You cross two heterozygous plants and observe the following numbers of the 320 offspring:       ObservedNumber AABB Long stems,wrinkled seeds 160 AAbb Long stems,round seeds 55 aaBb Short stems,wrinkled seeds 75 aabb Short stems,round seeds 30 The calculated chi square for this data is 11.39 and the critical chi-square is 7.81. Should you reject the null hypothesis? Group of answer choices 1. Yes, we reject the null hypothesis 2. No, we fail to reject the null hypothesis

Two different house Flys were crossbreed for 2 generations. The female was a wild fly and the male was a mutant with apterous wings and a black body. Given the data of the second generation: is this data consistent with the expected 9:3:3:1 ratio expected of a dihybrid cross?

Please solve question 5 and letter B

Please help with this Evolutionary Biology question

Pretend that you are comparing the actual genotype distribution for a population with the distribution of genotypes predicted by the Hardy-Weinberg theorem.  So your hypothesis is that the population is in Hardy-Weinberg equilibrium (i.e. that actual population data fit the Hardy-Weinberg expectations).  If you carry out a chisquare goodness of fit test and calculate a total chisquare value of 0.03 with 1 degree of freedom (see table), what does this mean? (select all true statements)a) The data do NOT fit the hypothesized distribution.b) The data do fit the hypothesized distribution well enough, so we accept the hypothesis at this time (i.e. we cannot reject the hypothesis). c) The probability that the data came from a population in Hardy-Weinberg equilibrium is too small, so we reject the hypothesis.d) The probability that the data came from a population in Hardy-Weinberg equilibrium is too big, so we reject the hypothesis.e) The data support Hardy-Weinberg expectations – there is no…

Can you show the cross on how to get the answers? What is the ratio that is expected with independent assortment? Can you explain anything else related to it ? Nonindependent assortment

John, noted acarologist (he studies mites) and fancy humming bird breeder hobbyist, is attempting to breed giant puffing humming bird that he can present at the fancy balls given in his glittering social circle. After years of experimenting with humming bird, John has found that the slope of the least squares regression of Midoffspring on Midparent values is 0.75. All the humming birds were allowed to breed, and a first batch of offspring was produced. The table below presents data on the puffing ability (in cm3) of 10 breeding families of humming bird in his dovecote: Family midparent mid offspring (first batch) 1 55 56 2 68 72 3 77 81 4 37 40 5 25 30 6 34 32 7 44 44 8 23 49 9 60 39 10 67 68 All the humming birds were allowed to breed, and a first batch of offspring was produced. These offspring were used to calculate the midoffspring values above, and when John calculated the slope of the least squares regression of Midoffspring on Midparent values, he…