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Melody  11 Feb 2022
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PROBLEM #1.
You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:
The frequency refers to the frequency in the population.

The frequency of the "aa" genotype.
The frequency of the "a" allele.
The frequency of the "A" allele.
The frequencies of the genotypes "AA" and "Aa."

Hi tjhill19,
I see that you have become a member. Welcome to Web2.0calc forum.
Your questions are interesting. I shall have some fun trying to help you.

I am not a biologist so I may not use the correct words or put them together in the correct ways. Sorry about that.
Now, these are probability questions. They are Bernoulli probability questions that use binomial techniques to solve. (Once again my words may not be quite correct, I have always just referred to them as binomial probability questions.)
Now.
Each strand of DNA is a double helix. I have not heard the word allele before but it must be the the gene on one side of the double helix.
This means that for any particular genotype (on just one side of the double helix) you either have a recessive a allele or you have the Dominant A allele.
You have one or the other, you can't have both, and you can't have anything else.
Since it is a double helix, these are the combinations that are possible
AA
Aa
aA
aa

Now I need to discuss probability.
All probabilities must be between 0 and 1
If the event cannot happen then the probability is 0
if the event is absolutely going to happen then the probability is 1
The probability of an event happening = the number of ways it can happen / the total number of things that can happen altogether.

"You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:"
If the genotype is homozygous it means that both sides of the double helix carry the same gene (One was inherited from the mother and the other from the father)
They are both the recessive gene which means that this trait (whatever it is) will tend to be unusual in the population. If only one was a and the other was A then the alternate trait would be the one in evidence.

Back to probability
say i have 7 b***s in a bag. 3 are blue and 4 are red
If i choose a ball at random then the probability that it will be blue is 3/7 the probability that it will not be blue is 4/7 (3/7 + 4/7 = 1 this must be so because I have to choose one or the other.)
This is a large part of what makes it a binomial probability. Either one thing happens or it doesn't. Nothing else is important.
If I put the ball back into the bag and choose again then on the second try
The probability that it will be blue is 3/7 the probability that it will not be blue is 4/7 (Exactly the same as before)
This is another large part of what makes it a binomial probability. Each trial is independent of all other trials, the odds (probability) of success stays the same.

Now the probability of selecting two Blue b***s is P(BB) = 3/7 * 3/7 = 9/47 (it is always multiply for questions like these)

Going backwards now, (I know that there are only blue and red b***s to choose from)
if i was told that the probability of choosing 2 blue b***s was 9/47 . I could know from that. that the probability of choosing a blue ball on a single trial must be sqrt( 9/49) = 3/7
If the probability of choosing a blue ball is 3/7 then the probability of NOT choosing a blue ball must be 1- 3/7 = 4/7
The probability of choosing 2 NON blue b***s must be 4/7 * 4/7 = 16/49
Since it is blue b***s that I want I am going to let the probability of choosing a blue be 'p'
and I am going to let the probability of not choosing a blue by 'q'
p+q = 1 ==> or alternatively q=1-p
also if I square both sides
(p+q) 2 = 1 2
(p+q) 2 = 1
p 2 + 2pq + q 2 = 1 (these were the formulas you were told that you must learn, but it would be better if you understood them rather than just memorized them)
If you don't understand how i did this expansion - make sure you ask and i will show you.

Back to your question
You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:
The frequency refers to the frequency in the population.

The frequency of the "aa" genotype. 36% It is in the question ===> p 2 from the formula
The frequency of the "a" allele. sqrt(36%)= sqrt(0.36) or 0.6 ==> 60%
The frequency of the "A" allele. 1-60% = 40%
The frequencies of the genotypes "AA" and "Aa."
P(AA) = 0.4 * 0.4 ==> 0.16 =======16% =====> q 2 from the forula
P(Aa)+P(aA) = 2* 0.4* 0.6 = 0.48 ==> 48% ===> this is 2pq (from the formula)

The frequencies of the two possible phenotypes if "A" is completely dominant over "a."
the phenotype must be the way in which the trait presents itself in the population.
The recessive gene will only present if the homozygous recessive genotype (aa) is present
Therefore the dominant phenotype will occur 16% + 48% = 64% of the time.
Since we already knew that the recessive phenotype would occur 34% of the time.
We could also have said that the recessive phenotype would occur (100-34)% = 64% of the time.

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P(A) + P(a) = p+q =1
P(AA) + [P(Aa)+P(aA)] + P(aa) = p 2 + 2pq + q 2 = 1

Try and digest what I have said and then get back to me.
Maybe you can try the next problem by yourself.
18 Feb 2014
 #1
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I have no idea, sorry.
17 Feb 2014
 #40
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Well now, that was two weird days,
I hope we never get another one like those !

I think that Andre Massow (creator of web2.0calc.com) heard all our cries. He sent me a very nice email.

Basically he said he is working very long hours and is self funding the site and it has been difficult for him.
He apologized for all today's technical difficulties but said that the servers are working to capacity and the site will have to be revamped in the not too distant future.
Andre said that a lot of processing capacity was used to keep spam posters out of the system. For this we should be very grateful. I have seen forums overrun by spam and I always wondered why this forum did not get similarly swamped. He was concerned though that someone was using an adblocker on the site. He needs genuine adds to generate funding. I know nothing about this, if anyone else does perhaps they could send me a private message.
He thanked me for all the time and effort that I have given to the forum. It is always nice to be appreciated. He said he had not answered any of my correspondence in the past because he gets swamped with emails, he has very little time and he simply had not seen mine.
Anyway, it was not his intent to stop this forum from functioning. I don't think that he realized that it was even being used effectively. He understands now and he says he will be considerate of our needs.

It was obviously a very quiet weekend, (except for the horror cries) but there were some great answers provided by Rom and Demogorgan.
Stu posted two more useful web sites. Thanks Stu.
1) Enrich Learning - this is an enriching maths site rather than a standard learning site, if that makes sense. It looks quite interesting.
2) Tiger maths - Algebra explained and illustrated. This also looks like a very comprehensive and worth while site.
Goodnight all,
I am glad that I am a lot happier now than I was this morning.
Melody.
17 Feb 2014