Answer number 3,4 and 5 by using punnett square 58B. Answer the following problems on a separate sheet of papel.1. A homozygous dominant yellow gumamela (YY) is crossed with a homozygousdominant red gumamela (RR). Determine the phenotypic and genotypicprobabilities of this cross.2. Codominance governs the hair color of short-horn cattles. Red alleles arecodominant with white alleles, and heterozygotes (roans) have red and whitehairs. Cross two roans and determine the genotypic and phenotypic probabilityof this cross.3. Make a Punnett square to show how two people who do not have color-blindness can have a child who is color-blind. What is the probability that thecolor-blind child will be male? What is the probability that the color-blind childwill be female?4. Using a Punnett square, demonstrate how two individuals with Type A bloodcan have a child with Type O blood.5. Cross a person with Type AB blood and a person with Type O blood. What arethe genotypic and phenotypic ratios of this cross?Init I Thi

Colour blindness It is a X-chromosome linked recessive disease.

8 aele 3. Make a Punnett square to show how two people who do not have color- blindness can have a child who is color-blind. What is the probability that the color-blind child will be male? What is the probability that the color-blind child will be female? 4. Using a Punnett square, demonstrate how two individuals with Type A blood can have a child with Type O blood. 5. Cross a person with Type AB blood and a person with Type O blood. What are the genotypic and phenotypic ratios of this cross? Init I Thi

8 aele 3. Make a Punnett square to show how two people who do not have color- blindness can have a child who is color-blind. What is the probability that the color-blind child will be male? What is the probability that the color-blind child will be female? 4. Using a Punnett square, demonstrate how two individuals with Type A blood can have a child with Type O blood. 5. Cross a person with Type AB blood and a person with Type O blood. What are the genotypic and phenotypic ratios of this cross? Init I Thi

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What if the couple wanted prenatal testing so that a normal fetus could be aborted?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
B What is the genotype ratio? What is the phenotype ratio? 3. What are the expected genotype and phenotype ratios in the following genetic conditions? Use scratch paper to do the Punnett square if needed but you do not need to draw it on the worksheet. a. Monohybrid cross between 2 heterozygous individuals (Aa x Aa) Genotype ratio: 1:2:1 Phenotype ratio: 3:1 G q 8:32 a b. Dihybrid cross between 2 heterozygous individuals (AaBb x AaBb) Genotype ratio: Phenotype ratio: 4. Both Mrs. Smith and Mrs. Jones had babies the same day in the same hospital. Mrs. Smith took home a baby girl, whom she named Sharon. Mrs. Jones took home a girl, whom she named Jane. Mrs. Jones began to suspect, however, that the child had been accidentally switched with Mrs. Smith baby in the nursery. Blood test were made; Mr. Smith was type A, Mrs. Smith was type B, Mr. Jones was type A, Mrs. Jones was type A. Sharon was type O, and Jane was type. B. Had a mix-up occurred? Use scratch paper, you do not need to…
Please consider the pedigree below. There are no cases of false paternity. I B II A 2 3 III AB (A IV в 1 a. Which individual/s definitely has/have Bombay phenotype in the descendants of I-1 and I-2? b. What are the genotypes of individuals II-2 and III-2 at the ABO and H loci?
Dihybrid Cross Practice In a breed of dog called a Doberman, black fur is dominant to brown fur and floppy ears are dominant to straight ears. These letters represent the genotypes and phenotypes of the dogs: EE = floppy ears Ee = floppy ears bb = brown fur ee = pointed ears BB = black fur %3D Bb = black fur 1. A female dog with the genotype BBee is crossed with a male dog with the genotype bbEe The square is set up below. Fill it out and determine the phenotypes and proportions in the offspring. Ве Ве Ве Ве bE How many out of 16 have: black fur and floppy ears? be black fur and pointed ears? brown fur and floppy ears? bE brown fur and pointed ears? be
eritance / 13 of 15 Black hair color is dominant to white hair color in mice. Interpret the Punnett squrare below to determine the expected phenotypic ratio for the offpring of a homozygous black mouse and a white mouse. В Bb Bb Bb Bb O A. 4 Black: 0 White O B. 3 Black: 1 White O C 2 Black: 2 White O D. O Black: 4 White acer
DIHYBRID CROSS Heterozygous Yellow and heterozygous round seed crossed with homozygous yellow and heterozygous round seed. Find the following: 1. Alleles of both parents (given problems) 2. Genotype 3. Phenotype ratio 4. Write the punnet square -Please answer this on the paper and explain the answer step by step. Thank you asap
BIU E I = E E 11 Section 2: Codominance 3. In shorthorn cattle, coat color may be red, white, or roan. Roan is an intermediate phenotype expressed as a patches of red and patches of white hairs. The following data are obtained from various crosses: Red x red all red White x white all white Red x white all roan Roan x roan 25 % red, 50% roan, 25% white a. From these results, determine the appropriate allele symbols and indicate which genotypes lead to which phenotypes. b. What are the genotypes of the parents and offspring in each cross above? C. Predict the results of a red x roan cross. d. Predict the results of a white x roan cross. acer
II. Solve the following problems. 1. Mendel crossed a plant that was heterozygous for height and heterozygous for pea shape with a plant that was homozygous dominant for height and heterozygous for shape. What are the possible genotypes and phenotypes of the offspring? Dominant for height= T (tall) Recessive for height =t (short) Dominant for pea shape = R (round) pea %3D Recessive for pea shape =r(wrinkled) P1: Genotypes: Phenotypes: Genotypic Ratio: Phenotypic Ratio
What will be the results of the following crosses, where N is black, n is brown, L is short hair and l is long hair. Make the tables and the corresponding explanation: 1. Crossing a Nnll woman dog with a Nnll dog. 2. Crossing a NnLl woman dog with a NnLl dog 3. Crossing a woman dog nnLl with a dog NNll
V. Two Gene Pairs (Extra Credit) 1... A. In peas, the allele for tall plants (T) is dominant over the allele for short plants (t). The allele for smooth peas (S) is dominant over the allele for wrinkled peas (s). Use this information to cross the following parents. I 1. heterozygous tall and smooth X heterozygous tall and smooth 2. heterozygous tall, wrinkled X short, wrinkled AL.