When 2 heterozygotes were crossed with each other i.e. #AaBb# x #AaBb#, the progeny showed: (i) A_B_ = 400 (ii) A_bb = 310 (iii) aaB_ = 290 (iv) aabb = 200 Does this prove Mendelian ratio? Find with a chi square test. (A and B- dominant)

The Mendelian ratio of a dihybrid cross is expected to create #16# genotypes in the ratio #"9 A-B-: 3 A-bb: 3 aaB-: 1 aabb"#.

To determine the expected numbers of genotypes in the progeny of the cross in question, multiply the number of each genotype times its expected ratio out of #16#. For example, the total number of progeny is #1200#. To determine the expected number of progeny with the #"A-B-"# genotype, multiply #9/16 xx 1200#, which equals #675#. Then perform the Chi-square equation.

The Chi-square #("X"^2")# equation is #("observed-expected")^2/"expected"#
Genotype: #"A-B-"#
Observed: #400#
Expected: #9/16xx1200=675#
#"X"^2# equation:#(400-675)^2/675=112#

Genotype: #"A-bb"#
Observed: #310#
Expected: #3/16xx1200=225#
#"X"^2# equation: #(310-225)^2/225=32#

Genotype: #"aaB-"#
Observed: #290#
Expected: #3/16xx1200=225#
#"X"^2# equation: #(290-225)^2/225=19#

Genotype: #"aabb"#
Observed: #200#
Expected: #1/16xx1200=75#
#"X"^2# equation: #(200-75)^2/75=208#

Determine the Chi-Square Sum

#"X"^2# Sum: #112+32+19+208=371#

Once you have the Chi-Square sum, you need to use the Probability table below to determine the probability that the results of the dihybrid cross is due to the Mendelian inheritance of independent assortment.

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The degree of freedom is the number of categories in the problem minus 1. In this problem there are four categories, so the degree of freedom is 3.

Follow Row #3# until you find the column closest to your sum of #"X"^2"#. Then move up the column to determine the probability that the results are due to chance. If #p>0.5#, there is a high probability that the results are due to chance, and therefore follow Mendelian inheritance of independent assortment. If #p<0.5#, the results are not due to chance, and the results do not represent Mendel's law of independent assortment.

The sum of #"X"^2"# is #371#. The greatest number in Row #3# is #16.27#. The probability that the results are due to chance is less than #0.001#. The results are not indicative of Mendelian inheritance of independent assortment.