If a recombination event occurs between the curly hair allele and the straight hair allele, what would be my possible gamete arrangements? Blue eyes Brown eyes Curly hair Straight hair
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- Consider this pedigree showing an autosomal dominant rare disorder. What is the degree of penetrance? Show your work. na оп 16 19 fa 16 R 9XDuchenne muscular dystrophy is a recessive disorder caused by a rare,loss-of-function allele that is located on the X chromosome in humans. Anunaffected woman (i.e., without disease symptoms) who is heterozygousfor the X-linked allele causing Duchenne muscular dystrophy has childrenwith a man with a functional (non-disease-causing) allele. What is theprobability that this couple will have an unaffected son?The alleles his-5 and lys-1, found in baker’s yeast, result in cellsthat require histidine and lysine for growth, respectively. A crosswas made between two haploid yeast strains that are his-5 lys-1and his+ lys+. From the analysis of 818 tetrads, the followingnumbers of tetrads were obtained:2 spores with his-5 lys+ + 2 spores with his+ lys-1 = 42 spores with his-5 lys-1 + 2 spores with his+ lys+ = 5021 spore with his-5 lys-1 + 1 spore with his-5 lys+ + 1 spore withhis+ lys-1 + 1 spore with his+ lys+ = 312A.What is the frequency of single crossovers between these twogenes?B. Based on your answer to part B, how many NPDs are expectedfrom this cross? Explain your answer. Is positive interferenceoccurring?
- The alleles his-5 and lys-1, found in baker’s yeast, result in cellsthat require histidine and lysine for growth, respectively. A crosswas made between two haploid yeast strains that are his-5 lys-1and his+ lys+. From the analysis of 818 tetrads, the followingnumbers of tetrads were obtained: 2 spores with his-5 lys+ + 2 spores with his+ lys-1 = 42 spores with his-5 lys-1 + 2 spores with his+ lys+ = 5021 spore with his-5 lys-1 + 1 spore with his-5 lys+ + 1 spore withhis+ lys-1 + 1 spore with his+ lys+ = 312 Compute the map distance between these two genes using firstthe method of calculation that considers double crossovers andthen the one that does not. Which method gives a higher value?Explain why?Using the pedigree below for an autosomal dominant disorder to determine the phase (which ones are recombinants and which are non-recombinants? Identify them by pedigree position (11, 112, etc.) (SLO4) 1 11 III 2/2 1/2 1/1 -O 2/12 2/2 2/2 2/2 2/2 1/2 2/2 1/2 2/2 1/2 2/2 a. 0.01 b. 0.1 c. 0.2 Calculate the LOD score for each theta below for the pedigree above. Note: keep lots of places behind the decimal until the very end for accuracy. (SLO4)ybrid Cross - Parental (P) vill be observing the F, offspring of the cross shown in nage. The purple colof (P) of the kernel is the result of ment called anthocyanin, which is dominant and not nked. /hat are the genotypes of the F, offspring if both nts shown are homozygous? F1 Mendel's dihybrid cross always showed a the same ratio ONLY for traits that are not linked on the same mosome and inherited together. If the genes are linked the ratio will not follow Mendel's dihybrid ratio. What is atio seen in dihybrid crosses that are not linked? F2 EXPERIMENTAL QUESTION: Are the genes for color (P) and shape (R) linked (on the same chromosome)?
- 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. What if the couple wanted prenatal testing so that a normal fetus could be aborted?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?
- A woman who sought genetic counseling is found to be heterozygousfor a chromosomal rearrangement between the second andthird chromosomes. Her chromosomes, compared to those in anormal karyotype, are diagrammed on the next page:(a) What kind of chromosomal aberration is shown?What pattern of Mendelian inheritance is represented in this pedigree and explain how you made this determination.(please refer to image attached)i. Chands syndrome (OMIM 214350) is an autosornal recessive condition characterized by very ourly hair, underdeveloped nails, and abnormally shaped eyelids. In the following pedigree, which individuals must be carriers?