Question

Testing the Hypothesis of Independent Assortment

A monohybrid cross considers the inheritance of a single trait. To better appreciate the amount of labor and ingenuity that went into Mendel’s experiments, proceed through one of Mendel’s dihybrid crosses.

Background: Consider that pea plants mature in one growing season, and you have access to a large garden in which you can cultivate thousands of pea plants. There are several true-breeding plants with the following pairs of traits: tall plants with inflated pods, and dwarf plants with constricted pods. Before the plants have matured, you remove the pollen-producing organs from the tall/inflated plants in your crosses to prevent self-fertilization. Upon plant maturation, the plants are manually crossed by transferring pollen from the dwarf/constricted plants to the stigmata of the tall/inflated plants.

Hypothesis: Both trait pairs will sort independently according to Mendelian laws. When the true-breeding parents are crossed, all of the F1 offspring are tall and have inflated pods, which suggests that the tall and inflated traits are dominant while the dwarf and constricted traits are recessive. A self-cross of the F1 heterozygotes results in 2,000 F2 progeny.

Test the hypothesis: Because each trait pair sorts independently, the ratios of tall:dwarf and inflated:constricted are each expected to be 3:1. The tall/dwarf trait pair is called T/t, and the inflated/constricted trait pair is designated I/i. Each member of the F1 generation therefore has a genotype of TtIi. Construct a grid analogous to Figure 12.16 found in the textbook, in which you cross two TtIi individuals. Each individual can donate four combinations of two traits: TI, Ti, tI, or ti, meaning that there are 16 possibilities of offspring genotypes. Because the T and I alleles are dominant, any individual having one or two of those alleles will express the tall or inflated phenotypes, respectively, regardless if they also have a t or i allele. Only individuals that are tt or ii will express the dwarf and constricted alleles, respectively. As shown in Figure 12.19 in your textbook, you predict that you will observe the following offspring proportions:

tall/inflated: tall/constricted: dwarf/inflated: dwarf/constricted in a 9:3:3:1 ratio.

Notice from the grid that when considering the tall/dwarf and inflated/constricted trait pairs in isolation, they are each inherited in 3:1 ratios as expected with a monohybrid cross.

Figure 12.19 in your textbook shows all possible combinations of offspring resulting from a dihybrid cross of pea plants that are heterozygous for the tall/dwarf and inflated/constricted alleles.

Compose a response to the following questions. Enter both question and answers in your Learning Journal.

Test the hypothesis: You cross the dwarf and tall plants and then self-cross the offspring. For best results, this is repeated with hundreds or even thousands of pea plants. What special precautions should be taken in the crosses and in growing the plants?

Analyze your data: You observe the following plant phenotypes in the F2 generation: 2706 tall/inflated, 930 tall/constricted, 888 dwarf/inflated, and 300 dwarf/constricted. Reduce these findings to a ratio and determine if they are consistent with Mendelian laws.

Form a conclusion: Were the results close to the expected 9:3:3:1 phenotypic ratio? Do the results support the prediction? What might be observed if far fewer plants were used, given that alleles segregate randomly into gametes? Try to imagine growing that many pea plants, and consider the potential for experimental error. For instance, what would happen if it was extremely windy one day?

Answer 1

In a Cross between dwarf and tall plants, we need to assure that the both parents are homozygous for that trait. They are cross fertilized, for which emusculation is done in one parent plant. Next precaution is to make self pollination in f1 plants which is made by proper covering of flowers. The progeny should be counted accurately.

F2 generation:

2706 tall/inflated,

930 tall/constricted,

888 dwarf/inflated,

300 dwarf/constrict

Total progeny is 4824,

According to 9: 3: 3: 1 ratio, it should be 2713: 905: 905: 301.

If fewer plants would be selwcted, the ratio should be same,

I. e, 9: 3: 3: 1.

On windy day there are chances of cross pollination, that would lead to failure of experiment, but as we are expecting self pollination therefore we had to cover the flowers to make them self pollinate.  On that occasion there is no role of wind.