These practice questions are from prior LS4 finals and are courtesy of Drs. Laski, Chen and Sagasti. A few short questions – no short answer questions on the exam but good practice 1. If a grandfather has a Y-linked trait, what is the probability that his grandson (his daughter's son) will have this trait? Zero. He does not pass on his Y chromosome to his daughter. 2. A male is affected with the X-linked recessive condition hemophilia. He marries a carrier woman. What percentage of their male children will be affected? 50%. The father’s status is unimportant because he does not pass his X-chromosome on to his male offspring. 3. A woman is the offspring of a father with the X-linked recessive condition hemophilia. What is the likelihood that her first child will be affected? 25%. The woman must be a carrier so there is a 50% chance she will pass on the mutant allele. There is also a 50% chance her child will be male. 4. The ability to roll the tongue is controlled by a single gene with two alleles. The allele for being able to roll is dominant over the allele for not being able to roll. In a given population of 10,000 individuals there are 7400 tongue rollers and 2600 non-rollers. Assuming the population is at Hardy-Weinberg equilibrium, how many heterozygous tongue rollers are there in the population? q2 = .26, q = .51, p = .49, 2pq = .4998 = 4,998 people out of 10,000 5. Phenylketonuria (PKU) is a disease in which people who are homozygous for the recessive PKU allele lack the ability to properly breakdown the amino acid phenylalanine. Build up of phenylalanine causes mental retardation in the children born with PKU. Given that PKU disease occurs in 1 out of 10,000 births and that the U.S. population is so large that Hardy-Weinberg equilibrium has been achieved for this gene, 1) what is the PKU allele frequency, and 2) what is the proportion of carriers in the population? q2 = 1/10,000, q = 1/100; 2pq = 2/100 = 1/50 6. The wild type zebrafish has short tail. From a genetic screen, you identified three recessive mutations that cause the long-tail phenotype (A, B and C). You cross the true breeding mutant A fish to the true breeding mutant B fish, and found that all progeny have normal tail. However, all progeny from the cross of true breeding mutant A fish to the true breeding mutant C fish have long tails. Provide a genetic explanation for this phenomenon. Circle the correct answer(s).1. A and B are allelic. 2. A and C are allelic. 3. B and C are allelic. 4. The allele strength is A > C > B. 7. E. coli strain B is doubly infected with two rII mutants of phage T4. 0.1 ml of a 106 dilution of the progeny is plated on E. coli B and 0.1 ml of a 104 dilution of the progeny is plated on E. coli K. 50 plaques appeared on strain B, 6 on strain K. Calculate the recombination frequency between these two mutations. 2 X 6 / 5000 = 0.0024 = 0.24% 8. Which termination codon(s) can be suppressed by a point mutation in the anticodon of trp tRNA? UAG and UGALonger problems 1. Manx cats are tailless and when crossed with one another produce on average 1 long-tailed (wild type) cat for every 2 Manx. The M (Manx) allele is lethal in homozygous condition due to problems arising during development. Thus, a MM genotype is lethal, a Mm cat is Manx (tailless), whereas a mm cat is wild type with a long tail. A. A large number of Manx cats (Mm) are put on an island and reproduce. The F1 progeny are composed of Manx and wild type cats in a 2:1 ratio, respectively. What is the allele frequency of the M allele in this F1 population. (circle correct answer) 1 3/4 2/3 1/2 1/3 1/4 0 none of above If none of above, the correct answer is: B. The F1 population breeds randomly among themselves, producing an F2 population. What fraction of the F2 population are Manx (have no tails). (circle correct answer) 1 3/4 2/3 1/2 1/3 1/4 0 If none of above, the correct answer is: Keep in mind that the MM offspring die so you have to correct for this in the surviving population. 2. Ten percent of the males of a large and randomly mating population are colorblind, a recessive X-linked trait. A representative group of 1000 from this population migrates to a South Pacific island,where there are already 1000 inhabitants and where 30 percent of the males are colorblind. Assuming that Hardy Weinberg equilibrium applies throughout (in the two original populations before emigration and in the mixed population immediately following the arrival of the immigration), what fraction of males and females can be expected to be colorblind in the generation immediately following the arrival of the immigrants? percent females colorblind: 4% percent males colorblind: 20%3. Below is shown the RNA sequence from the imaginary protein coding region of the rII gene of bacteriophage T4. As you can see the gene encodes a protein that is 7 amino acids long. Previous analysis shows that the Val-Val-Val amino acids at the C terminus of the protein (underlined below) are all that is required for rII+ activity. You isolate 4 mutations in the gene. Mutation #1 and #2 are both positive frameshift mutations, the base inserted is shown in bold. Mutations #3 and #4 are both minus frameshift mutations, the base missing is shown as a gap in the sequence. rII phenotype Wild Type 5'UUAUGCCUGGUAAAGUCGUCGUCUGAUACUAA rII+ MetProGlyLysValValValStop #1 5'UUAGUGCCUGGUAAAGUCGUCGUCUGAUACUAA3' rII- #2 5'UUAUGCGCUGGUAAAGUCGUCGUCUGAUACUAA3' rII- #3 5'UUAUGCCU GUAAAGUCGUCGUCUGAUACUAA3' rII- #4 5'UUAUGCCUGGUAA GUCGUCGUCUGAUACUAA3' rII- A. The following double mutants are made. For each double mutant write out the sequence of the rII protein it will make. Predict (circle) whether the phage will be rII+ or rII-. Double mutant Sequence #1, #3 rII+ or rII- 5’UUAGUGCCUGUAAAGUCGUCGUCUGAUACUAA3’ No Met codon so no protein #2, #3 rII+ or rII- 5’UUAUGCGCUGUAAAGUCGUCGUCUGAUACUAA3’ MetArgCysLysValValValStop (could be rII-if ArgCys can’t functionally substitute for ProGly #2, #4 rII+ or rII- 5’UUAUGCGCUGGUAAGUCGUCGUCUGAUACUAA3’ MetArgTrpStop4. The human gene for hemophilia is on the X chromosome. Below is a pedigree from a family affected with hemophilia. Blackened symbols indicate that the person has hemophilia. To help with genetic diagnosis, a probe that detects an RFLP