(Genetic Variation) Can you please provide me an example solution using this formula: (1-m)t = qt - qm / q0 - qm Give 2nd generation of q (q2) Please provide step-by-step process using your given figures. Thank you very much
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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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Can you please provide me an example solution using this formula:
(1-m)t = qt - qm / q0 - qm
Give 2nd generation of q (q2)
Please provide step-by-step process using your given figures. Thank you very much
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- Possible equations to use: PID,] = 1- (PBa + 3/4 Pab + 1/2Pbb)" !3! log (1-PID,) log (Pa8 + 3/4 P8b+ 1/2Pbb) n = 8. A breeder is concerned that his best purebred Angus bull may be a carrier for Curly Calf Syndrome. He decides to conduct some test matings by taking this bull and mating him to 15 of the bulls own daughters to determine if he is a possible carrier of this and any other undesirable recessive genes. Show your work. a. How sure would we be that he is not a carrier if all the calves are born normal? b. How many would you need to be 99% sure he was not a carrier?Answer the following genetic problems.(Problem 65a) In the plant Arabidopsis, the loci for pod length (L, long; I, short) and fruit hairs (H, hairy; h, smooth) are linked 16 m.u. apart on the same chromosome. The following crosses were made: (i) L H/L H × 1 h/l h F1 (ii) L h/L h × 1 H/I H F1 If the F1's from cross i and cross ii are crossed, what proportion of the progeny are expected to be I h/l h? Oa. 16.00% Оb. 8.00% Oc. 4.00% Od. 3.36% Oe. 1.28%
- Normal wife Affected husband As a genetic counsellor you are presented with a married couple where one of them has a family history of this disease. The husband is affected by this disease and the wife is normal. The couple, who are thinking of starting a family, would like to know what their chances are of having a child born with this condition. They would also like to know what the probabilities are of having an affected boy or affected girl. Use the symbols above to complete the diagram right and determine the probabilities stated below (expressed as a proportion or percentage). Parents 4. Determine the probability of having: Gametes (a) Affected children: Possible (b) An affected girl: fertilizations (c) An affected boy: O O O O Children Affected aite if 耳Why B is correct?Name Sofia Falcione P Pedigree Analysis Practice - for each pedigree, write the genotypes of the individuals The disorder shown on the pedigree is Maple Syrup Urine Disease (MSUD) which is a metabolic disorder that affects the body's ability to process certain proteins. It was named after a distinctive odor of a baby's urine. 1. What is the inheritance pattern of this gene? a) autosomal dominant b) autosomal recessive c) X-linked recessive 2. Provide at least one piece of evidence for your claim. This pedigree shows the inheritance Leber congenital amaurosis (LCA) which is a type of hereditary blindness. Individuals with this disease lose their vision during childhood. 3. What is the inheritance pattern shown? 4. Highlight one individual whose genotype is unknown. What additional information would you need to determine his/her genotype? Marfan syndrome affects the connective tissue and causes individuals to have long, thin, arms, legs, fingers and toes. 5. What is the inheritance…
- 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) 6I know I need to plot a Lineweaver-Burk plot, but the manual way I'm trying doesn't make sense. I'm seeing elsewhere that the x-intercept is 0.8 and the slope is 0.32 but I have no idea how those are being found. Could I get a step by step walkthrough of how I plot this in order to get the line equation? Or is there a way to plot it on a graphing calculator?Determine whether the 3 phenotypes will be in either a 9:3:4 or 12:3:1 ratio. (Yellow/Red/Purple) Observe the red and yellow kernels (up to 100) and use this to see the ratio and determine the deviation (number of observed minus the number of expected). I am having trouble on how to exactly get this. Please help.
- Complete the problem below and upload your work here. Partial credit will be considered. In corn, purple kernels (P) are dominant over yellow kernels (p) and round kernels (R) are dominant to wrinkled kernels (r). In a cross of PpRr (purple, round) with a pprr (yellow, wrinkled) plant, the following progeny were observed: Phenotype Observed Expected (O-E) (O-E)2/E Purple, Round 117 Yellow, Round 126 Purple, Wrinkled 130 Yellow, Wrinkled 127 Total=500 Fill out the expected values in the chart above and calculate the Chi-Square value for your table, look up the p value associated with the Chi-Square. How many degrees of freedom do you have in this cross? 2. What is your Chi Square value? 3. What is the p value for this cross and what does your value mean? Make sure to explain the p value clearly in a couple of sentences.Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. After crossing the F1 generation of the cross between mutant strains 1 and 3, you count and determine the phenotypes of 1,000 colonies (here a colony is equivalent to an individual): 563 colonies that can grow on minimal medium alone; 437 colonies that require adenine…Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other? B. What phenomenon is occurring in the cross between mutant strains 1 and 3? After crossing the F1 generation of the cross between mutant strains 1…