Explain how discrete genes can lead to continuous phenotype distributions. Your answer should include discussion about both additive alleles and environmental effects.
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Genetic Variation
Genetic variation refers to the variation in the genome sequences between individual organisms of a species. Individual differences or population differences can both be referred to as genetic variations. It is primarily caused by mutation, but other factors such as genetic drift and sexual reproduction also play a major role.
Quantitative Genetics
Quantitative genetics is the part of genetics that deals with the continuous trait, where the expression of various genes influences the phenotypes. Thus genes are expressed together to produce a trait with continuous variability. This is unlike the classical traits or qualitative traits, where each trait is controlled by the expression of a single or very few genes to produce a discontinuous variation.
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- Below in Figure 1 is a pedigree for a family afflicted by a genetic disorder. In some populations, Cystic fibrosis has an incidence of 1 in 2500 newborns. The carrier frequency calculated from this is 1/25. Analyze the pedigree below assuming the disease is similar to cystic fibrosis in incidence and carrier frequency. However this disease may not have the same type of genetic transmission as cystic fibrosis. What type of genetic transmission is most consistent with the pedigree? Label all the individuals that you can determine, with hypothesized genotypes. Label the individuals that have unknown genotypes with possible genotypes. Assuming her father is known to NOT be a carrier, calculate the probability that IV1 is a carrier for disease. Use the Χ2 test to determine whether your proposed transmission fits this data.How do you think consanguinity is increasing the risk factors of genetic disorders? Explain with examples (at least two). It won’t be possible to change the tradition all of a sudden. If you are a clinical geneticist, how you will be managing this type of community?Consider two blood polymorphisms that humans have in addition to the ABO system. Two alleles LM and LN determine the M, N, and MN blood groups. The dominant allele R of a different gene causes a person to have the Rh+ (rhesus positive) phenotype, whereas the homozygote for r is Rh− (rhesus negative). Two men took a paternity dispute to court, each claiming three children to be his own. The blood groups of the men, the children, and their mother were as follows:From this evidence, can the paternity of the children be established?
- Assumptions: . If the parent expresses the trait, they are homozygous dominant If the child/children express the trait but the parents do not, the parents are heterozygous. Traits selected: Brown Hair/Eyes And Widows Peak Write the Letters you will be using n the Punnet square for each trait: _Ww Bb_and_ww Bb_ Father Genotype: WW Bb Mother Genotype: ww Bb Parent Gametes possible gametes: possible gametes: Do you have the phenotype of either of the traits in the cross? If you do, which one? Are there any other offspring with either trait? Which trait? Which phenotypic probability did you fall into? Which phenotypic probability did other offspring fall into? ICystic fibrosis is an autosomal recessive disease characterized by two copies of a mutated CFTR gene. If one in 100 (hypothetical scenario, not reality) people in the United States have cystic fibrosis, calculate the p and q frequency for the normal allele (p) and the mutated allele (q). Based on those calculations, what percentage of individuals would be expected to be homozygous dominant?Consider the following scenario: A man without freckles (freckles are a dominant trait, determined by the dominant allele “F”) is a carrier of cystic fibrosis (recall that CF is a recessive trait, determined by the recessive allele "a"), mates with a woman whose genotype is heterozygous for freckles and is also a carrier of CF. Assume the two genes in question are in different chromosomes and, therefore, assort independently. Complete the following Punnett square to generate the offspring probabilities from this couple, by entering the genotypes of the parents, the gametes, and the offspring for the two traits described above. (4) Father’s genotype ffAa Mother’s genotypeFfAa ● Sperm: fA ● Sperm: fa ● Sperm: fA ● Sperm: fa ● Egg: FA ● Egg: fA ● Egg: Fa ● Egg: fa What is the probability for this couple to have a child with freckles? What is the…
- Six single-nucleotide polymorphism (SNP) loci are known to contribute to the development of type 2 diabetes. Diabetes is certain if eight or more of the alleles at these six loci are of the “contributing” variety. An uppercase letter indicates an allele that contributes to diabetes and a lowercase symbol indicates a noncontributing allele. A married couple wants to know the probability of producing a child who is at severe risk of type 2 diabetes. Their genotypes were determined by microarray analysis and are as follows: AaBbccDDEEFf × AaBbCCDdEeFfCystic fibrosis (CF) is an inherited chronic disease that affects the lungs and digestive system of about 30,000 children and adults in the United States. It is caused by a recessive allele that encodes a defective chloride channel. Genetic screening has determined that two parents are carriers (heterozygous). What is the probability that they have a child with CF? Rabbits with the genotype CC° are black, heterozygous rabbits CCW appear brown, and rabbits that are CWC" have white fur. VWhat is the phenotypic and genotypic ratio of the offspring of a cross between a brown and a white rabbit? Using your book, determine the probability of each blood type in the offspring of a heterozygous man who has type B blood, and a heterozygous woman who has type A blood. A woman who is type O gives birth to a baby who is also type 0. In a paternity case, a man claims that he cannot be the father because he has type A blood. Is he right? Explain.The data are generated from an initial parental cross. One parent displays the disease phenotype and one displays the wild-type (WT) phenotype. The WT parent always has a homozygous genotype. Your task is to perform a chi-square goodness of fit test on each of two F2 data sets, and make a decision, based on your statistical analyses as to which F2 data set provides greater evidence for indicating the correct mode of inheritance. In this problem, the true mode of inheritance is autosomal dominant. Evidence is measured in the following ways: the p-value is greater than 0.05, so we do not reject the null hypothesis, and the p-value is closer to 1. A few things of which to be mindful. 1. In the parental generation, the WT parent always has a homozygous genotype. 2. For the autosomal dominant mode of inheritance, the disease gene always homozygous for the disease allele in the parental generation. 3. For the autosomal dominant, homozygous lethal any person with the disease…
- The population risk of NIDDM is highly dependent on the population under consideration; in most populations, this risk is 1%-5%, although it is 6% to 7% in the United States. If a patient has one affected sibling, the risk increases to 10%; an affected sibling and another first-degree relative, the risk is 20%; an affected monozygotic twin, the risk is 50%-100%. 1) Is this a single gene inheritance or multifactorial disease? 2) Is there a strong genetic or environmental cause to the development of this disease? If both genetic and environmental causes are implicated, you have to indicate each of them separately.Most 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% chanceIn tomato, the purple color of the stem is determined by an autosomal dominant "A" allele. The recessive allele "a" determines a green colored stem. An independent gene controls the shape of the leaf: the dominant allele "C" determines leaf with a trimmed border and the recessive allele "c" determines leaf with a full border. Perform Genotype and Phenotype giving probability percentages.