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- Sample problems related to non Mendelian inheritanceRead, analyze and answer completely the following problems: 1. 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? 2. Agouti (A) is wild type and produces alternating bands of pigment on each hair. Black (a) is recessive to agouti. A mutation on gene B (recessive b) can eliminate all color. In a cross between agouti (AABB) and albino (aabb) mice, what genotypes, phenotypes, and proportions are expected in the offspring in F1 and F2 generations?TRAIT: BENT FINGER 1. What are the roles of the DNA, genes and proteins in a given trait?2. How would you relate the individual trait or characteristics to proteins,genes and DNA?3. Would the manifestation of a trait be affected once the DNA nor the genesare altered?Question:- Based on your selected mode of inheritance, show the genotypes for the following individuals. [Use these symbols for alleles: if it is autosomal, then use the symbols B - dominant, b - recessive (e.g. BB, bb etc.) if it is X-Linked, then X(B) - dominant, X(b) - recessive, and Y for Y-chromosome (e.g. X(B)X(B), X(B)Y etc.) ] I-1 I-2 II-7 II-8 III-10 III-11 III-12 IV-8 IV-9
- Complementation tests of distinct recessive mutants, 1 through 8, produce the data in the matrix below. A plus (+) indicates complementation, meaning the phenotype of the combined alleles is wild type, and a minus (-) indicates a failure to complement meaning that a mutant phenotype results. Assume that the missing mutant combinations would yield data consistent with the entries that are shown. How many complementation groups are formed by these eight mutants? (Picture attached) A) 2 B) 3 C) 4 D) 5 E) 6Solve for the genetic structure of a population with 12 homozygous recessive individuals (yy), 8 homozygous dominant individuals (YY), and 4 heterozygous individuals (Yy).D) Genetics Problem show the table and word In certain breeds of chickens, the allele "B" is responsible for black feathers whereas the contrasting allele "b" produces feathers that are white. Another r pair of alleles influences the shape of the feathers. "F" produces straight feathers whereas the contrasting gene for "f results in the frizzled condition. Give the genotypic and phenotypic ratios to be expected from the following cross: white, frizzled hen with a heterozygous black, homozygous straight rooster. Genotypic Ratio: Phenotypic Ratio:
- polygenic trait mating 2 An AaBBCcdd male mates with an AaBbCCDD female. 1. What is the maximum number of ridge-producing genes possible in one of the children? 2. What would be the TRC for this child if it is a male? 3. What is the minimum number of ridge-producing genes possible in a child of this couple? 4. If this child were a female, would she have a higher or lower TRC than the parent with the lower ridge count? A. lower B. higher C. equal11) In the Hardy-Weinberg equation, which term refers to the heterozygous genotype? A) 2pq B) Q squared C) P squared D) 2pq squaredSubject: Genetic problems 6. how would you recognize a line of garden peas that had become genotypically pure for a given trait? 7. A cross of two pink-flowered plants produces offspring whose flowers are red, pink, or white. Defining your genetic symbols, give all the different kinds of genotypes involved, and the phenotypes they represent. 9. In snapdragons, red flowers (R) are incompletely dominant to white (r), the hybrid being pink; narrow leaves (N) are incompletely dominant to broad leaves (n), the hybrid being intermediate in width ("medium"). show the genotypes and phenotypes for the progeny of a cross between a. red medium and pink medium plant b. a pink medium and white narrow c. two identical dihybrids 12. In guinea pigs, short is dominant to long. A short-haired guinea pig was mated to a long-haired one. What proportions of the offspring (F1) will be expected to be: a. homozygous short-haired b. homozygous long -haired c. heterozygous short-haired d. heterozygous…
- Instruction: Solve for the genetic problems. a. Genotype of the parentsb. Phenotype of the parentsc. Punnett squared. Genotypic ratioe. Phenotypic ratio Question: Two individuals with widow’s peaks want to have a child with a continuous hairline. Is this possible?Punnet square problems A=Codominant; B=Codominant; O=Recessive Mary is homozygous for type A blood. Steve is homozygous for type O blood. If they have children, what are the possible phenotypes and genotypes of their children, and what is the probability of each? Mary and Steve have a son, Brad. Brad’s wife, Samantha is heterozygous for type B blood. If they have children, what are the possible phenotypes and genotypes of their children, and what is the probability of each? Stella loves roses and decides to cross her red rose with her white rose. All of the resulting offspring of this cross are pink roses. What can you say about the red and white alleles as a result of this cross? Stella decides to cross two of the pink roses. What are the possible genotypes and phenotypes of the offspring and the probabilities of each? DNA replication, Transcription and Translation problems It is S phase of the cell cycle, and time to replicate the cell’s DNA. Using the following strand of DNA…Give typed full explanation The Genes you are examining exhibit a continuous pattern of variation. For these Genes, a Dominant Allele represents a 1 foot increase in overall height. If you cross the following individuals, AaBb and AaBb, what percentage of the offspring will be 7 feet tall. Assume an individual with no Dominant Alleles (aabb) is 4 feet tall. Report your answer as a percentage i.e., 22.1%.