The parental generation is the first set of parents crossed.
The F1 (first filial) generation consists of all the offspring from the parents - their children.
The F2 (second filial) generation consists of the offspring from allowing the F1 individuals to interbreed - the grandchildren of the parental generation.
2007-11-04 13:22:36
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answer #1
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answered by ecolink 7
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F1 Generation
2016-09-28 09:32:30
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answer #2
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answered by ? 4
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1. We don't know which is dominant yet, so let's represent wild-type as Wi, and white-eyed as Wh. All females are Wh in the first generation, so they are WhWh. All males in the first generation as Wi, but all of the males in the second generation are Wh, while all of the females are Wi. Wi is dominant E, and Wh is recessive e. The gene is X-linked, so males only have 1 copy. The presence of a brown-eyed female is confusing here--are these lab results, or results your teacher handed to you? As lab results, you would most likely explain that as another fruit fly getting into the tube, and it wouldn't usually show up. You could make a Punnett square showing genotype crosses if the gene is autosomal (each parent has two copies), and then show the genotype crosses if the gene is X-linked. For the autosomal cross, you'd show WiWi x WhWh crossing. Notably, every single offspring ends up with the same genotype in this cross--WiWh--and so should have the same phenotype. For the X-linked cross, you'd show WhWh x Wi Y crossing. In this case, all females are heterozygous (WiWh) and all males are homozygous (WhY). If Wh was dominant, everyone would have the same phenotype. Because all females are wild-type and all males are white-type, you know that Wi is the dominant E, and Wh is the recessive e. Parental females are ee, and parental males are EY. F1 females are Ee, and F1 males are eY. It's useful to do the F1->F2 cross to see if the prediction holds. You'll see in the Punnett square that you get equal mixing of phenotypes for F2, and that basically holds with that you see. 2. To use the chi-square test, you'll need to figure out the predicted values for each F2 phenotype. We can use the cross we did at the end of 1 for this. We expect 25% of the F2 generation to have each phenotype. There were a total of (31+23+24+22) 100 offspring in F2, so we would expect 25 of them to have each phenotype. For each phenotype, do (Observed - Expected)^2 / Expected. Then add up all of the values. This is your chi-squared value. ((31-25)^2)/25 + ((23-25)^2)/25+ ((24-25)^2)/25+ ((22-25)^2)/25 = 2 Now, you'd use a chart to figure out the probability that your prediction is correct. You have 4 terms, so there are 3 degrees of freedom (degrees of freedom is just the number of terms minus 1). It's basically an examination of how many categories of things you are testing. According to the chart in my old genetics textbook (the links at the bottom give other charts--I had to do a quick review of Chi-Square), a chi-square value of 2 with 3 degrees of freedom is somewhere between 80% and 50%. The critical value for rejecting a hypothesis is 50%; your data surpasses this value, so the prediction is correct.
2016-03-26 21:28:36
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answer #3
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answered by Anonymous
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how do i know that either i have to stop a cross at F1 generation or at f2 ??????
2015-11-14 02:14:23
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answer #4
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answered by aftab 1
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parental is the first generation. use a punnett square and cross them
y y
x xy xy
x xy xy
now the resulting generation, al xy's are the F1 generation. if you cross them, you get this.
x y
x xx xy
y xy yy
The F2 generation has a 25% chance of xx, 50% chance of xy, and 25% chance of yy.
2007-11-04 13:24:03
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answer #5
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answered by Anonymous
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so if getting a puppy would would people say is better to get F1 or F2? thanks
2015-06-14 18:37:44
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answer #6
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answered by Alicia 1
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Ecolink has gotten everything
2014-10-13 01:41:19
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answer #7
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answered by Juma james 1
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