# 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)

Aug 28, 2015

The results of the dihybrid cross in question do not indicate Mendel's law of independent assortment.

#### Explanation:

The Mendelian ratio of a dihybrid cross is expected to create $16$ genotypes in the ratio $\text{9 A-B-: 3 A-bb: 3 aaB-: 1 aabb}$.

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 $\text{A-B-}$ genotype, multiply $\frac{9}{16} \times 1200$, which equals $675$. Then perform the Chi-square equation.

The Chi-square $\left(\text{X"^2}\right)$ equation is ("observed-expected")^2/"expected"
Genotype: $\text{A-B-}$
Observed: $400$
Expected: $\frac{9}{16} \times 1200 = 675$
${\text{X}}^{2}$ equation:${\left(400 - 675\right)}^{2} / 675 = 112$

Genotype: $\text{A-bb}$
Observed: $310$
Expected: $\frac{3}{16} \times 1200 = 225$
${\text{X}}^{2}$ equation: ${\left(310 - 225\right)}^{2} / 225 = 32$

Genotype: $\text{aaB-}$
Observed: $290$
Expected: $\frac{3}{16} \times 1200 = 225$
${\text{X}}^{2}$ equation: ${\left(290 - 225\right)}^{2} / 225 = 19$

Genotype: $\text{aabb}$
Observed: $200$
Expected: $\frac{1}{16} \times 1200 = 75$
${\text{X}}^{2}$ equation: ${\left(200 - 75\right)}^{2} / 75 = 208$

Determine the Chi-Square Sum

${\text{X}}^{2}$ 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.

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 $\text{X"^2}$. 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 $\text{X"^2}$ 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.