Albino Gene Investigation

The purpose of this investigation is to determine the inheritance pattern of the albino trait and whether the albino gene in corn seedlings is dominant or recessive. One group of lab partners in a lab class examine a tray of corn seedlings showing both green and albino plants. The students record their findings and compare their tray of corn seedlings’ results to five other lab groups in the class. The lab students used this information to test all of the possible gene combinations the parents that produced partial green and albino offspring may have had, using Punnett squares. Based on their findings, the groups determine the hypothetical genotype and phenotype ratios of the corn seedlings and use this to form a hypothesis. Their hypothesis:

Next, the amount of seedlings in the tray that expressed a green or white phenotype were both counted. The group observed that there were eleven total seedlings; eight of those seedlings were green and the three remaining were white. The next step in the experiment was to add up the class totals of green to albino seedlings, as shown in table 1. After determining a group phenotype ratio of 8:3 and a class ratio of 42:10, the group questioned what defect the white plants could have that might make the seedlings appear albino instead of green. The students concluded that a lack of chloroplast could explain why some of the seedlings were albino. The lab partners also decided that for a recessive, albino phenotype to be expressed, GG, Gg, and gg had to be the genotypes of the seedlings. Subsequently, the lab group answered: “Would it be possible to have had sexually mature parents with either of the traits expressed in the offspring? Explain (Inheritance of a Single Trait - Lab 9, p. 131).” The group decided that yes, it would be possible for the parents to have been sexually mature and still possess the traits necessary to produce an albino offspring. However, the only way this would have been possible was if the parent’s genotype carried the recessive g gene, but the phenotype of said parent or the actual parent was green. If the parent had been albino, and thus lacked

Since the degree of freedoms equals one minus the number terms in the ratio being analyzed and the ratio being analyzed is 3:1 (with two terms), the students decided that the degrees of freedom was one. The last step was to find the p-value on the chi-square chart in Table 7. This is the most important step in the experiment because if the p-value number is less than .05, the hypothesis must be rejected (Chi-Square Test). Using Table 7, the group went to 1 degree of freedom in the y axis and the chi-square number on the x axis that was closest to .92. The students then conclude that the approximate p-value was .0158 or