During glycolysis, for each molecule of glucose, two molecules of pyruvate are produced ( glucose+ NAD+ + 2 ADP + 2Pi-> 2 pyruvate+ 2 ATP + 2NADH+. These 2 molecules of pyruvate then undergo the pyruvate decarboxylation reaction: 2(pyruvate+ CoA-SH+ NAD+ -> NADH+ CO2+ acetyl CoA).

During the pyruvate decarboxylation reaction , a thioester bond links the acetyl group of pyruvate with coenzyme A to produce acetyl CoA.

The pyruvate decarboxylation reaction is pyruvate+ CoA-SH+ NAD+ -> NADH+ CO2+ acetyl CoA.

First, determine the expected ratios using a Punnet Square. Given that red is dominant to white, the genotype of a true-breeding white flower can be denoted rr, and the genotype of a heterozygous flower can be denoted Rr. This cross will produce 50% Rr, 50% rr. Then, converting these percentages to decimals (50%= 0.50), and multiplying by the total population size of the observed population (0.50*100), gives expected values of 50 Rr (red flowers) and 50 rr (white flowers). These expected values and the observed values can then be plugged into the chi square equation . The equation will be . Degrees of freedom is n-1, so 2-1= 1. To determine p-value, use critical values table. The chi square value of 6.76 with 1 degrees of freedom will fall between a critical value corresponding with a p-value of 0.05 and 0.025. Thus, p>0.025

 A larger difference between observed and expected values will result in a larger chi square value. Using the critical values table, for a given degrees of freedom, as chi square value increases, p-value decreases.

When the p-value is equal to or below the significance level (alpha), the null hypothesis is rejected. For the p-value to be low, the chi square value would need to be large (large difference between observed and expected values). The null hypothesis would not be accepted (under no circumstance is a null hypothesis “accepted”; onlay rejected or failed to reject). The alternative hypothesis would be accepted if the p-value is equal to or below the significance level (alpha).

A larger difference between observed and expected values will result in a larger chi square value. Using the critical values table, for a given degrees of freedom, as chi square value increases, p-value decreases. A small p-value results in rejection of the null hypothesis.

Critical values cannot be negative. The chi squared formula  will not result in a negative value as the numerator is squared.

If the chi square value is large, this indicates a large difference between the observed and expected values. This will subsequently result in a small p-value when using the critical value table. When the p-value is equal to or smaller than the significance level (alpha), the null hypothesis is rejected.

Epistasis describes the interaction of genes, where the epistatic locus masks the effects of a gene at another locus. In this example, locus E is epistatic. As this is stated to be recessive masking epistasis, when the E locus is homozygous recessive (ee), this locus will mask the effect of the B locus (color). Thus, any combination of B/b with ee will result in yellow fur. When the E locus is heterozygous (Ee) or homozygous dominant (EE), the effect of the B locus will not be masked. Thus, BbEe will result in black fur (as black B is dominant to brown b), and the E locus does not mask the B locus here, as the E locus is heterozygous in this individual (Ee).