Construct a grid in which you cross two TtIi individuals. F1 = TtIi Gametes T, t, I, i. F2 TI Ti tI ti TI TTII TTTi TtII TtIi Ti TTIi TTii TtIi Ttii tI TtII TtIi ttII ttIi ti TtIi Ttii ttIi ttii 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? We should make sure that the F1 generation is allowed to self-pollinate. We should make sure that the F1 generation do not pollinate with a true breeding plant/pure breed pant. 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. tall/inflated: tall/constricted: dwarf/inflated: dwarf/constricted in a 9:3:3:1 ratio 2706/300 = 9.02 930/300 = 3.1 888/300 = 2.96 300/300 = 1 9.02: 3.1: 2.96: 1 Our Ratio is consistent to Mendelian law. 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? Yes, our results 9.02: 3.1: 2.96: 1 is close to the expected 9:3:3:1 phenotypic ration. Yes our results support the prediction as our ratios are close to the expected one. I don’t think anything could go wrong on a windy day if there are no true breeding plants around but it there are, the wind could carry the true breeding plant’s pollen and pollinate with the F1 generation and give errors in the results since the true breeding flowers won’t be TtIi. In accidents like these we would have to start the whole experiment over again, staring with two true breeding plants.

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