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Ah...Thanks a lot. My heart is now (relatively) at ease for States.

does anyone have any helpful hints for heredity

Seal_of_Disapproval wrote:No. Hardy-Weinberg is really the only genetics topic that requires math, and it is not tested in Heredity. Of course, basic math is required to do mono-, di-, and trihybrid crosses, but you knew that already.

Interestingly, Hardy-Weinberg actually did show up on my state test... But other states may be different.

Does anyone have any last minute suggestions for studying, for Heredity, and in general? States is tomorrow, and I can't help but think that I'm overestimating myself. My notes have quite a bit of blank space on them, and I don't know what topics to keep studying. What details do you guys think they could possibly ask for?
Thank you for your help in advance. Also, please don't bother responding after today/tomorrow night.

Does anyone have any suggestions for solving punnett squares with 5 (or more) traits?
Several practice tests I have done have multiple questions having to do with 5 trait crosses, i.e.: AaBbCCDdEe x aaBbccDdee. Some questions are easy, like what fraction is expected to have all recessive traits. (Obviously, none because of the CC on one of the parents.) But questions like what fraction could have all dominant traits, or all heterozygous....would take like 10 hours! 5 x 5 punnett squares have 1,024 boxes!!! D:

Half-Blood-Princess wrote:Does anyone have any suggestions for solving punnett squares with 5 (or more) traits?
Several practice tests I have done have multiple questions having to do with 5 trait crosses, i.e.: AaBbCCDdEe x aaBbccDdee. Some questions are easy, like what fraction is expected to have all recessive traits. (Obviously, none because of the CC on one of the parents.) But questions like what fraction could have all dominant traits, or all heterozygous....would take like 10 hours! 5 x 5 punnett squares have 1,024 boxes!!! D:

Oh my god. Using a punnett square to solve this is pretty much impossible, haha.
Use probability for each individual "cross", then multiply each.
I'll use your example, but CC in the first parent is changed to cc.
For "AaBbccDdEe x aaBbccDdee" with an offspring of "aabbccddee", you'd do Aa x aa * Bb x Bb * CC x cc * Dd x Dd * Ee x ee.
For A, the probability of having homozygous recessive individual is .5. For B, it's .25. For C, it's 1. For D, it's .25. For E, it's .5.
.5 * .25 * 1 * .25 * .5 = .015625 = probability of getting aabbccddee
I hope my calculations were correct. Have fun studying!

Half-Blood-Princess wrote:Does anyone have any suggestions for solving punnett squares with 5 (or more) traits?
Several practice tests I have done have multiple questions having to do with 5 trait crosses, i.e.: AaBbCCDdEe x aaBbccDdee. Some questions are easy, like what fraction is expected to have all recessive traits. (Obviously, none because of the CC on one of the parents.) But questions like what fraction could have all dominant traits, or all heterozygous....would take like 10 hours! 5 x 5 punnett squares have 1,024 boxes!!! D:

There's a trick you can use to solve them very easily, and you can even use it on any number of traits; definitely easier than making a Punnett square for two or more traits, which is time consuming.

Remember, alleles are independently assorted. When you cross AaBb x AaBb, the As and the Bs don't affect each other, which is why you can get possibilities of AABB or Aabb or whatever. So what you can do is find the probability of each trait separately, and just multiply the probabilities together.

Let's say you want to cross AaBbCcDdEE with aaBbCcddEE, and you want to know the probably of getting an offspring with aaBBCcDdEE. Do the following probabilities:

Aa x aa = 1/2 aa
Bb x Bb = 1/4 BB
Cc x Cc = 1/2 Cc
Dd x dd = 1/2 Dd
EE x EE = 1 EE

So to get aaBBCcDdEE, you do 1/2 x 1/4 x 1/2 x 1/2 x 1, which is 1/32.

This works for phenotypes too. Let's say that you have a cross of AaBbCc x AaBbCc, and you want an offspring that shows all three dominant phenotypes. So you're resulting genotype is A_B_C_, because it doesn't matter what that second allele is. You do the crosses:

Aa x Aa = 3/4 A_
Bb x Bb = 3/4 B_
Cc x Cc = 3/4 C_

Then you do 3/4 x 3/4 x 3/4 and you get 27/64.

So you can just break down these big crosses into simple single gene crosses and just multiple them, it saves a ton of time and simplifies everything greatly.

That works perfectly and so much faster!!!! Thank you both so much! I thought about doing that once but for some reason I didn't think it would work... Yay! Thank you

@CS:
Yeah, Hardy Weinberg showed up on the test. I asked the proctor if that wasn't a Division C topic, but he said it was, and I quote, "a bit of a stretch, but it's fine". -.- So my partner and I ended up putting: "1. Assuming we are Div. C 2. Assuming we know this" and some other stuff.. xD

PicturePerfect wrote:@CS:
Yeah, Hardy Weinberg showed up on the test. I asked the proctor if that wasn't a Division C topic, but he said it was, and I quote, "a bit of a stretch, but it's fine". -.- So my partner and I ended up putting: "1. Assuming we are Div. C 2. Assuming we know this" and some other stuff.. xD

Isn't that a nats topic? I mean so is transcription and translation but we had that on our test. I haven't seen H-W yet