At least two pairs of genes control eye color. Both pairs influence the production of the pigment, melanin, but act independently. One pair of alleles is B (Brown color; dominant) and b (blue color); the other pair is A (pigment production; dominant) and a (no pigment production; albino). The gene pair aais epistatic to (masks) B and b and produces the nonpigmented eyes of the albino. What is the typeof gene interaction that exists between the two gene pairs?
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At least two pairs of genes control eye color. Both pairs influence the production of the pigment, melanin, but act independently. One pair of alleles is B (Brown color; dominant) and b (blue color); the other pair is A (pigment production; dominant) and a (no pigment production; albino). The gene pair aais epistatic to (masks) B and b and produces the nonpigmented eyes of the albino. What is the typeof gene interaction that exists between the two gene pairs?
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- Junctional epidermolysis bullosa (JEB) is a severe skin disorder that results in blisters over the entire body. The disorder is caused by autosomal recessive mutations at any one of three loci that help to encode laminin 5, a major component in the dermal–epidermal basement membrane. Leena Pulkkinen and colleagues described a male newborn who was born with JEB and died at 2 months of age (L. Pulkkinen et al. 1997. American Journal of Human Genetics 61:611–619); the child had healthy, unrelated parents. Chromosome analysis revealed that the infant had 46 normal-appearing chromosomes. Analysis of DNA showed that his mother was heterozygous for a JEB-causing allele at the LAMB3 locus, which is on chromosome 1. The father had two normal alleles at this locus. DNA fingerprinting demonstrated that the male assumed to be the father had, in fact, conceived the child. Q. Assuming that no new mutations occurred in this family, explain the presence of an autosomal recessive disease in the child…Junctional epidermolysis bullosa (JEB) is a severe skin disorder that results in blisters over the entire body. The disorder is caused by autosomal recessive mutations at any one of three loci that help to encode laminin 5, a major component in the dermal–epidermal basement membrane. Leena Pulkkinen and colleagues described a male newborn who was born with JEB and died at 2 months of age (L. Pulkkinen et al. 1997. American Journal of Human Genetics 61:611–619); the child had healthy, unrelated parents. Chromosome analysis revealed that the infant had 46 normal-appearing chromosomes. Analysis of DNA showed that his mother was heterozygous for a JEB-causing allele at the LAMB3 locus, which is on chromosome 1. The father had two normal alleles at this locus. DNA fingerprinting demonstrated that the male assumed to be the father had, in fact, conceived the child. Q. How might you go about proving your explanation? Assume that a number of genetic markers are available for each chromosome.1) Perception of a bitter taste from phenylthiocarbamide (PTC) is dominant (T) to not being able to taste PTC. Albinism is recessive to normal pigmentation (A). The genes controlling these traits are on separate chromosomes. a. A normally pigmented woman who cannot taste PTC has a father who is an albino taster, marries a many homozygous for normal pigmentation who is a PTC taster, but whose mother did not taste PTC. What are the genotypes of the woman and man listed here? b. Assuming the man and woman in the above section have children, what percentage of these children will be albino? c. What percentage of these children will have the same phenotype as their mother?
- The term epistasis is used to refer to a situation in which the expression of a gene is influenced by another independently inherited gene. In labs, the gene B determines how much pigment is made. The dominant allele B results in black fur and the recessive allele b results in brown (chocolate) fur. A separate gene, E, codes for a protein that determines whether the pigment is deposited in the hair. (E = pigment deposited; e = pigment not deposited). What genotypes are possible for a black lab? What genotypes are possible for a chocolate lab? What genotypes are possible for a yellow Lab? In a cross between a black lab that is homozygous for both alleles and a yellow lab that is homozygous for both alleles, what would the genotype and phenotype of the offspring be? Imagine you cross two heterozygous parents. What are the genotypic and phenotypic ratios of the offspring?Red–green color blindness is caused by an X-linked recessive genetic defect. Hence females rarely exhibit the red–green colorblind phenotype but may be carriers of the defective gene. When a narrow beam of red or green light is projected onto some areas of the retina of such a female carrier, she can readily diff erentiate the two colors, but on other areas she has diffi culty in doing so. Explain.Red-green color blindness, in humans, is a sex-linked trait controlled by alleles on the X chromosome. Normal color vision (X+) is dominant to colorblindness (Xc). (NOTE: You could write this as XC for the normal allele and Xc for the recessive colorblind allele. Unfortunately, capital C and lowercase c are hard to tell apart as superscripts, especially in type. I will use C -- "X-plus" -- for the normal allele.) If a colorblind man marries a woman with normal vision and they have a colorblind son, what are the genotypes of the father and the mother?
- In human beings, the gene for red‑green colorblindness (r) is sex‑linked and recessive to its allele for normal vision (R), while the gene for freckles (F) is autosomal and dominant over its allele for nonfreckled (f). A nonfreckled, normal‑visioned woman whose father was freckled and colorblind, marries a freckled, colorblind man whose mother was nonfreckled. What is the genotype of the woman's father? What is the probability that the couple's first child will be a non-freckled, normal visioned girl? What is the probability that the first two children born to the couple will be freckled and colorblind girls?In human beings, the gene for red‑green colorblindness (r) is sex‑linked and recessive to its allele for normal vision (R), while the gene for freckles (F) is autosomal and dominant over its allele for nonfreckled (f). A nonfreckled, normal‑visioned woman whose father was freckled and colorblind, marries a freckled, colorblind man whose mother was nonfreckled. What is the probability that the first child born to the couple will either be a freckled, colorblind boy or a non‑freckled, normal visioned girl or a non-freckled, normal visioned boy? What is the probability that the first four children born to the couple will be freckled and normal visioned girls?In humans, hair color is controlled by two interactinggenes. The same pigment, melanin, is present in bothbrown-haired and blond-haired people, but brown hairhas much more of it. Brown hair (B ) is dominant to blond(b). Whether any melanin can be synthesized depends onanother gene. The dominant form of this second gene (M )allows melanin synthesis; the recessive form (m) preventsmelanin synthesis. Homozygous recessives (mm) are albino.What will be the expected proportions of phenotypes in thechildren of the following parents?a. BBMM * BbMmb. BbMm * BbMmc. BbMm * bbmm
- In man, muscular dystrophy is a condition in which the muscles waste away during early life and may result in a shorter life expectancy. It is due to a sex-linked, recessive gene. A certain couple has five children – three boys (ages 1yr, 3yrs, and 10yrs old) and two girls (ages 5yrs and 7yrs old). The oldest boy shows the symptoms of this disease. You are their family physician and they come to you for advice. What would you tell them about the chances of their other children developing the disease?Color-blindness is a sex-linked trait. A boy, whose parents and grandparents had normal vision, is color-blind. What are the genotypes for his mother and his maternal grandparents? Show all workA recessive epistasis cross gives in the F2 a 9:3:4 phenotypic ratio. What is the expected phenotypic ratio when the F1 (BbEe) is crossed with the double recessive homozygote (bbee)? Keeping in mind ee masks the expression of B.