1. Map Units Snapdragons with genotypes CCDD and ccdd are crossed. The CcDd progeny were then crossed with ccdd plants and we obtain 30% CcDd, 20% ccD 20% Ccdd and 30% ccdd offspring. We conclude a. Genes C and D are linked, separated by 20 map units b. Genes C and D are linked, separated by 30 map units c. Genes C and D are linked, separated by 40 map units d. Genes C and D are unlinked

2. Answer: CD= 30% cd= 30% Cd= 20% cD= 20% 20% cD + 20% Cd = 40% 40% = 40 map units

3. Snapdragons with genotypes DDEE and ddee are crossed. The DdEe progeny are then crossed with ddee plants and we obtain 45% DdEe, 5 % ddEe, 5% Ddee and 45% ddee. How far apart are genes D and E? a. 5 map units b. 10 map units c. 45 map units d. Genes D and E are not linked

4. Answer: DE= 30% de= 30% De= 5% dE= 5% 5 % dE + 5% De = 10% 10% = 10 map units

5. A bit tougher: In fruit flies, there are 3 recessive mutations. s = Scute (no thorax bristles) S = wild type e = Echinus (rough eye surface) E = wild type v = vestigial wings V = wild type P: We cross homozygous recessive flies (sseevv) with homozygous dominant wild type flies (SSEEVV)

6. F1: All F1’s have the wild type phenoytype, and genetically would be: SsCcVv

7. To produce the F2: Cross the F1 flies (SsCcVv) with recessive flies (ssccvv) and you get an F2 with 6 different phenotypes:

8. Out of 1000 flies: + = wild type ScuteEchinusVestigialTotal scuteechinusvestigial420 +++435 scute++46 +echinusvestigial44 scuteechinus+20 ++vestigial25 scute+vestigial6 +echinus+4

9. Assignment: From the genetic data, determine the recombination frequencies between each pair of genes (s/e, e/v, s/ v). Using these frequencies, construct a genetic map:

10. ANSWER: s-e recombination 46+44+6+4= 100/1000= 10% s-v recombination 46+44+20+25= 135/1000= 13.5% e-v recombination 20+25+6+4= 55/1000= 5.5% sv e 10 map units 5.5 map units