 50% heterozygous 25% homozygous recessive Phenotype: 75% purple, 25% white Third generation Cross : Gg (heterozygous) x gg (homozygous recessive) G g Gg gg g G g Gg gg Genotype: 50% heterozygous 50% homozygous recessive Phenotype: 50% purple, 50% white Fourth generation Cross : gg (homozygous recessive) x gg (homozygous recessive) g g gg gg g G g gg gg Genotype: 100% homozygous recessive Phenotype: 100% white First generation dihybrid cross: GGTT (homozygous purple tall) vs ggtt (homozygous white short GT GgTt GT GgTt gt gt GT GgTt GgTt GgTt gt GgTt GgTt GgTt gt GgTt GgTt GgTt GgTt GT GgTt GgTt GgTt GgTt Genotype: 100% heterozygous Phenotype: 100% green tall Second generation dihybrid cross: GgTt (heterozygous) vs GgTt (heterozygous) GT GT Gt gT gt GGTT Gt GGTt GGTt GGTt GgTT GgTt GgTt Ggtt gT GgTT GgTt ggTT ggTt gt GgTt Ggtt ggTt ggtt First generation Cross : XHXH (Normal Female) x XhY (hemophiliac male) Genotype: 50% hetero, 50% homo dominant Phenotype: 50% normal male, 50% carrier female XH XH XHXh XHXh Xh G Y XHY XHY Second generation Cross : XHXh (Normal Female) x XhY (hemophiliac male) XH Xh XHXh Xh Xh Xh G Y XHY Xh Y Genotype: 25% hetero, 50% homo recessive, 25% homo dominant Phenotype: 25% carrier female, 25% hemophiliac female, 25% normal male, 25% hemophiliac male …Name ______________________________ GENETICS ASSIGNMENT-FALL 2020 1. In cats, black coat color is dominant over gray. A female black cat whose mother is gray mates with a gray male. If this female has a litter of kittens, what is the probability that the mating will produce gray kittens? Show your work in a Punnett square and give the probability. 2. In foxes, silver-black coat color is governed by a recessive allele “b” and red color by the dominant allele “B”. Determine the genotypic and phenotypic ratios expected from the following matings. Be sure to show your work and give the ratios for each mating. A. Pure red x carrier B. Silver-black x Pure red 3. Any measurable or distinctive characteristic or trait possessed by an individual is referred to as the individual’s _________________________________. 4. If an individual is the result of the union of gametes carrying different alleles one would refer to that individual’s genotype as being ___________________________. 5. The normal cloven-footed (a hoof which is split into two toes) condition in swine is produced by the homozygous recessive genotype “mm”. A mule-footed condition is produced by the dominant gene “M”. White coat color is governed by the dominant allele “B” while the black coat color is governed by its recessive allele “b”. Suppose a white, mule-footed sow (female) is mated to a black, cloven-footed boar (male) and produces several litters. Of the 26 offspring produced from this mating, all were found to be white with mule feet. A. What is the most probable genotype of the sow?______________________ B. The next litter produced 8 white, mule-footed offspring and 1 white, cloven-footed pig. Now, what is the most probable genotype of the sow?____________________ 6. Hemophilia is a sex linked recessive trait. A. Is it possible that two normal parents could produce a girl with hemophilia?_________ B. A boy with hemophilia? ___________________ Explain each answer and support your reasoning with a Punnet Square 7. Piebald spotting is a condition found in humans in which there are patches of skin that lack pigmentation. The condition results from the inability of pigment producing cells to migrate properly during development. Two adults with piebald spotting have one child who has the trait and a second child with normal skin pigmentation. A. Is the piebald trait dominant or recessive?Temple University Gray Male Is Homozygous Recessive Discussion _____________________________________ B. Explain how you came to this conclusion? In other words, prove this by showing what the situation would have to be in order to be the opposite of what you chose. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ C. What are the genotypes of the parents? ________________________________________ 8. Shown below is one pair of chromosomes from a female and male cat. Each carries 4 genes as shown. E s l c chromosome from egg of female cat e S l C chromosome from sperm of male cat A. Given the information below, describe what this kitten will look like ___________________ __________________________________________________________________________ B. (E) is dominant for electric green eyes and (e) is recessive for brown eyes (S) is dominant for silky hair and (s) is recessive for hairless (L) is dominant for long hair and (l) is recessive for straight hair (C) is dominant for calico color and (c) is recessive for solid color 9. A rare recessive allele inherited in a Mendelian manner, causes the disease Cystic Fibrosis. Suppose a man who is phenotypically normal, and whose mother was normal but his father had cystic fibrosis, marries a woman who has Cystic Fibrosis. The parents of the woman are both normal. They are concerned that their children may inherit this disorder and decide to come to you, their genetic counselor. You begin by drawing a pedigree chart to show them the inheritance pattern in their family thus far. In the space below draw the pedigree chart which you would show them. Label the man, woman and each of their parents giving their possible genotypes. 10. Bronze turkeys have at least one dominant allele “R”. Red turkeys are homozygous for its recessive allele and have the genotype “rr”. Another dominant allele “H” produces normal feathers, and the recessive genotype “hh” produces feathers lacking webbing resulting in a condition known as “hairy”. Show the resulting F2 progeny from a cross between a homozygous bronze, hairy bird and one which is a homozygous red, normal-feathered bird. Be sure you read the question carefully. Show all of your work using Punnett Square(s) and be sure to give both the genotypic and phenotypic ratios showing all of the genotypes and phenotypes clearly in your answer. First generation Cross : GG (homozygous dominant) x gg (homozygous recessive) G G Gg Gg g G g Gg Gg Genotype: 100% hetero Phenotype: 100% purple Second generation Cross : Gg (heterozygous) x Gg (heterozygous) G g GG Gg G G g Gg gg Genotype: 25% homozygous dominant 50% heterozygous 25% homozygous recessive Phenotype: 75% purple, 25% white Third generation Cross : Gg (heterozygous) x gg (homozygous recessive) G g Gg gg g G g Gg gg Genotype: 50% heterozygous 50% homozygous recessive Phenotype: 50% purple, 50% white Fourth generation Cross : gg (homozygous recessive) x gg (homozygous recessive) g g gg gg g G g gg gg Genotype: 100% homozygous recessive Phenotype: 100% white First generation dihybrid cross: GGTT (homozygous purple tall) vs ggtt (homozygous white short GT GgTt GT GgTt gt gt GT GgTt GgTt GgTt gt GgTt GgTt GgTt gt GgTt GgTt GgTt GgTt GT GgTt GgTt GgTt GgTt Genotype: 100% heterozygous Phenotype: 100% green tall Second generation dihybrid cross: GgTt (heterozygous) vs GgTt (heterozygous) GT GT Gt gT gt GGTT Gt GGTt GGTt GGTt GgTT GgTt GgTt Ggtt gT GgTT GgTt ggTT ggTt gt GgTt Ggtt ggTt ggtt First generation Cross : XHXH (Normal Female) x XhY (hemophiliac male) Genotype: 50% hetero, 50% homo dominant Phenotype: 50% normal male, 50% carrier female XH XH XHXh XHXh Xh G Y XHY XHY Second generation Cross : XHXh (Normal Female) x XhY (hemophiliac male) XH Xh XHXh Xh Xh Xh G Y XHY Xh Y Genotype: 25% hetero, 50% homo recessive, 25% homo dominant Phenotype: 25% carrier female, 25% hemophiliac female, 25% normal male, 25% hemophiliac male …Temple University Gray Male Is Homozygous Recessive Discussion First generation Cross : GG (homozygous dominant) x gg (homozygous recessive) G G Gg Gg g G g Gg Gg Genotype: 100% hetero Phenotype: 100% purple Second generation Cross : Gg (heterozygous) x Gg (heterozygous) G g GG Gg G G g Gg gg Genotype: 25% homozygous dominant 50% heterozygous 25% homozygous recessive Phenotype: 75% purple, 25% white Third generation Cross : Gg (heterozygous) x gg (homozygous recessive) G g Gg gg g G g Gg gg Genotype: 50% heterozygous 50% homozygous recessive Phenotype: 50% purple, 50% white Fourth generation Cross : gg (homozygous recessive) x gg (homozygous recessive) g g gg gg g G g gg gg Genotype: 100% homozygous recessive Phenotype: 100% white First generation dihybrid cross: GGTT (homozygous purple tall) vs ggtt (homozygous white short GT GgTt GT GgTt gt gt GT GgTt GgTt GgTt gt GgTt GgTt GgTt gt GgTt GgTt GgTt GgTt GT GgTt GgTt GgTt GgTt Genotype: 100% heterozygous Phenotype: 100% green tall Second generation dihybrid cross: GgTt (heterozygous) vs GgTt (heterozygous) GT GT Gt gT gt GGTT Gt GGTt GGTt GGTt GgTT GgTt GgTt Ggtt gT GgTT GgTt ggTT ggTt gt GgTt Ggtt ggTt ggtt First generation Cross : XHXH (Normal Female) x XhY (hemophiliac male) Genotype: 50% hetero, 50% homo dominant Phenotype: 50% normal male, 50% carrier female XH XH XHXh XHXh Xh G Y XHY XHY Second generation Cross : XHXh (Normal Female) x XhY (hemophiliac male) XH Xh XHXh Xh Xh Xh G Y XHY Xh Y Genotype: 25% hetero, 50% homo recessive, 25% homo dominant Phenotype: 25% carrier female, 25% hemophiliac female, 25% normal male, 25% hemophiliac male … First generation Cross : GG (homozygous dominant) x gg (homozygous recessive) G G Gg Gg g G g Gg Gg Genotype: 100% hetero Phenotype: 100% purple Second generation Cross : Gg (heterozygous) x Gg (heterozygous) G g GG Gg G G g Gg gg Genotype: 25% homozygous dominant 50% heterozygous 25% homozygous recessive Phenotype: 75% purple, 25% white Third generation Cross : Gg (heterozygous) x gg (homozygous recessive) G g Gg gg g G g Gg gg Genotype: 50% heterozygous 50% homozygous recessive Phenotype: 50% purple, 50% white Fourth generation Cross : gg (homozygous recessive) x gg (homozygous recessive) g g gg gg g G g gg gg Genotype: 100% homozygous recessive Phenotype: 100% white First generation dihybrid cross: GGTT (homozygous purple tall) vs ggtt (homozygous white short GT GgTt GT GgTt gt gt GT GgTt GgTt GgTt gt GgTt GgTt GgTt gt GgTt GgTt GgTt GgTt GT GgTt GgTt GgTt GgTt Genotype: 100% heterozygous Phenotype: 100% green tall Second generation dihybrid cross: GgTt (heterozygous) vs GgTt (heterozygous) GT GT Gt gT gt GGTT Gt GGTt GGTt GGTt GgTT GgTt GgTt Ggtt gT GgTT GgTt ggTT ggTt gt GgTt Ggtt ggTt ggtt First generation Cross : XHXH (Normal Female) x XhY (hemophiliac male) Genotype: 50% hetero, 50% homo dominant Phenotype: 50% normal male, 50% carrier female XH XH XHXh XHXh Xh G Y XHY XHY Second generation Cross : XHXh (Normal Female) x XhY (hemophiliac male) XH Xh XHXh Xh Xh Xh G Y XHY Xh Y Genotype: 25% hetero, 50% homo recessive, 25% homo dominant Phenotype: 25% carrier female, 25% hemophiliac female, 25% normal male, 25% hemophiliac male … Is it possible that two normal parents could produce a girl with hemophilia?_________ B. A boy with hemophilia? ___________________ Explain each answer and support your reasoning with a Punnet Square 7. Piebald spotting is a condition found in humans in which there are patches of skin that lack pigmentation. The condition results from the inability of pigment producing cells to migrate properly during development. Two adults with piebald spotting have one child who has the trait and a second child with normal skin pigmentation. A. Is the piebald trait dominant or recessive? _____________________________________ B. Explain how you came to this conclusion? In other words, prove this by showing what the situation would have to be in order to be the opposite of what you chose. ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________ C. What are the genotypes of the parents? ________________________________________ 8. Shown below is one pair of chromosomes from a female and male cat. Each carries 4 genes as shown. E s l c chromosome from egg of female cat e S l C chromosome from sperm of male cat A. Given the information below, describe what this kitten will look like ___________________ __________________________________________________________________________ B. (E) is dominant for electric green eyes and (e) is recessive for brown eyes (S) is dominant for silky hair and (s) is recessive for hairless (L) is dominant for long hair and (l) is recessive for straight hair (C) is dominant for calico color and (c) is recessive for solid color 9. A rare recessive allele inherited in a Mendelian manner, causes the disease Cystic Fibrosis. Suppose a man who is phenotypically normal, and whose mother was normal but his father had cystic fibrosis, marries a woman who has Cystic Fibrosis. The parents of the woman are both normal. They are concerned that their children may inherit this disorder and decide to come to you, their genetic counselor. You begin by drawing a pedigree chart to show them the inheritance pattern in their family thus far. In the space below draw the pedigree chart which you would show them. # Label the man, woman and each of their parents giving their possible genotypes. 10. Bronze turkeys have at least one dominant allele “R”. Red turkeys are homozygous for its recessive allele and have the genotype “rr”. Another dominant allele “H” produces normal feathers, and the recessive genotype “hh” produces feathers lacking webbing resulting in a condition known as “hairy”. Show the resulting F2 progeny from a cross between a homozygous bronze, hairy bird and one which is a homozygous red, normal-feathered bird. Be sure you read the question carefully. Show all of your work using Punnett Square(s) and be sure to give both the genotypic and phenotypic ratios showing all of the genotypes and phenotypes clearly in your answer. First generation Cross : GG (homozygous dominant) x gg (homozygous recessive) G G Gg Gg g G g Gg Gg Genotype: 100% hetero Phenotype: 100% purple Second generation Cross : Gg (heterozygous) x Gg (heterozygous) G g GG Gg G G g Gg gg Genotype: 25% homozygous dominant 50% heterozygous 25% homozygous recessive Phenotype: 75% purple, 25% white Third generation Cross : Gg (heterozygous) x gg (homozygous recessive) G g Gg gg g G g Gg gg Genotype: 50% heterozygous 50% homozygous recessive Phenotype: 50% purple, 50% white Fourth generation Cross : gg (homozygous recessive) x gg (homozygous recessive) g g gg gg g G g gg gg Genotype: 100% homozygous recessive Phenotype: 100% white First generation dihybrid cross: GGTT (homozygous purple tall) vs ggtt (homozygous white short GT GgTt GT GgTt gt gt GT GgTt GgTt GgTt gt GgTt GgTt GgTt gt GgTt GgTt GgTt GgTt GT GgTt GgTt GgTt GgTt Genotype: 100% heterozygous Phenotype: 100% green tall Second generation dihybrid cross: GgTt (heterozygous) vs GgTt (heterozygous) GT GT Gt gT gt GGTT Gt GGTt GGTt GGTt GgTT GgTt GgTt Ggtt gT GgTT GgTt ggTT ggTt gt GgTt Ggtt ggTt ggtt First generation Cross : XHXH (Normal Female) x XhY (hemophiliac male) Genotype: 50% hetero, 50% homo dominant Phenotype: 50% normal male, 50% carrier female XH XH XHXh XHXh Xh G Y XHY XHY Second generation Cross : XHXh (Normal Female) x XhY (hemophiliac male) XH Xh XHXh Xh Xh Xh G Y XHY Xh Y Genotype: 25% hetero, 50% homo recessive, 25% homo dominant Phenotype: 25% carrier female, 25% hemophiliac female, 25% normal male, 2