In garden pea, resistance to a certain disease is controlled by a single locus with dominance for resistance (R) and recessive susceptible (r). Another locus governs seed color with yellow being dominant (Y) to green (y). A pea plant which is heterozygous for both gene pairs is crossed with a susceptible and green- seeded plant. The following data was observed in the progeny. Phenotype                                Number Resistant yellow                           43 Susceptible green                         49 Resistant green                             16 Susceptible yellow                         12 Total                                               120 a. What is the calculated chi-square value based on an independent assortment assumption? Compute for the df and write the conclusion. (NOTE: The cross will not yield a 9:3:3:1 ratio since not both of the parental genotypes are heterozygous for the 2 pairs of alleles. Thus, perform the cross first, show the parental genotypes, the gamete types that will be produced from the parents and the GR and PR (Phenotypic ratio). Use the computed PR in calculating the expected ratio/value. The observed ratio is given in the table above. b. Assuming linkage, which of the test cross classes are the results of crossing-overs? c. What is the map distance between these two loci? Show the computations.

Human Heredity: Principles and Issues (MindTap Course List)
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Chapter3: Transmission Of Genes From Generation To Generation
Section: Chapter Questions
Problem 15QP: More Crosses with Pea Plants: The Principle of Independent Assortment Two traits are examined...
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In garden pea, resistance to a certain disease is controlled by a single locus with dominance for
resistance (R) and recessive susceptible (r). Another locus governs seed color with yellow being dominant (Y)
to green (y). A pea plant which is heterozygous for both gene pairs is crossed with a susceptible and green-
seeded plant. The following data was observed in the progeny.

Phenotype                                Number
Resistant yellow                           43
Susceptible green                         49
Resistant green                             16
Susceptible yellow                         12
Total                                               120

a. What is the calculated chi-square value based on an independent assortment assumption? Compute for
the df and write the conclusion. (NOTE: The cross will not yield a 9:3:3:1 ratio since not both of the
parental genotypes are heterozygous for the 2 pairs of alleles. Thus, perform the cross first, show the
parental genotypes, the gamete types that will be produced from the parents and the GR and PR
(Phenotypic ratio). Use the computed PR in calculating the expected ratio/value. The observed ratio is
given in the table above.

b. Assuming linkage, which of the test cross classes are the results of crossing-overs?

c. What is the map distance between these two loci? Show the computations.

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