Explain your answer. Below is a pedigree showing the inheritance of colorblindness inAkoto family. Çolorblindness is a recessive and X-linked trait (X-). The allele for normalvision is dominant and is represented by Xe.2IIIV1. What are the genotypes of the founding parents (I-1, I-2)?2. What is the percentage of the affected offspring (II)?3. What is the phenotype of III-2?4. Is the inheritance autosomal or sex-linked?

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Description: Explain your answer. Below is a pedigree showing the inheritance of colorblindness inAkoto family. Çolorblindness is a recessive and X-linked trait (X-). The allele for normalvision is dominant and is represented by Xe.2IIIV1. What are the genotypes

Colorblindness is a X-linked recessive disorder. X-linked recessive disorders are those which are…

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Explain your answer. Below is a pedigree showing the inheritance of colorblindness in Akoto family. Colorblindness is a recessive and X-linked trait (X-). The allele for normal vision is dominant and is represented by X•. 모오 1 2 2 IV 2 1. What are the genotypes of the founding parents (I-1, I-2)? 2. What is the percentage of the affected offspring (II)? 3. What is the phenotype of III-2? 4. Is the inheritance autosomal or sex-linked?

Explain your answer. Below is a pedigree showing the inheritance of colorblindness in Akoto family. Colorblindness is a recessive and X-linked trait (X-). The allele for normal vision is dominant and is represented by X•. 모오 1 2 2 IV 2 1. What are the genotypes of the founding parents (I-1, I-2)? 2. What is the percentage of the affected offspring (II)? 3. What is the phenotype of III-2? 4. Is the inheritance autosomal or sex-linked?

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 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?
Explain your answer. Below is a pedigree showing the inheritançe of colorblindness in Akoto family. Colorblindness is a recessive and X-linked trait (X'). The allele for normal vision is dominant and is represented by X°. 2 II 2 3 5 II 1 2 5 6 IV 1. What are the genotypes of the founding parents (I-1, I-2)? 2. What is the percentage of the affected offspring (II)? 3. What is the phenotype of III-2? 4. Is the inheritance autosomal or sex-linked?
Explain your answer. Below is a pedigree showing the inheritance of colorblindness in Akoto family. Colorblindness is a recessive and X-linked trait (Xb). The allele for normal vision is dominant and is represented by XB. 1 2 II 2 3 4 5 6 II 1 2 5 6 IV 1 1. What are the genotypes of the founding parents (I-1, I-2)? 2. What is the percentage of the affected offspring (II)? 3. What is the phenotype of III-2? 2.
PEDIGREES: Problem (continued) This pedigree shows the inheritance of cystic fibrosis in this family. I • QUESTIONS ••. 5. What is the genotype of individual II-3? Use the letter "f" to 1 2 represent the disease allele. II 1 2 3 6. Individuals II-I and II-2 are sisters. Explain how it is possible for one sister to have cystic fibrosis but NOT the other. III 1 2 3
Trivla Game Show _Make Your Own Tri ngston.schoology.com/common-assessment-delivery/start/4789189591?action=onresume&submissionld=463322566 Dillon WF g Aa v Done In guinea pigs, black hair (B) is dominant to white hair (b) and rough hair (R) is dominant to smooth hair (r). What are all the possible genotypes of a guinea pig that has black, rough hair? (Select all that apply.) O BBRR BBRr BBrr BBRR BbRr O bbRR O bbRr O bbrr O Black O White O Rough OSmooth O Rough O Smooth
Answer the following. Letters only. 1. In mice, black color (B) is dominant over brown (b), and a solid pattern (S) is dominant over white spotted (s). Color and spotting are controlled by genes that assort independently. A homozygous black, spotted mouse is crossed with a homozygous brown, solid mouse. All the F1 mice are black and solid. What are the genotypes of the parents? * a. BBSs and bbss b. BBSs and bbSS c. BbSs and bbSs d. BBss and bbss e. BBss and bbSS f. None of the choices 2. In mice, black color (B) is dominant over brown (b), and a solid pattern (S) is dominant over white spotted (s). Color and spotting are controlled by genes that assort independently. A homozygous black, spotted mouse is crossed with a homozygous brown, solid mouse. All the F1 mice are black and solid. What will be the genotypic probability of the first filial (F1) generation offspring? * a. 100% BBSS b. 100% BbSs c. 50% BBSS and 50% BbSs d. 25% BBSS; 50% BbSs and 25% bbss e. 100% black, solid mice…
2/7 - <. Hair texture is an incompletely dominant trait in humans. The three phenotypes are curly, wavy and straight. Straight only occurs when both parents have straight hair. Curly hair also tends to follow this pattern. Only two wavy haired parents produce all three phenotypes. Please explain this observation using punnett squares. Curly x Curly straight x straight Wavy x wavy
Pedigree 2: A. What is the most likely mode of inheritance of this disease? Choose from: autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive. B State the genotypes of individuals # 1 #4. C If individual #3 has another daughter with the same partner, what is the probability that this daughter will be affected (show the disease)?
Topic: Penetrance. Petal number is controlled by a single gene in merigonias. The gene has a completely dominant wild type allele F that makes a plant have five petals and a mutant recessive six petal allele(f). However the six petal trait is only 50% penetrant. You do the cross Ff x Ff. What fraction of the progeny do you have the 6 petals? what is the meaning for 50% penetrant.
Consider the following pedigree. 하 3 10 (5 3 2 (a) What pattern of transmission is most consistent with this pedigree? (1) autosomal recessive, (2) autosomal dominant, (3) X-linked recessive, (4) X-linked dominant. (b) If individual V-2 marries a normal individual, and if the condition has a pene-trance of 85 percent, what is the probability that their second child will express the trait? (c) On the third line, what does the diamond with a 10 in the middle mean?
I. White eye cross White eye color is a recessive trait found on the X chromosome. Also called Sex-linked. This is called sex-linked. Gene symbols Xw White eye color X normal red eye color Y it’s a boy The parents were White eye color female XwXw and red eye color male XY Diagram of P1 cross: You are crossing XwY and XXw (figure out the phenotype) Diagram of F1 cross: Give expected F2 phenotypic ratio: Give expected F2 genotypic ratio: