We want to map the distance between genes A (green color), B (rough leaf), and C (normal fertility). Each gene has a recessive allele (a= yellow, b=glossy and c=variable).Results from the mating are as follow:  Phenotype combinations and counts Combination Combination 1) Green, rough, normal: 85 5) Green, glossy, normal: 600 2) Yellow, rough, normal: 45 6) Yellow, glossy, normal: 5 3) Green, rough, variable: 4 7) Green, glossy, variable: 50 4) Yellow, rough, variable: 600 8) Yellow, glossy, variable: 90   The parental progeny can be observed in the phenotype #s 4 (Yellow, rough, variable) with its corresponding genotype and 5 (green, glossy, normal) with its corresponding genotype . The actual configuration of the parental chromosomes in the trihybrid plant is .

What is F for V-1? Hint: F takes into consideration all the alleles IBD. Probability of an allele IBD , where n= # of transmission events. Image Long Description A pedigree chart showing five generations labeled with Roman numerals I through V on the left side. The chart uses squares and circles connected by horizontal and vertical lines. In generation I, there is a circle labeled “A₁A₂” with number 1, connected to a square labeled “A₃A₄” with number 2. Generation II shows a square labeled 1, a circle labeled 2, a circle labeled 3, and a square labeled 4, all connected by lines. Generation III displays a square labeled 1, a circle labeled 2, a square labeled 3, a square labeled 4, and a circle labeled 5. Generation IV contains only a circle labeled 1, which has a diagonal line connecting it to square 3 in generation III. Generation V shows a single circle labeled 1 at the bottom of the chart.       Select the right answer and show your work on your scratch paper for credit.

You perform a cross (RrSs x rrss) and obtain the following results. According to these results, which map is correct? Phenotype counts Phenotype Counts RS 438 rs 432 Rs 63 rS 67 Genetic Map Options Map A: R — 8 cM — T — 5 cM — S Map B: R — 3 cM — T — 5 cM — S Map C: R — 5 cM — T — 3 cM — S   Select the right answer and show your work on your scratch paper for credit.

The genetic profile of the embryos in a population is indicated in the table below. Following natural selection, the deleterious phenotype disappears and only a portion of the population reaches adulthood. Think about the values needed to complete the table, this will help you find the values needed for this question and the following two. Embryo genotype counts, frequencies, and fitness Embryos AA Aa aa Total Number 100 200 100 400 Frequency 1 Fitness 1 0.50 0 Natural selection occurs Adult genotype counts after natural selection Adult AA Aa aa Total Number   What is the frequency of allele A in the reproductive adults? Select the right answer and show your work on your scratch paper for credit.

The following table shows the recombination frequencies of 7 gene loci (a, b, c, d, e, f, and g). Recombination frequencies between loci Loci Recombination Frequency Loci Recombination Frequency Loci Recombination Frequency Loci Recombination Frequency a and b 0.50 b and c 0.12 c and d 0.20 d and f 0.50 a and c 0.50 b and d 0.08 c and e 0.50 d and g 0.16 a and d 0.50 b and e 0.50 c and f 0.50 e and f 0.20 a and e 0.15 b and f 0.50 c and g 0.04 e and g 0.50 a and f 0.05 b and g 0.08 d and e 0.50 f and g 0.50 a and g 0.50   Which of the following correctly shows the order of alleles in one of the linkage groups?

You are conducting research on a tree lizard population in Arizona and found that the lizards leg length is controlled by a single gene where dominant S1 allele and recessive S2 allele code for short or long hind legs, respectively. Table below shows the current and expected future genotype frequencies in this population:   Tree Lizards in Arizona Genotype Current frequencies Expected future frequencies S1S1 0.705 0.690 S1S2 0.225 0.215 S2S2 0.070 0.095 Based on these numbers, which phenotype of tree lizards will be favored by natural selection if the environment doesn’t change in future generations? 

The following question will only be shown in the actual exam, so it is important to collect all the data/information before submitting/disconnecting from your exam. NOTE: The essay question will be graded separately and it is worth 20 points. Complete your work on a blank piece of paper. You will have 30 minutes after you finish your test to finish and submit all the answers for this question and the calculations to the indicated multiple choice questions to the “Exam 2 Part 2” assignment on the “Exams” link. Submit BOTH SIDES of each paper as a single PDF file (preferred) or jpeg file. A population of fish called the founder generation is introduced into the freshwater lake of Karu. The population consists of 200 Blue (BB) individuals, 400 Green (Bb) individuals, and 400 Yellow (bb) individuals. Assume the population mates randomly, the genotypes are selectively neutral, and mutation and migration are negligible. Questions: (a) What are the frequencies of alleles B and b in the founder generation? Show your calculations. (4 points). Frequency of allele B = __________________ Frequency of allele b = __________________ (b) Calculate p2, 2pq, q2 and the expected number of BB, Bb, and bb individuals in the population, assuming Hardy-Weinberg equilibrium (HWE). Show your calculations. Hint: use the p and q values you found above. (6 points). p2 = 2pq = q2 = Expected number Blue (BB) = Expected number Green (Bb) = Expected number Yellow (bb) = Is the founder generation at Hardy-Weinberg equilibrium? Find ALL the values needed to complete the table below and write them down on your sheet. Last row can be left as fractions. The χ2 value was kindly provided by Dr. Holechek’s Grad TA. Hint: The expected frequencies and expected values are the ones you estimated in (b). (4 points, 0.25 per square). Observed and expected genotype counts by phenotype Phenotype Blue Green Yellow Total Genotype BB Bb bb – Observed 200 400 400 Observed Frequency Expected Frequency Expected (O − E)2 / E χ2 = 27.77   Use the provided χ2 table below to determine whether this population is in HWE at the B locus. (c) What is the P value that corresponds to this Chi-square (χ2) value? (1 point). ______________ (d) Is the population in HWE? Briefly explain your answer. (2 points). ___________________________________________ (e) Mention 3 reasons why most populations are not in Hardy-Weinberg equilibrium. (3 points). 1. 2. 3. Chi-square critical values by degrees of freedom (k) and probability (P) value k Probability (P) Value 0.99 0.975 0.95 0.90 0.10 0.05 0.025 0.01 1 0.000 0.001 0.004 0.016 2.706 3.841 5.024 6.635 2 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 3 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.34 4 0.297 0.484 0.711 1.064 7.779 9.488 11.14 13.28 5 0.554 0.831 1.145 1.610 9.236 11.07 12.83 15.09 6 0.872 1.237 1.635 2.204 10.64 12.59 14.45 16.81 7 1.239 1.690 2.167 2.833 12.02 14.07 16.01 18.48 8 1.646 2.180 2.733 3.490 13.36 15.51 17.54 20.09 9 2.088 2.700 3.325 4.168 14.68 16.92 19.02 21.67 10 2.558 3.247 3.940 4.865 15.99 18.31 20.48 23.21 Show both sides of each of the papers you used to the camera before submitting and exiting this part of the Exam. Do not forget to show your work for your questions on the scratch papers. You have up to 30 minutes to complete the essay question and submit your calculations and essay through Exam 2 Part 2.

Find the percentage by mass of O in Ga2O3 if it is 74.4% Ga by mass:

When the following equation is balanced, what is the coefficient in front of the Pb(NO3)2?Pb(NO3)2 → PbO + NO2 + O2

What is the chemical name for the compound N2O4?