Problem Set #2 BioS 101 Spring 2008

10 possible points of credit (1 point for each number.)

SOLUTIONS

1. How many different loci do you infer there are when you see the following genotypes?

A. HHMmoo 3, H, M, O

B. AbCD/aBCD 4

C. PUSU/pusu 2, PU & SU

D. I^AI^Brh^-rh^- 2

E. VGVGsese 2 (numbers are all that is needed, I provided loci names in 2 cases for purposes of illustration.)

2. For each of the genotypes in #1 list ALL loci that are homozygous.

A. both H & O are homozygous B. C & D C. none D. RH

E. VG & SE loci are homozygous

3. If the probability of an offspring being YELLOW is 0.75 and the probability of it being WHITE is 0.25, calculate the Chi-squared value of the observing 80 YELLOW and 40 WHITE peas? Do you accept or reject the hypothesis?

N = 120 so expected YELLOW = 90 and expected WHITE = 30.

χ² = (80-90)^2/90 + (40-30)^2/30 = 1.11 + 3.33 = 4.44. 4.44 > 3.84 (the critical value for a 5% chance of rejecting when it is true and 1 degree of freedom) so the hypothesis that the sample is compatible with ¾ and ¼ should be REJECTED.

4. Alleles found in haploid organisms cannot be evaluated as dominant or recessive. Why? (This is question #3 on p.298 of Freeman. The answer (p.302) is given as “a. Dominance and recessiveness describe interactions between two alleles of the same gene in the same individual”. That answer is fine, but a good alternative is “The characterization of alleles as dominant or recessive can be evaluated only in the heterozygous condition (limited to diploids).

5. List every different gametic genotype that could be formed by an individual with the ADg/adG genotype. There are 3 heterozygous loci, so 2³ or 8 types are possible. Be systematic in generating the states. I will do the four staring with A first.

ADg, ADG, Adg, AdG, aDg, aDG, adg, adG.

6. Consider the ADg/adG individual again.

a. List the two parental gametic genotype(s) (those that would result if there were no crossovers). The two parental types are the two types separated by a /, ADg and adG

b. List the gametic genotypes that would be produced if there was a crossover between the A and D loci. A crossover between the A and D loci means there was a break on both chromosomes at the same location and the fixing of the break by attaching part from the mother chromosome to the father chromosome. The gametes that would result are AdG and aDg.

c. List the gametic genotypes that would be produced if there was a crossover between the A and D loci AND between the D and G loci. The crossver between the A and D loci produces Ad and aD, the break between the D and G loci changes the D,G arrangement so the final answer is Adg and aDG.

7. Answer parts a & b based on human ABO blood types.

A. Which phenotype(s) must be heterozygous? AB

B. Which phenotype(s) are you sure is homozygous? O

8. In peppers a cross between a true-breeding BROWN line and a true-breeding YELLOW line yielded F1 individuals everyone of which was RED. A simple hypothesis is that the heterozygote has the RED phenotype. This hypothesis predicts that the F2 will have 25% BROWN, 50% RED and 25% YELLOW. The observed results included a new phenotype, GREEN. A new phenotype is not possible with a one locus model. Thus you are forced to consider the next level of model, a two locus model.

Propose a model (hypothesize genotypes and for each genotype tell its phenotype) that can explain the observed F2 results: RED 53 BROWN 22 YELLOW 18 GREEN 7.

This problem is featured in the Freeman text, Fig 13.18, p291. The 4^th phenotype requires one hypothesize a second locus. Assume two alleles at each locus. Since the GREEN phenotype is rare and only appears in the F2, it must be the double recessive genotype. As the RED phenotype appears first in the F1, it requires dominant alleles at both loci. Using the symbols from Freeman (not necessary to be correct) the original cross was RRyy by rrYY. The F1 is RrYy. The F2 is 9/16 R_Y_, 3/16 rrY_, 3/16 R_yy, and 1/16 rryy.

9. In less than 60 words describe the salient phenotypic features of individual’s with Down syndrome.

Google the words, read the text at http://en.wikipedia.org/wiki/Down_syndrome , express the major points in your own words.

10. If you expect each pea to have a 3/4ths chance of being YELLOW and 1/4th chance of being GREEN, what percentage of the pods with 5 peas do you expect NOT to have 3 YELLOW peas and two GREEN peas? (Hint: Use ‘not’ logic = 1 –probability of condition described)

Figure out the probability of having 3 YELLOW and 2 GREEN peas in a pod of five. The probability of YYYGG is ¾∙ ¾∙¾∙¼∙¼ = 27/1024. The hard part is figuring out all the possible 3 Y & 2 G arrangements.

YYYGG, YYGYG, YYGGY, YGYYG, YGYGY, YGGYY, GYYYG, GYYGY, GYGYY, GGYYY are all different and seem to exhaust possibilities. Each of the ten has the probability of 27/1024 so all ten add up to 270/1024. But the question asked for everything BUT these so 1 – 270/1024 = 754/1024 = 0.736.