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- A genetic disorder is caused by a LOF mutation (Bm) and epigenetic imprinting of gene B. Through pedigree analysis of many families, researchers have observed the following results: Female carrier B+Bm x B+B+ --> B+Bm and Mother's genotype Father's genotype Children's genotype Children's phenotype Cross 1: B+Bm B+B+ B+Bm , B+B+ some affected, some unaffected Cross 2: B+B+ B+Bm B+Bm , B+B+ always unaffected Based on these results, gene B is imprinted on the: a. maternal b. paternalMost forms of albinism are inherited in an autosomal recessive pattern. Using a Punnett square, determine the chance that a child would phenotypically express albinism if the genotypes for both parents is Aa, where "A" indicates the dominant unaffected allele and "a" indicates the recessive affected allele. O 75% chance O 50% chance O 0% chance O 100% chance O 25% chanceColored aleurone in the kernels of corn is due to the dominant allele R. The recessive allele r, when homozygous, produces colorless aleurone. The plant color (not kernel color) is controlled by another gene with two alleles, Y and y. The dominant Y allele results in green color, whereas the homozygous presence of the recessive y allele causes the plant to appear yellow. In a testcross between a plant of unknown genotype and phenotype and a plant that is homozygous recessive for both traits, the following progeny were obtained: colored, green 88colored, yellow 12colorless, green 8colorless, yellow 92Explain how these results were obtained by determining theexact genotype and phenotype of the unknown plant, including the precise association of the two genes on the homologs (i.e., the arrangement).
- Colored aleurone in the kernels of corn is due to the dominant allele R. The recessive allele r, when homozygous, produces colorless aleurone. The plant color (not the kernel color) is controlled by another gene with two alleles, Y and y. The dominant Y allele results in green color, whereas the homozygous presence of the recessive y allele causes the plant to appear yellow. In a testcross between a plant of unknown genotype and phenotype and a plant that is homozygous recessive for both traits, the following progeny were obtained: colored, green 88 colored, yellow 12 colorless, green 8 colorless, yellow 92 Explain how these results were obtained by determining the exact genotype and phenotype of the unknown plant, including the precise arrangement of the alleles on the homologs.In this case a family history revealed a genetic basis for the disorder. The pedigree is shown in Fig. 1 Below. Key Ø Female: affected Female: unaffected || IV V 5600 orize 077808 15 10 9 10 CHO વ Male: affected Male: unaffected Deceased Disease status not given Dizygotic twins Monozygotic twins Fig. 1 Disease pedigree. Five generations I, II, III, IV, V are shown. Females are represented by circles, males by squares, dizygotic (non-identical) twins by diagonal lines originating from the same point, Monozygotic (identical) twins by diagonal lines originating from the same point and joined symbols and deceased by a diagonal line through the symbol. Filled symbols indicate that the individual displays the disease phenotype. Unfilled symbols indicate that the individual does not display the disease phenotype. Carriers of the disease are not indicated. Information on disease status is not known for generation I and is omitted for the individuals represented by a symbol with an asterisk.…ABO blood type is determined by three alleles: *, 18, and i. On a separate chromosome, anotherset of alleles determines the Rhesus factor: D and d.A woman with blood type A, who is heterozygous for both blood type and Rhesus factor, had achild with a man with blood type B, who is heterozygous for both blood type and Rhesus factor. What is the percentage chance that this child is blood type B and heterozygous for both traits? Record your answer as a percentage rounded to one decimal place.
- A scientist is investigating the inheritance of two autosomal recessive genes in rat, ear length (long is dominant to short) and fur color (gray is dominant to white). The table below provides the number of observed off-spring for each phenotype, when heterozygous rats are crossed. What are the expected values for each phenotype? Phenotype Observed Expected Long, gray 189 Long, white 56 Short, gray 74 Short, white 23 Total 342A proband female with an unidentified disease seeks the advice of a genetic counselor before starting a family. Based on the following data, the counselor constructs a pedigree encompassing three generations: (1) The maternal grandfather of the proband has the disease. (2) The mother of the proband is unaffected and is the youngest of five children, the three oldest being male. (3) The proband has an affected older sister, but the youngest siblings are unaffected twins (boy and girl). (4) All the individuals who have the disease have been revealed. Duplicate the counselors featAs it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progeny
- Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?Achondroplasia is a rare dominant autosomal defect resulting in dwarfism. The unaffected brother of an individual with achondroplasia is seeking counsel on the likelihood of his being a carrier of the mutant allele. What is the probability that the unaffected client is carrying the achondroplasia allele?Examine the following pedigrees. Which is the most likely mode of inheritance of each disorder? (a) autosomal recessive (b) autosomal dominant (c) X-linked recessive (d) a, b, or c (e) a or c 10.