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°. 1 2 3 5 5 6 IV 2 1. What are the genotypes of the founding parents (I-1, l-2)? 2. What is the percentage of the affected offspring (II)? 3. What is the phenotype of Il-2? 4. Is the inheritance autosomal or sex-linked? 2.
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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 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.
- Explain your answer. Below is a pedigree showing the inheritance of colorblindness in Akoto family. Çolorblindness is a recessive and X-linked trait (X-). The allele for normal vision is dominant and is represented by Xe. 2 II 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?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 SmoothPEDIGREES: 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
- 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 wavyTopic: 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.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:
- Brown eyes are dominant over blue. This is NOTA sex-linked trait. If a blue-eyed colorblind woman marries a normal visioned man who is homozygous for brown eye color, what kind of children might they expect with respect to these two traits? 5. Genotype of woman _Genotype of man What two genotypes are possible in the children? If one of the sons in turn marries a heterozygous brown-eyed, normal visioned woman, not a carrier, what kinds of children might they expect? Genotypes PhenotypesO e. Parent 2: Parent 1: Parent 2: ¡Ai QUESTION 9 Bi QUESTION 8 Let's assume that, in dragons, red scales (B) are dominant to green scales (b), and long tongues (S) are dominant to short tongues (s). The genes that determine these characteristics assort independently. A homozygous red, long-tongued dragon is crossed with a homozygous green, short-tongued dragon. If an F1 dragon is crossed to a homozygous green and homozygous short-tongued dragon, what phenotypes and proportions are expected in the offspring? O a. 100% green and long-tongued O b. ½ green and short-tongued & ½ red and long-tongued O c. ½ red and short-tongued & ½ green and long-tongued O d.% red and long-tongued, % red and short-tongued, ½ green and long-tongued, % green and short-tongued O e. 9/16 red, long-tongued, 3/16 green, long-tongued, 3/16 red, short-tongued, 1/16 green, short-tongued Save and Submit to save and submit. Click Save All Answers to save all answers. 000 MacBook AirPEDIGREES: Problem 7 This pedigree shows the inheritance of atype of X-linked color blindness. It is a recessive trait. Carriers have NOT been half-shaded in this pedigree. • QUESTIONS • • • I ODO0 I. Work out who is definitely a carrier & drag the correct shape onto them. 1 2 3 4 II 1 3 2. Is it possible to work out the geno- type of individual II-I? Explain III your reasoning. 1 2 3 IV 1