Suppose a man is heterozygous for heterochromia, an autosomal dominant disorder which causes two different-colored eyes in an individual, produced 25-offspring with his normal-eyed wife. Of their children, 15 were heterochromatic and 10 were-normal. Calculate the chi-square value for this observation.
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- Suppose a man is heterozygous for heterochromia, an autosomal dominant disorder which causes two different‑colored eyes in an individual, produced 25‑offspring with his normal‑eyed wife. Of their children, 16 were heterochromatic and 9 were‑normal. Calculate the chi‑square value for this observation.152 Phenylketonuria (PKU) is a disorder caused by a recessive allele. Two carrier individuals have progeny. Answer the following questions in order and show solutions whenever relevant. If they have a normal child, what is the probability that he or she will be heterozygous? If they have three children, what is the probability of having 2 affected children and one normal child?E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescentwhite seedlings, and 260 yellow seedlings . Q. Explain how color is determined in these seedlings.
- In addition to the allelic pair determining pattern baldness in man (B,b), consider early baldness to be due to another autosomal allele (E) on a different pair of chromosomes and also dominant in males but recessive in females. The phenotype for ee may be late or nonbaldness depending on sex and the genotype for B, b alleles. Two doubly heterozygous persons marry. What is the phenotype of the male parent? What is the phenotype of the female parent? Give the phenotypic ratio expected among male children of couples such as this one. Show corresponding genotypes for each phenotype mentioned in your phenotypic ratio. Give the phenotypic ratio expected among female children of couples such as this one. Show corresponding genotypes for each phenotype mentioned in your phenotypic ratio.Assuming brown eyes (B) are dominant over blue eyes (b), determine the genotypes of all the following individuals. The blue-eyed son of two brown-eyed parents marries a brown-eyed woman whose mother was brown eyed and whose father was blue eyed. Their child is blue eyed.my widows peak phenotype is aa. what is my widows peak genotype.
- Explain why it is possible for the proband in the following pedigree to have children of blood types A, B, and AB. Considering epistatic genes, what are the possible genotypes of II-2?All the non-shaded individuals are wild type apart from III.1. III.1 has been proven to have the causative mutation for this Autosomal Dominant condition, but they exhibit no symptoms.What is the percentage level of penetrance for the condition in the diagram?Pompe disease is a glycogen storage disorder caused by a lack of ⍺-glucosidase, the enzyme that converts glycogen to glucose in the muscles. Babies born with this disorder die by the age of two. Pompe disease is a genetic disorder caused by the presence of two recessive autosomal alleles. A man and a woman heterozygous for the condition have two female unaffected children. Determine the probability, expressed as a whole number percentage, of a third baby being born a male with Pompe disease. Your answer must include the use of a Punnett square as well as a legend indicating the allele symbols used. (2 marks)
- In a dihybrid cross of two bi-allelic Mendelian genes, A (two alleles – completely dominant A and recessive a), and B (two alleles – completely dominant B and recessive b), such that the parental generation comprises of pure-bred homozygotes (i.e. AABB with aabb), what are the expected genotype and phenotype ratios in the F1 and F2, if the F1 cross is (a) an intercross and if the F1 cross is (b) a testcrossAlong with the trait in the pedigree, individual IV-6 and the woman are also both heterozygous for the autosomal dominant allele causing Huntington's disease. If they have a child, what is the probability that it will be affected by at least one of these traits? Remember to include both the trait in the pedigree and Huntington's disease in your calculations. Enter your answer to two decimal places (e.g., 0.55).In a problem involving albinism (see Problem 4), which of Mendel’spostulates are demonstrated? Problem No. 4 Albinism in humans is inherited as a simple recessive trait.Determine the genotypes of the parents and offspring for thefollowing families. When two alternative genotypes are possible,list both.(a) Two parents without albinism have five children, four withoutalbinism and one with albinism.(b) A male without albinism and a female with albinism havesix children, all without albinism.