Let's say an autosomal gene undergoes mutation in a population of 28 barn cats to become a new neutral allele. Calculations are required to answer these questions with numbers (Random/spam answers will be reported to chegg) 1. What would be the probability that the allele becomes fixed? 2. What is the probability that it becomes lost overtime? 3. What would the average time to fixation be, assuming that it is eventually fixed?
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- The XG locus on the human X chromosome has twoalleles, XG+ and XG. The XG+ allele causes the presence of the Xg surface antigen on red blood cells,while the recessive XG allele does not allow antigento appear. The XG locus is 10 m.u. from the STSlocus. The STS+ allele produces normal activity ofthe enzyme steroid sulfatase, while the recessive STSallele results in the lack of steroid sulfatase activityand the disease ichthyosis (scaly skin). A man withichthyosis and no Xg antigen has a normal daughterwith Xg antigen. This daughter is expecting a child.a. If the child is a son, what is the probability he willlack Xg antigen and have ichthyosis?b. What is the probability that a son would have boththe antigen and ichthyosis?c. If the child is a son with ichthyosis, what is theprobability he will have Xg antigen?As 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 progenyAn individual has the following genotype. Gene loci (A) and (B) are 15 m.u. apart. What are the correct frequencies of some of the gametes that car be made by this individual? Bl a O A. Ab = 7.5%; AB = 42.5% B. ab = 25%; aB = 50% O C. AB = 7.5%; aB = 42.5% O D. aB = 15%; Ab = F0% E. aB = 70%; Ab = 15% Reset Selection OMark for Review What's This?
- A DNA variant has been found linked to a rare autosomal dominant disease in humans and can thus beused as a marker to follow inheritance of the diseaseallele. In an informative family (in which one parentis heterozygous for both the disease allele and the DNA marker in a known chromosomal arrangementof alleles, and his or her mate does not have the samealleles of the DNA variant), the reliability of such amarker as a predictor of the disease in a fetus is related to the map distance between the DNA markerand the gene causing the disease.Imagine that a man affected with the disease(genotype Dd) is heterozygous for the V1and V2forms of the DNA variant, with form V1on the samechromosome as the D allele and form V2on the samechromosome as d. His wife is V3V3dd, where V3isanother allele of the DNA marker. Typing of the fetusby amniocentesis reveals that the fetus has the V2andV3variants of the DNA marker. How likely is it thatthe fetus has inherited the disease allele D if thedistance…The Turner Kieser Syndrome (or nail patella syndrome) is a dominant human disease (prevalence about 1/100.000), which impairs the normal development of, nails and kneecaps (patella). In the following study the mutant allele for this gene is noted "M" and the wild type allele "m". On the family tree below black symbols (circles for women and squares for men) indicate people suffering from KSS and open symbols represent healthy persons. Moreover, all individuals have been genotyped at the locus controlling blood type ([O] : ii ; [A] : Iª_ ; [B] : /B_ ; [AB] : /A/B) 1 2 Reminder 3 ilB iIB 2 3 O iIB IB iIB 7 O ii iIB IB IA ii ii Z(0) = Log10 6 4 What to conclude from this value of Z ? 5 ilA ilA ii 8 9 probability if linkage probability if independant 10 2 ilB Calculate the LOD-score Z (Logarithm Of Odds) for a genetic distance set to 10 cM (hence a recombination frequency of 0 = 0,1). 11 12 13 = ii iIB ii ii ii iIB IB 14 15 16 Log10 L (0<0,5) L (0=0,5)We often speak of diseases such as phenylketonuria (PKU) andachondroplasia as having a genetic basis. Explain whether the followingstatements are accurate with regard to the genetic basis ofany human disease (not just PKU and achondroplasia).A. An individual must inherit two copies of a mutant allele to havedisease symptoms.B. A genetic predisposition means that an individual has inheritedone or more alleles that make it more likely that she or he willdevelop disease symptoms than other individuals in a populationwill.C. A genetic predisposition to develop a disease may be passedfrom parents to offspring.D. The genetic basis for a disease is always more important thanthe environment.
- Figure 19-18a shows a plot of P values (represented bythe dots) along the chromosomes of the dog genome.Each P value is the result of a statistical test of association between a SNP and body size. Other than the clusterof small P values near IGF1, do you see any chromosomalregions with evidence for a significant association between a SNP and body size? ExplainConsidering the genetics cross andobserved phenotypes pictured:a. Which genetic cross(es) led to themost number of healthy progeny?b. What type of model organism islikely being used in these experiments?c. Explain what is likely causing the embryonic lethality based on the observed percentages of embryoniclethality.d. What form of genetic material (maternal, paternal, or both) is leading to the embryonic lethalphenotype?e. Why do you think the second bar has the highest “n” number? What does the “n” value indicate?Null mutations are valuable genetic resources becausethey allow a researcher to determine what happens to anorganism in the complete absence of a particular protein. However, it is often not a trivial matter to determinewhether a mutation represents the null state of the gene.a. Geneticists sometimes use the following test forthe nullness of an allele in a diploid organism: If theabnormal phenotype seen in a homozygote for theallele is identical to that seen in a heterozygote(where one chromosome carries the allele in question and the homologous chromosome is known tobe completely deleted for the gene) then the alleleis null. What is the underlying rationale for thistest? What limitations might there be in interpreting such a result?b. Can you think of other methods to determinewhether an allele represents the null state of a particular gene?
- In individuals affected by cystic fibrosis, salt crystals may appear afterperspiration dries up. In addition, the disease causes respiratory disorderswhich can be both debilitating and lethal. It occurs in individuals homozygousfor the recessive gene. Two normal parents had a daughter with thesymptoms of this disease, and a normal son who marries a normal womanwith an afflicted A test (salt concentration in perspiration of heterozygotes ishigher than normal) disclosed that both are indeed carriers of the gene. If thefirst child born to the mating in (b) was defective, what is the probability thatthe 2nd child would also be defective?Express answer in fraction formPo es) 20 F1 KEY CROSS- RR X IT RESULTS- Rr GENOTYPE- RR IT Ơ 天 2 T 1. To determine the results of this experiment, complete the Punnett squares below by writing the correct allele(s) in the space provided P Generation PHENOTYPE- Round seeds x Wrinkled seeds F4 R 1, a r Rr F5 Rr RR Rr RR R X Rr F6 Rr Rr AaBbCcDdE AaBbCcDdE Aa BbCcD Aa BbCcD Heading 1 Normal No Spacing IT IT I F1 F2 F2 KEY CROSS-RrxRr RESULTS ace provid GENOTYPE 2. hat phenotypes are present in the F₁ generation? What gen'otypes are present? F7 3. What phenotypes are present in the F2 generation? In what ratio are they present? PHENOTYPE 4. Does your analysis support or refute Mendel's hypothesis of dominant and recessive inheritance? Explain. DII F8 Title F9 Aa BbCcDdEe Focus Subtitle E E 1Tay-Sachs disease is caused by loss-of-function mutations in a gene on chromosome 15 thatencodes a lysosomal enzyme. Tach-Sachs is inherited as a autosomal recessive conidition.Among Ashkenazi Jews of central European ancestry, about 1 in 3600 children is born withthe disease. What fraction of the individuals in this population are carriers? (Assume thepopulation is in Hardy-Weinberg equilibrium)