I need a hypothesis based on this lab. Hypothesis : Write a possible solution fo

Question

I need a hypothesis based on this lab.

Hypothesis: Write a possible solution for the question or problem being investigated. The hypothesis statement should be testable and is often in an if-then-because format to illustrate the manipulated variable, the expected results, and the reason(s) for those expectations.

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As question says that there are total 100 beans in a bad. In which 75 are light colored (FF) and 25 are dark colored (ff). If the gene pool which is produced by mating between homozygous (FF) and and heterozygous (Ff) determine then we can determine the number of dominant alleles in the population and the number of reces let we picked FF-28 Ff-19 f-3 Then we can calculate the allele frequency We can calculate dominant allele number of offspring with genotype FF (two light beans) 28 × 2-56 F alleles number of offspring with genotype Ff(one dark one light) 19 × 19 F allcles Total F alleles = 75 p- Total number of F alleles Total number of alleles in population (Total number of bcans you used) p= 75/100-0.75 We can calculate recessive allele by using the same formula number of offspring with genotype ff (two dark beans) 3-2-6 F alleles number of offspring with genotype Ff (one dark one light) 19 x 1 -25 F alleles Total f alleles = 25 q Total number of F alleles Total number of alleles in population (Total number of beans you used) q25/100-0.25 Predictions for dominant allele frequency Generation Start sive alleles in the ulation Number 75 28(FF) 19(Ff) 32(FF) 1(Ff) Frequency 075 percentage 75 28x2-56 19x1-19 32x2-64 11x1-11 34x2-68 07x1=7 30x2-60 15x1=15 27x2-54 21x1-21 75/100 0.75 75/100-0.75 75/100-0.75 6(FF) 23(Ff) 75 /1 00-0.75 75/100-0.75 31(Ff) Predictions for recessive allele frequency Generation Start umbe 25 3(ff) 19(Ff) Percentage 25 3(ff x2-6 Frequency 0.25 19x1-19 25/100-0.25 7(ff) 11(Ff) 9(ff) 7(Ff) S(ff) 15(Ff) 2(ff) 1(F0 7x2-14 11x1-11 9x2-18 7x1-7 5x2=10 15x1=15 2x2-4 21x1-21 25/100-0.25 25/100-0.25 25/100 0.25 25/100-0.25

Explanation / Answer

Hypothesis-

In a large, randomly mating population where mutations, migration, and selection do not occur, allele and genotype frequencies will remain at equilibrium.

Let frequency of the dominant allele is p and the frequency of the recessive allele is q.
then the expected genotype frequencies using the Hardy Wienberg equation p2 + 2pq + q2 = 1

as p=.75 and q=0.25

p2+2pq+q2=1

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