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1
OBSERVATIONS:
Table 1. Results of testcross of male heterozygous for three enzymes
(Locus 1, 2 and 3).
GENOTYPE
# OF PROGENY
LOCUS
LAB SLOT _________
1
2
3
TEST FOR EQUAL SEGREGATION OF ALLELES
Table 2. Comparison of the number of individuals showing homozygous and heterozygous
genotypes for each gene.
ENZYME
LOCUS 1
LOCUS 2
LOCUS 3
GENOTYPE
NUMBER
HYPOTHESIS: (ie., if the alleles of a gene are segregated equally, what should the results
show in terms of numbers of each genotype(ie., homozygous and heterozygous)?):
Table 3. Chi Square test for equal segregation of the alleles in the gene for enzyme LOCUS 1.
GENOTYPE OBSERVED EXPECTED (O – E)
(O – E)2
(O – E)2/E
TOTAL
X2 =
Table 4. Chi Square test for equal segregation of the alleles in the gene for enzyme LOCUS 2.
GENOTYPE OBSERVED EXPECTED (O – E)
(O – E)2
(O – E)2/E
TOTAL
X2 =
2
Table 5. Chi Square test for equal segregation of the alleles in the gene for enzyme LOCUS 3.
GENOTYPE OBSERVED EXPECTED (O – E)
(O – E)2
(O – E)2/E
TOTAL
X2 =
INTERPRETATION (one interpretation can include all 3 tests from Tables 3, 4 and 5):
TEST FOR INDEPENDENT ASSORTMENT
Table 6. Comparison of the numbers found for each genotype when considering each gene pair
separately.
GENOTYPE
LOCUS 1 & 2
LOCUS 2 & 3
LOCUS 1 & 3
HYPOTHESIS ( ie., what ratio of the four possible genotypes should occur if the 2 genes are
independently assorted?):
Table 7. Chi Square test for independent assortment of genes of LOCUS 1 & 2.
GENOTYPE OBSERVED EXPECTED (O – E)
(O – E)2
(O – E)2/E
TOTAL
X2 =
Table 8. Chi Square test for independent assortment of genes of LOCUS 2 & 3.
GENOTYPE OBSERVED EXPECTED (O – E)
(O – E)2
(O – E)2/E
TOTAL
X2 =
3
Table 9. Chi Square test for independent assortment of genes of LOCUS 1 & 3.
GENOTYPE OBSERVED EXPECTED (O – E)
(O – E)2
(O – E)2/E
TOTAL
X2 =
INTERPRETATION (one interpretation can cover all 3 test from Tables 7,8 and 9):
MAP DISTANCE
If there is evidence of linkage, calculate percentage recombination to estimate map distance
(show calculations ):
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