Question

6. In guinea pigs, it is hypothesized that the allele L governs short hair and its recessive allele / governs long hair. Codominant alleles at an independently assorting locus are hypothesized to govern hair color such that CC = yellow, CYCW = cream, and CWCW = white. From the cross LI CYCW X LI CYCw the following progeny were obtained: 50 short cream 21 short yellow 23 short white 21 long cream 7 long yellow 6 long white. Are the data consistent with what would be predicted by the allele and/or gene interactions hypothesized above? A chi square table is given at left: degrees of 0.05 freedom prob 2 3 3.841 5.991 7.815 9.488 11.070

Answer 1

cowe white CYcw = cream Y CY ₂ L Yellow Short l Long L dominant over l. خوب Wall Long white 6 tent Given classes of progeny and their number - Short cream 50 a 50/128 = 0.39 H Data is Short yellow 21 = 21/128 = 0.164 Nuy consistent Short white 23 = 23/128 a 0.18 hypothe with Long cream 21 2 21/128 = 0.164 -šis day Loug yellow = 7/128 a 0.055 t Data is = 6/128 = 0.049. wiconsis- 128 with expected values Cross → Le cyco & Lecy cw Genotypic classes oo Loy Low Tech lcw. LICYCY LLCYCW Leccy Legal a Lecycy D 4 Le cow Low LLL LL cu Legyew Lech 2 l l c wage 9 2 tlou wo ect recky Lee Yow lesy llcy w 4 other e t #4 different classes Lecw how low lleory elcro orys peleases. Classes 0 / احلااا و Lcy Heteroxy soud at both locus. Short Yellow Short cream Short cream. Short Short cream Short cream Short white. Shast yellow Short cream Long, Short white long cream Dreaba yellow Short cream Short white Long cream Long white. Short yellow - 3/16 a 0.19 Short cream - 6/16 = 0.375 Short white 3/16 = 0.19 Long cream- 3/16 = 0.125 Long white - 1/16 = 0.0625 Loug yellow - 1/16 = 0.0625

For phenotypic classes, CO- E CO-E) 6-E2 T cream (0.39-0.375) (.015) 0.0006 Short yellow (0.164-0.19) (60 026) 2 0.0036 short white. (0.18-0.19 0.0001 0.0005 Long cream (0.1.64 0.125 0.001521 0.012168 Long yellow Cookso- 0.0625 0.000056 0.0009 Long white. (0.049 - 0.0625) 0.00018 0.002916 & 2 - 5 Co-E) 2 X = 0.020684. = 2 Degrees of freedom. & No of genotypic class No of phenotypic classes z 9- 6 = 3 df 3 CHA. 1 HD) $0.05 - 7.815 x2 = 0.02068 cas poos & = 7.815 so, Nell hypothesis is accepted and Alternate hypothesis is rejected. Data given or obserwed data is consistent with expected values.