In a plant, fruit color is either red or yellow, and fruit shape is either oval or long. Red and oval are the dominant traits. Two plants, both heterozygous for these traits, were testcrossed, with the results shown in the following table. Determine the location of the genes relative to one another and the genotypes of the two parental plants. $$\begin{array}{lcc} & & \text { Progeny } \\ \text { Phenotype } & \text { Plant A } & \text { Plant B } \\ \text { red, long } & 46 & 4 \\ \text { yellow, oval } & 44 & 6 \\ \text { red, oval } & 5 & 43 \\ \text { yellow, long } & 5 & 47 \\ \text { Total } & 100 & 100 \end{array}$$

In this exercise, we have two traits: fruit color and fruit shape. Let's assign symbols for these traits: - R: red (dominant) - r: yellow - O: oval (dominant) - o: long Note that both parental plants are heterozygous for these traits, which means their genotypes are RrOo.

For a dihybrid cross where both plants are heterozygous for both traits, we would expect a phenotypic ratio of 9:3:3:1 for the progeny if the genes are independently assorted. That is, the expected genotypic ratios would be: - 9 red, oval (R- O-): DOMINANT- DOMINANT - 3 red, long (R- oo): DOMINANT- RECESSIVE - 3 yellow, oval (rr O-): RECESSIVE- DOMINANT - 1 yellow, long (rr oo): RECESSIVE- RECESSIVE To apply these ratios to our problem and use them on a Chi-square test, we must adjust them to represent the percentages of each phenotype. To do this, we can consider that the ratios add up to 16 (9+3+3+1), and find the percentage for each phenotype by dividing each ratio by 16 and multiplying by the total number of progeny for each parental plant, which is 100.

We now calculate the expected number of progeny for each phenotype considering the 9:3:3:1 ratio: - Red, long (DOM-REC): Expected = (3/16) * 100 = 18.75 - Yellow, oval (REC-DOM): Expected = (3/16) * 100 = 18.75 - Red, oval (DOM-DOM): Expected = (9/16) * 100 = 56.25 - Yellow, long (REC-REC): Expected = (1/16) * 100 = 6.25 Now, we perform the chi-square test using the formula: $$\chi ^{2}=\sum \frac{\left(\text {observed}-\text {expected}\right)^{2}}{\text {expected}}$$ For Plant A: $$\chi _{A} ^{2}=\sum \frac{\left(\text {observed}-\text {expected}\right)^{2}}{\text {expected}} = \frac{(46-18.75)^2}{18.75} + \frac{(44-18.75)^2}{18.75} + \frac{(5-56.25)^2}{56.25} + \frac{(5-6.25)^2}{6.25} = 29.07$$ For Plant B: $$\chi _{B} ^{2}=\sum \frac{\left(\text {observed}-\text {expected}\right)^{2}}{\text {expected}} = \frac{(4-18.75)^2}{18.75} + \frac{(6-18.75)^2}{18.75} + \frac{(43-56.25)^2}{56.25} + \frac{(47-6.25)^2}{6.25} = 29.07$$

Both chi-square values are higher than the critical value for 3 degrees of freedom (7.815), which means that the null hypothesis (independent assortment) is rejected. It indicates that there is a linkage between the fruit color and fruit shape genes. The progeny numbers show that the dominant traits are mainly appearing together or along with the recessive traits, which further strengthens the assumption that the genes are linked. Now, we can report the genotypes of the parental plants and the gene linkage. Since both parental plants were heterozygous for these traits, their genotypes are RrOo. The high chi-square values demonstrate that the genes are linked, and the fruit color and fruit shape genes are not independently assorted.