Pairwise genetic diversity in humans is â = 0.0005/bp and in cockroaches it is ✩ = 0.01/bp. You can assume that the mutation rate is μ = 2 x 10-8/bp/generation in both species. A deleterious mutation with s = 10-6 arises in both humans and cockroaches. A. What is the probability that the allele reaches fixation in humans? Assume it is a new mutation with frequency 1/2N. B. What is the probability that the allele reaches fixation in cockroaches? Assume it is a new mutation with frequency 1/2N. C. Explain in 2-3 sentences why the probability of fixation differs between humans and cockroaches.
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- A recent estimate of the rate of base substitutions atSNP loci is about 1 × 10−8 per nucleotide pair pergamete.a. Based on this estimate, about how many de novomutations (that is, mutations not found in the genomes of your parents) are present in your own genome?b. Where and when did these de novo mutations inyour genome most likely occur?stion 6 of 18 Suppose that a geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation quiver, qu, and determines that it is due to an autosomal recessive gene. She wants to determine whether the gene encoding quiver is linked to the recessive gene for vestigial wings, vg. She crosses a fly homozygous for quiver and vestigial traits with a fly homozygous for the wild-type traits, and then uses the resulting F, females in a testcross. She obtains the flies from this testcross. Phenotype Number of flies vg* qu+ 230 vg qu 224 vg qut vg* qu 97 99 Test the hypothesis that the genes quiver and vestigial assort independently by calculating the chi-squared, X², for this hypothesis. Provide the X2 to one decimal place. X2 = Does the X value support the hypothesis that the quiver and vestigial genes assort independently? Why or why not? the partial table of critical values for X2 calculations to test this hypothesis.Suppose you are constructing a human genomic library in BAC vectors where the human DNA fragments are on average 100,000 bp. a. What is the minimum number of different recombinant BACs you need to construct in order to havea greater than zero chance of having a completelibrary—meaning one in which the entire genomeis represented?The simple statistical equation that follows allows youto determine the size that a genomic library needs tobe (that is, the number of independent recombinantclones you need to make) for a given likelihood thatthe entire genome is represented in the library.N = ln (1 − P)ln (1 − f )In the equation, N is the number of independent recombinant clones; P is the probability that any particular partof the genome is represented at least one time; f is thefraction of the genome in a single recombinant clone.(Note: ln is the natural log, sometimes written as loge.)b. Calculate f for the genomic library described in part (a).c. How many different recombinant BAC…
- A pair of paralogous repeats, A and B, have 96% sequence similarity and therefore can promote non-allelic homologous recombination (NAHR). They exist in four possible arrangements in a genome, illustrated below as arrangements 1 – 4. What is the result of NAHR between repeats A and B in arrangement 1? A.Translocation between chromosomes 1 and 2 resulting in monocentric chromosomes B.Deletion or duplication of the region between A and B C.Translocation between chromosomes 1 and 2 resulting in acentric and dicentric chromosomes D.Inversion of the region between A and BWith the information that a Wright- Fisher population have a size of 10,000 and assume that mutation rate is 2.5x10-8 per site per generation. What is the probability of difference between the two sequences that were randomly taken from the population? Show all working and explanation.You are studying four linked genes located on chromosome 2 in the fruit fly Drosophila melanogaster: adp (which contributes to obesity), b (which contributes to body color), pr (which contributes to metabolism), and vg (which contributes to wing formation). A series of crosses between pair-wise combinations of these mutations yielded the following recombination frequencies between the indicated loci: pr and adp 9% adp and b 6% pr and b 3% vg and b 10% vg and pr 7% What is the genetic distance in map units (cM) between the adp and vg loci? 16 9 6 4
- If you compare the frequency of the sixteen possible dinucleotide sequences in the E. coli and human genomes, there are no striking differences except for one dinucleotide, 5ʹ-CG-3ʹ. The frequency of CG dinucleotides in the human genome is significantly lower than in E. coli and significantly lower than expected by chance. Why do you suppose that CG dinucleotides are underrepresented in the human genome? (hint: The C in the CG pair is often methylated). Explain how this observation has an impact on the cells immune response.Following the analysis of a pedigree, a genetic link at 4cM is considered between a mutation leading to a pathology and the molecular marker HUMTH01. The study counts 14 "parental" and 3 "recombinant" individuals. We call p(theta=0.04) is the probability of obtaining such a pedigree in case of a 4cM genetic linkage, p(theta=0.5) is the probability of obtaining such a pedigree in case of independence between the mutation and the marker, Z(theta=0.04) the value of the Lod-score under the assumption of 4cM genetic linkage. Tick all the correct answers: p(theta=0,04)=1,79.10E-9 and Z(theta=0,04)=0,47 p(theta=0,5)=7,18.10E-12 and Z(theta=0,04)=0,77 and Z(theta=0,04)=0,67 p(theta=0,04)=2,75.10E-10 p(theta=0,5)=6,04.10E-10 and Z(theta=0,04)=0,47 p(theta=0,5)=9,36.10E-12 and Z(theta=0,04)=1,33 p(theta =0,5)=5,82.10E-11 and Z(theta=0,04)=0,67 no correct answer p(theta=0,04)=4,31.10E-11 p(theta=0,04)=2,01.10E-10 and Z(theta=0,04)=0,77 and Z(theta=0,04)=1,33Consider two maize plants:a. Genotype C/cm ; Ac/Ac+, where cm is an unstableallele caused by a Ds insertionb. Genotype C/cm, where cm is an unstable allele causedby Ac insertionWhat phenotypes would be produced and in whatproportions when (1) each plant is crossed with a basepair-substitution mutant c/c and (2) the plant in part a iscrossed with the plant in part b? Assume that Ac and care unlinked, that the chromosome-breakage frequencyis negligible, and that mutant c /C is Ac+.
- You have the following DNA coding sequence of a wild-type allele: 5’-ATG TTC CAG CTA GAT GAT ATG CTG GTA ATT GGG GAA CGC GCG CGG TAA-3’ For each of the following mutations: A. State whether the mutation is missense, nonsense, frameshift, or silent. B. Write the codon change that occurs for the missense, nonsense, and silent mutations (ex. GAA -- GAT). C. For frameshift mutations, write out the entire mutant sequence with each codon clearly indicated (if the frameshift creates a new stop codon, end the sequence at the new stop). Using the wild type DNA sequence above as a guide : Write the amino acid sequence of the mutants. Mutant 1: transition at nucleotide 23 Mutant 2: T --> G transversion at nucleotide 29 Mutant 3: an insertion of “A” after nucleotide 14 Mutant 4: transition at nucleotide 7 Mutant 5: An insertion of GG after nucleotide 40 Mutant 6: transition at nucleotide 15 Mutant 7: a deletion of nucleotide 25A gene has length of 10000 nucleotides, some of them are ancestral (A) and others are mutant letters (M). Every generation 3% of the ancestral letters turn mutate and become M and 1% of the mutant letters revert to the ancestral state. Write a discrete time model for the number of the mutant nucleotides M (Hint: A = 10000 - M). If initially there are 0 mutant letters (all 10000 are ancestral), predict how many there will be in 10 generationsImagine a 9-armed, hairy monster, who lives in Monster Land. It has a genome that is very similar to a mammal genome (diploid, same type of sex determination, etc). There are three different colours of this monster: Purple, Yellow, and Purple and Yellow mixed. Here is a table of 20 individuals and 10 SNPs, where the name of the SNP is the chromosome and the position (chromosome.position). Using this table, do a GWAS with only these 10 SNPs. How many markers of the genome are significantly associated with the trait? Phenotype chr1.1 chr2.8 chr3.12 chr4.12 chr4.55 chr7.88 chr9.55 chr10.1 chrX.1 chrX.5 Purple AA AT F GG CC CC 8 TT TT TT CT Purple AA AT TT GC CC CC AT TT TT CT Yellow AA AT TT GC CC CC AT AT GG CT Purple and AA TT TT GC TT CC AT AT GT CT Yellow Purple AA AT AA GG Purple AA AT 순 E TT GC AT TT TT CT AA GG TT GC AT TT TT CT Purple AA AT AA CC TT GG AT TT TT TT Purple AA AT AA GG CC GG AT TT TT TT Purple and AA TT AT CC CC GG AT AT GT TT Yellow Purple AA AT AT CC CC GC AT TT TT…