The latest “crossover event” does provide some much desired Bailey and Ben moments.
Give peas a chance greys anatomy putlocker update#
I’ll update you on firefighter drama where it seems necessary, but can we all just agree to cool it with the crossovers for a while now? The doctors over at Grey Sloan demand our full attention. If you don’t watch Station 19, however, I have to imagine this episode doesn’t hit as hard. As someone who watches both shows, let me tell you: This one’s a doozy it’s one of the more emotionally affecting Seattle crossovers in a long time. Thus, we may tentatively accept the idea that these genes assort independently.It doesn’t seem like the best sign that Grey’s Anatomy had to borrow the firefighters from Station 19 in order to get the emotional heft this season has been lacking, does it? There have been complaints about the number of crossovers between the shows for a while now, and this one in particular relies heavily on your connection to the folks over at SFD. Because the computed value of the chi-square statistic (0.52) is much less than the critical value (7.815 see table 3.2), there is no evidence to reject the hypothesis of independent assortment of the vine length and tower colour genes.
This statistic must then be compared to the critical value of the chi-square frequency for 3 degrees of freedom (calculated as the number of phenotypic classes minus one). To compute the chi-square statistic, we must obtain the difference between each observation and its predicted value, square these differences, divide each squared difference by the predicted value and then sum the results: Under the assumption that the two genes assort independently, the four phenotypic classes in the F2 should each be 25 percent of the total (200) that is, each should contain 50 individuals. To obtain this statistic, the results must be compared to the predictions of the genetic hypothesis. The hypothesis of independent assortment of the vine length and flower colour genes must be evaluated by calculating a chi-square test statistic from the experimental results. For each gene, the frequency of heterozygous offspring is 1/2 thus, the frequency of triple heterozygotes should be (1/2) x (1/2) x (1/2) = 1/8 (e) Offspring that are not heterozygous for all three genes occur with a frequency that is one minus the frequency in D, Thus the answer is 1 - 1/8 = 7/8 Thus, the frequency of triple recessive homozygotes is (1/4) x (1/4) x (1/4) = 1/64 (c) To obtain the frequency of offspring that are either triple dominant homozygotes or triple recessive homozygotes (these are mutually exclusive events) we sum the results of (a) and (b): 1/64 + 1/64 = 2/64 = 1/32 (d) To obtain the frequency of offspring that are ripple heterozygotes, again, we multiply probabilities. For each gene the frequency of recessive homozygotes among the offspring is 1/4. Thus, we can calculate the frequency (that is, the probability) of AA BB CC as (1/4) x (1/4) x (1/4) = 1/64 (b) The frequency of aa bb cc individuals can be obtained using similar reasoning. (a) When Aa individuals are selfed, 1/4 of the individuals will be AA likewise for the B and C genes, 1/4 of the individuals will be BB and 1/4 will be CC.
Because the genes assort independently, we can analyze them one at a time to obtain the answers to each of the questions.
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