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- If you cross a mouse with Ff genotype with another mouse with Ff genotype, what is theprobability of exactly 2 out of 6 of their offspring also having Ff genotype? Show your work using binomial and (p+q)^nUsing the HardyWeinberg Law in Human Genetics Suppose you are monitoring the allelic and genotypic frequencies of the MN blood group locus (see Question 2 for a description of the MN blood group) in a small human population. You find that for 1-year-old children, the genotypic frequencies are MM = 0.25, MN = 0.5, and NN = 0.25, whereas the genotypic frequencies for adults are MM = 0.3, MN = 0.4, and NN = 0.3. a. Compute the M and N allele frequencies for 1-year-olds and adults. b. Are the allele frequencies in equilibrium in this population? c. Are the genotypic frequencies in equilibrium?Do the data that Mendel obtained fit his hypotheses?For example, Mendel obtained 315 yellow round, 101yellow wrinkled, 108 green round, and 32 greenwrinkled seeds from the selfing of Yy Rr individuals(a total of 556). His hypotheses of segregation andindependent assortment predict a 9:3:3:1 ratio in thiscase. Use the chi-square test to determine whetherMendel’s data are significantly different from whathe predicted. (The chi-square test did not exist inMendel’s day, so he was not able to test his own datafor goodness of fit to his hypotheses.)
- Considering the Mendelian traits round versus wrinkledand yellow versus green, consider the crosses belowand determine the genotypes of the parental plants byanalyzing the phenotypes of their offspring.Parental Plants Offspring(a) round, yellow * round, yellow 3/4 round, yellow1/4 wrinkled, yellow(b) wrinkled, yellow * round, yellow 6/16 wrinkled, yellow2/16 wrinkled, green6/16 round, yellow2/16 round, green(c) round, yellow * round, yellow 9/16 round, yellow3/16 round, green3/16 wrinkled, yellow1/16 wrinkled, green(d) round, yellow * wrinkled, green 1/4 round, yellow1/4 round, green1/4 wrinkled, yellow1/4 wrinkled, greenShown are F₂ results of a monohybrid cross performed by Mendel. a) Calculate the expected numbers of each type of pods. Full pods Constricted pods Total 882 298 1180 0.84 b) What do these p-values mean with regards to your null hypothesis? (Please choose either reject or fail to reject.)Work out the Chi-Squared Goodness of Fit for Mendel's actual data on stem length: "Expt. 7: Length of stem. –– Out of 1,064 plants, in 787 cases the stem was long, and in 277 short. Hence a mutual ratio of 2.84:1. In this experiment the dwarfed plants were carefully lifted and transferred to a special bed. This precaution was necessary, as otherwise they would have perished through being overgrown by their tall relatives. Even in their quite young state they can be easily picked out by their compact growth and thick dark–green foliage." F2: O E O-E (O-E)^2 (O-E)^2/E # short 277 266 11 121 0.455 # tall 787 798 -11 121 0.152 sums: 1,064 1064 0.607 degrees of freedom = 2-1=1 Chi-squared value = .607 p-value: 0.25 < p < 0.50 Work out the Chi-Squared Goodness of Fit for Mendel's actual data on flower position: "Expt. 6: Position of flowers. –– Among 858 cases 651…
- Mendel describes subjecting each of the 34 varieties of peas he obtained to a two-year trial. During this time he let the plants self-fertilize and observed their offspring. What was he looking for, and what was the purpose of doing this two-year trial? Explain what Mendel means when he writes that the 3:1 ratio observed in the first generation from the hybrids "resolves itself" into a ratio of 2:1:1In which type of cross(es) can we apply and demonstrate the law of segregation and law of independent assortment? Why can’t we apply the 2 Mendelian laws on monohybrid crosses? Explain briefly. How can one use a pedigree chart to hypothesize how a certain condition is transmitted? Can a pedigree chart show probability of occurrence more accurately than the Punnett square? Why or why not?The following are F2 results of two Mendel mono hybrid crosses. For each cross state a null hypothesis to be tested using x2 analysis. a) full pods 882 constricted pods 299 b) violet flowers 705 white flowers 224. Calculate the X2value and determine the p value for both. Interpret the p values. Can the deviation in each case be attributed to chance or not? Which of the two crosses shows a greater amount of deviation?
- What can you conclude based on the value of the computed Chi-square? How can you relate the two principles of Mendel to Chi-Square Values?Which of the following statements is true about the molecular basis of Mendel's second law? OThe random attachment of the sister chromatids and the split of them into different daughter cells during the second round of cell division in Meiosis ensure Mendel's second law of independent assortment of alleles. OThe random attachment of the sister chromatids and the split of them into different daughter cells during the first round of cell division in Meiosis ensure Mendel's 2nd law. O The molecular basis of independent assortment of alleles is the segregation of sister chromatids during the second round of cell division in Meiosis. O The molecular basis of independent assortment of alleles is the segregation of homologous chromosomes during the second round of cell division in Meiosis. O The molecular basis of independent assortment of alleles is the segregation of homologous chromosomes during the first round of cell division in Meiosis.In classical Mendelian genetics, how can one check the genotype of a parent (A) expressing the characters of a dominant allele? Select one: a. By performing a back cross with a recessive homozygote parent (B). If the A parent is homozygote for the dominant allele, then all the individuals from the F1 will display the dominant character. If the parent A was, instead, a heterozygote, then 50% of the F1 progeny will express the recessive character (homozygote recessive) and 50% the dominant one (heterozygotes). b. It is impossible to check such genotype without using specific molecular assays. c. By performing a back cross with a dominant homozygote parent (B). If the A parent is homozygote for the dominant allele, then all the individuals from the F1 will display the dominant character.