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- In a series of three experiments, M. G. Addo sought to use chi-square test to determine the goodness of fit at 5% for each of his results below. Carry out the calculations and determine whether the results obtained in each case are consistent with the 3:1 or 1:1:1:1 ratio he predicted in his hypothesis. Comment on your results. CROSS Tall x short Purple x white Round yellow (F1) x Wrinkled green (F1) RESULTS HΥΡΟΤHESIS 712:352 3:1 705 : 224 3:1 31:26:27:26 1:1:1:1The basis for rejecting any null hypothesis is arbitrary. The researcher can set more or less stringent standards by deciding to raise or lower the p value used to reject or not reject the hypothesis. In the case of the chi-square analysis of genetic crosses, would the use of a standard of p = 0.10 be more or less stringent about not rejecting the null hypothesis? Explain.What is probability, and how is it applied in genetic analysis?
- Regarding the analysis of single marker STR results used in forensic science. Tick all the correct statements: o if a suspect's alleles are different from those collected at a crime scene, then the suspect is possibly innocent if a suspect's alleles are identical to those collected at a crime scene, then the suspect is possibly guilty two unrelated individuals could have a similar genetic profile monozygotic twins may have different alleles at an STR locus O if a suspect's alleles are identical to those collected at a crime scene, then the suspect is definitely guilty O O no correct statement O if a suspect's alleles are different from those found at a crime scene, then the suspect is definitely innocent dizygotic twins can have similar alleles at an STR locus monozygotic twins cannot have different alleles at an STR locus dizygotic twins cannot have similar alleles at an STR locus O OWhat does codominance mean in genetics? Select one: O a. Both alleles are expressed only when they are homozygous O b. Each allele is both dominant and recessive O c. The alleles are neither dominant nor recessive O d. Both alleles make products that are expressed when present /quiz/attempt.php?attempt=1173673&cmid=3837312&page=13# MacBook ProI have seen that this was answered as C, Why is the answer C, how is that not evidence of it being genetic? Shouldnt it be none of the above? Question: Of the following, which supports the idea that alcoholism has no genetic or a low genetic component? a) Some strains of mice select alcohol over water 75% of the time, whereas others shun alcohol. b) The concordance value is 55% for MZ twins and 28% for DZ twins. c) Biological sons of alcoholic men who have been adopted have a rate of alcoholism more like that of their adoptive fathers. d) There is a 20% to 25% risk of alcoholism in the sons of alcoholic men. e) None of these.
- Transmission of genetic information from parent to offspring follows the rule of probability. The possible genotypes of offspring are dependent on the alleles present in parent organism. The lake could of these possibilities can be represented as a fraction(1/4), decimal (0.25), or percentage (25%). Probability ranges between 0% and 100%. When predicting the likelihood of a combination of possibilities the individual probabilities may be added together and multiplied depending on the type of combination of calculating the probability of possibly A or possibility A and  possibility B The two probabilities are multiplied together. These rules continue to apply for calculating for more than two possibilities.  A) given the probabilities of the events listed above what is the probability of X and Y and Z? P(X)= 0.5, P(Y)= 0.25, P(Z)= 0.25. B) what is the probability of X and Y but not Z? 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.The S-s antigen system in humans is controlled by two codominant alleles S and s. In a group of 3146 individuals the following genotypic frequencies were found:188 SS, 717 Ss, and 2241 ss. The chi-square value of this test is?The degrees is freedom value is ?Using a significance level of 0.5 the chi-critical value is? from this chi- squar chi-square test is it significant or non significant is the nul hypothesis of HWE accepted or rejected
- In the F2 generation, 306 rabbits with red eyes and 71 with a white eye phenotype suppose the calculated x2 value is 0.35. Find the x2 range using the distribution chart. What is the p-value range? using these information do you accept or reject the null hypothesis? The distribution chart is attached below.Which of the following statements are true regarding linkage? Select ALL that apply. When the map distance between two genes is less an 100 map units they are genetically linked When the map distance between two genes is less than 50 map units they are genetically linked genes on the same chromosome are always genetically linked. When two genes occur on the same chromosome they are physically linked When the map distance between two genes in less than 50 cM they are genetically linked.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)=1,33