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Cock feathering in chickens is an autosomal recessive trait that is limited to males. List all possible genotypes for the chicken shown in pic
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- Cock feathering in chickens is an autosomal recessive trait that is limited to males. List all possible genotypes for the chicken shown in pic (male hen)Take the example of B-thalassemia, an autosomal recessive genetic disease that particularly affects people from around the Mediterranean. This disease is associated with an anomaly of hemoglobin, a protein essential for the transport of oxygen, which is composed of four chains: two alpha (a) and two beta (B). In case of B-thalassemia, the ẞ chains are produced in insufficient or no quantity in an individual homozygous recessive resulting in insufficient production of overall hemoglobin leading to anemia and other physiological challenges. The gene that controls the synthesis of the ẞ chains is located on chromosome 11. Here is part of the coding portion of this gene (which controls a total of 146 amino acids and of which you only see the portion 36 to 41) and one of the targeted mutations: 1. Give the sequence of amino acids from the template and mutated strands. 2. What type of point mutation is it? 3. Using the principles of the theory of evolution, explain briefly and generally why…In Drosophila, an X-linked recessive mutation, Xm causes miniature wings. List the F2 phenotypic ratios if: a miniature-winged female is crossed with a normal male and a miniature-winged male is crossed with a normal female. What would the phenotypic ratio from (a) be if the miniature-winged gene were autosomal? Assume in all cases that the P1 individuals are true-breeding.
- 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?The following pedigree illustrates the inheritance of Nance–Horan syndrome, a rare genetic condition in which affected people have cataracts and abnormally shaped teeth.This pedigree traces the inheritance of a rare disease in humans. a. Based on this pedigree, is the allele for this disease dominant or recessive? Explain. b. What genotypes are possible for the individuals labeled 1, 2, and 3?
- Anhidrotic ectodermal dysplasia is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. This trait is usually seen in men, but women who are heterozygous carriers of the trait often have irregular patches of skin with few or no sweat glands (see the illustration below). a. Explain why women who are heterozygous carriers of a recessive gene for anhidrotic ectodermal dysplasia have irregular patches of skin lacking sweat glands. b. Why does the distribution of the patches of skin lacking sweat glands differ among the females depicted in the illustration, even between the identical twins?IN DROSOPHILA, AN X-LINKED RECESSIVE MUTATION, Xm CAUSES MINIATURE WINGS. LIST THE F₂ PHENOTYPIC RATIOS IF: A MINIATURE-WINGED FEMALE IS CROSSED WITH A NORMAL MALE AND A MINIATURE-WINGED MALE IS ● ● CROSSED WITH A NORMAL FEMALE. WHAT WOULD THE PHENOTYPIC RATIO FROM (A) BE IF THE MINIATURE- WINGED GENE WERE AUTOSOMAL? ASSUME IN ALL CASES THAT THE P1 INDIVIDUALS ARE TRUE-BREEDING.In Alaskan Malamute dogs, an autosomal recessive allele causes a condition in which the cartilage in the front legs fails to develop properly. This results in a type of dwarfism. Dogs with the dominant wild-type allele are of normal height. In the same breed of dogs, there is an autosomal dominant allele that codes for a curled tail and a recessive allele that codes for an uncurled tail. Show the possible genotypes for the offspring of a dog heterozygous for height and tail curl crossed with a dog recessive for both traits. Your answer must include the use of a Punnett square and a legend indicating the allele symbols used.
- Albinism in humans is inherited as a simple recessive trait.Determine the genotypes of the parents and offspring for the following 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.A recessive maternal effect mutant in zebrafish, called ichabod, results in embryos lacking heads that are non-viable. You have been instructed to identify females that are homozygous for the ichabod mutant allele. At your disposal are a tank of wild-type fish (males and females), a tank of male and female parental fish that are all heterozygous for the ichabod mutant allele (ichabodl+), and a tank of F1 fish derived from a cross between a heterozygous male and heterozygous female (ichabodl+). Which of the following would be a way to identify females that are homozygous mutant, i.e. ichabodlichabod? Select all answers that would work. Cross F1 females to F1 males and observe their offspring. Crosses that produce headless offspring came from a homozygous female. a. Cross F1 males to females from the parental tank and observe their offspring. 25% of these crosses should produce headless offspring. b. Cross F1 females to F1 males to make the F2 generation. Cross F2 females to F2 males and…The DNA of every individual in the pedigree shown below has been sequenced at the causative locus. 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. Based on this small pedigree, what is the level of penetrance for the condition? Please give your answer as a WHOLE percentage, give the number only, no percentage symbol. Answer: The level of penetrance for the condition shown in the pedigree below is Blank 1 percent. 1:1 1:2 Il:1 I1:2 I1:3 Il:4 I1:5 I1:6 II:1 I:2 III:3 III:4 III:3 III:6 III:7 III:8 III:9 III:10 III:11 III12 II:13 III:14 IV:1 | IV:2 IV:3 IV:4 IV:5 IV:6 IV:7 IV:8 IV:9 IV:10 IV:11 IV:12 IV:13 IV:14 IV:15 IV:16 IV:17 IV:18 IV:19 V:1 V:2 V:3 V:4 V:5 V:6 V:7 V:8 V:9 V:10 V:11 V:12