FA23 week 1 problem set KEY

1 BICD100 FA23 week 1 problem set KEY Blue highlighting shows answers scored for correctness 1). In Mimulus (monkeyflower), plant stems can be either smooth or hairy, and this difference is determined by a single gene, H (call H the dominant allele and h the recessive allele). a. Examine the results of the following crosses. For each cross, indicate (in the table) the genotypes of the parents and progeny: cross# parents progeny 1 hairy X hairy all hairy HH HH HH (one of these parents could be Hh, but not both) 2 hairy X hairy 3/4 hairy, 1/4 smooth Hh Hh HH & Hh hh 3 hairy X smooth 1/2 hairy, 1/2 smooth Hh hh Hh hh 4 hairy X smooth all hairy HH hh Hh b. What is the phenotype of a “tester” for this trait (a plant that can be used in a cross determining whether the other parent’s genotype is HH or Hh?) smooth c. If plant of genotype Hh is crossed with a tester, what proportion of the progeny will have the same phenotype as the tester? 1/2 2.) Textbook problem 5.15 a-c only (pg. 215) a. short hair b. If s represents the recessive allele and S the dominant allele: angora female and angora male are both ss. Short-haired male is SS and short-haired female is Ss c. 100% 3.) Flour beetles may be red or ebony, and this difference in body color is due to a single gene. A flour beetle with a red body is crossed with another beetle that has an ebony body. Approximately half the F 1 progeny have red bodies, and the other half have ebony bodies. What cross could you do with F 1 progeny of the first cross to determine which phenotype is dominant? Specify just one cross in A, which gives different results in b and c depending on which phenotype is dominant. a. Cross (indicate phenotypes of parents): red x red OR ebony x ebony b. Predicted results if red is dominant to ebony (phenotypes & proportions): if you said red x red: ¾ red and ¼ ebony if you said ebony x ebony: all ebony c. Predicted results if ebony is dominant to red (phenotypes & proportions): if you said red x red: all red if you said ebony x ebony: ¾ ebony and ¼ red

2 4.) When Mendel crossed true-breeding green pod plants with true-breeding yellow pod plants, the F1 progeny all had green pods. When these F1s were crossed to the true-breeding yellow pod line, 428 of the progeny had green pods and 408 had yellow pods. Conduct a chi squared test of the hypothesis that the difference between green and yellow pod color is controlled by a single gene with dominant and recessive alleles. Include in your answer each of the following items in the space provided: i. your expected values for the chi squared calculation (show how you got them): Given the total of 836 progeny, we expect 418 in each progeny class. ii. the value of chi squared (show your work): Chi 2 = ∑ [(O – E) 2 /E] = (428-418) 2 /418 + (408-418) 2 /418 = 0.478 iii. your p-value (use Chi squared table on pg. 206 of textbook): At one degree of freedom (#progeny classes minus one), p approx. = 0.5 iv. your conclusion (reject hypothesis or not?): Cannot reject the hypothesis > conclude that the data do fit the hypothesis that this trait is controlled by a single gene with dominant and recessive alleles. 5). C. elegans worms (a valuable model system for genetics research) normally crawl around pretty quickly - they are true-breeding for “normal movement”. A variant was isolated that moves sluggishly. A true-breeding sluggish line was made; worms from this line were crossed to ones from a true-breeding normal movement line. The F 1 progeny all moved normally. a. Which phenotype is dominant? Normal movement is dominant. b) F 1 worms were crossed together to generate an F 2 generation of 300 worms. 231 F 2 progeny moved normally and 69 moved sluggishly. Use Chi squared analysis to test the hypothesis that normal vs. sluggish movement is determined by alternate alleles of a single gene. Include in your answer each of the following items in the space provided: i. your expected values for the chi squared calculation (show how you got them): Since this F2 population comes from a cross between two heterozygotes, we expect 3 normal movement to 1 sluggish. Expected values: 0.75 x 300 = 225 normal movement and 0.25 x 300 = 75 sluggish movement. ii. the value of chi squared (show your work): Chi 2 = ∑ [(O – E) 2 /E] = (231-225) 2 /225 + (69-75) 2 /75 = 0.16 + 0.48 = 0.64 iii. your p-value (use Chi squared table on pg. 206 of textbook): At one degree of freedom (#progeny classes minus one) p is between 0.5 (50%) and 0.3 (30%) iv. your conclusion (reject hypothesis or not?): Cannot reject the hypothesis; conclude that the data fit the prediction for a trait controlled by a single gene with dominant and recessive alleles.

4 8). Textbook problem 7.13 (pg. 301) a. Crossveinless is X-linked and recessive b. Males: half crossveinless; females all normal c. If X + represents the normal allele and X c the crossveinless allele, then one of the F 2 females is X + X + and the other is X + X c . Half the offspring of this female, when mated with a X c Y male, will be crossveinless and the other half normal (true for both male and female progeny). 9). Textbook problem 7.11 (pg. 301), but skip part c.iv. Also, although not stated in problem language, assume that the trait is rare. a. Yes it could occur, but it’s very unlikely if the trait is rare because II-1, II-3, and II-8, all with no known family history of their own, would have to be carriers. b. No c. i. 1/2 or 50% c.ii. 1/4 or 25% c.iii. 0 10). Textbook problem 7.21 (pg. 303) Note that it’s important to consider that the trait is rare for answers to all questions! a. Autosomal dominant is most likely. Autosomal recessive and X-linked recessive are both unlikely. X-linked dominant and Y-linked can both be ruled out. b. Autosomal recessive is most likely and all others can be ruled out. c. Y-linked dominant is the most likely but autosomal dominant is not unlikely (it could easily be a coincidence that both the affected offspring are XY’s – would need more offspring and families to discriminate). Autosomal and X-linked recessive are both unlikely. X-linked dominant can be ruled out. d. In a: 1/2 or 50%. In b: 0 (because trait is rare, we assume the egg parent of child #1 is not a carrier). In c: 1 or 100% if child is XY, and 0 if XX. Overall, 1/2 or 50% chance of being affected. e. 1 or 100% of being unaffected in all 3 pedigrees (note that in pedigree b we assume that the sperm parent of child #2 is not a carrier because the trait is rare)