Calculating P and Q Frequencies in Three Generations of Bacteria Populations

[komal12]

hi can someone help with this problem please..im having a hard time understanding how to do it. thanks

calculate the p and q frequencies for each of the three generation of bacteria populations, assuming the resistant allele is recessive.

1st generation: 3 resistant individuals and 53 non-resistant

2nd generation; 13 resistant individuals and 43 non-resistant

3rd generation: 53 resistant individuals and 3 non-resistant

 

[Astronuc]

Here are some notes on population genetics.
http://www.life.uiuc.edu/ib/201/lectures/PopGen.pdf

Now of one has resistant R and non-resistant N alleles, and R is recessive, does that not imply the resistant population must have RR, and the non-resistant population has RN or NN.

Then let p² be the frequency of RR, and the non-resitant population is 2pq+q² which are NR and NN. If one knows p then one can find q, since p²+2pq+q² = 1.


I don't do populations genetics, but that seems consistent with the notes - I hope.

 

[Oaksinstructor]

Yeah, this is all hardy-weinberg stuff, (which requires the utopia of hardy-weinberg equilibrium) Astronuc's lettering style is a bit unconventional but essentially correct. Usually p is the alleleic frequency of the dominant allele and q is the frequency of the recessive. Since it takes two dominant alleles to be homozygous dominant, the number of homozygous dominant organisms is p^2, same for homozygous recessive, so the number of recessive organisms is q^2, and our heterozygotes become 2pq (becuase of the 1:2:1 ratio of the standard punnett square).
Now, part of the problem here is, (since we are assuming complete dominance) we can't tell the difference in our Homozygous Dominant (which I will call 'NN'), and our Heterozygous (which I will call 'Nn') becuase they both express non-resistant. Only our Homozygous recessive (hereafter 'nn') express resistance and their number relates directly to q^2.Ok let's put this into practice: In your question, we know that 3 out of 56 total organisms are dominant recessive, this is roughly 5.4%; so q^2 = .054.
To get q (frequency of the recessive allele in the population), just take the square-root of .054 and you get q=.23 (23%) and since (p + q = 1) using algebra you can say (p = 1 - q)...

{Remainder of complete solution deleted by Moonbear. This is enough to get started.}[/color]

 

[Astronuc]

Oaksinstructor, welcome to PF and thanks for the contribution. I'm a nuclear engineer, and the last time I did anything like HW was 35+ years ago, so I defer to Oaksinstructor.

 

[komal12]

Moonbear is an idiot for deleting the solution because that isn't enough to get started and I am doing this through correspondence so i don't even have a teacher who i can ask for help.

 

[Dick]

Moonbear is an idiot for deleting the solution because that isn't enough to get started and I am doing this through correspondence so i don't even have a teacher who i can ask for help.

You can ask for help here. A complete solution is not help. It's a crutch. That's why it was deleted. Just ask the next question to help you take the next step. You've gotten a really amazing amount of help so far. Especially considering you gave absolutely zero input aside from stating the question.

 

[Ouabache]

...In your question, we know that 3 out of 56 total organisms are dominant recessive, this is roughly 5.4%; so q^2 = .054.

Typo? Don't you mean homozygous recessive?

Kamal, Dick is correct, complete solutions are a crutch. If you don't understand the concepts, you can easily get burned on an exam. Perhaps you can let us know where you are getting stuck.

 

[komal12]

hey everybody i'd like to apologize for what i said about moonbear..its just that i have a lot of courses that I am doing through correspondence and i don't have anyone to help me so i get frustrated when i don't get how to do things..

 

[fallenrose]

this doesn't make any sense to me..how do u get p=1-q and what does it equal?

 

[fallenrose]

hi I am learning the same thing in school. so if i understand this it would be a big help ..so would the p then be 77%?

 

[Moonbear]

this doesn't make any sense to me..how do u get p=1-q and what does it equal?

Let's try backtracking a bit. Do you know what p and q each represent? My suspicion is that you do not if you asked this question, but you might and just are confused by the algebra part, so I'm asking this question to find out where your confusion begins in order to know where to start helping.

 

[Oaksinstructor]

Yes, that was a typo, and yes it was supposed to be homozygous recessive.

to answer the latter question:
q = frequency of recessive alleles expressed as a percent or decimal. in this case 0.23 therefore, what I mean by the algebraic (p = 1 - q) is exactly that: (p = 1 - 0.23) and yes fallenrose is correct, .77 or 77%

to get teh rest of the answers, simply rinse and repeat, follow the same steps as described above only using the different numbers provided in the question.

 

[chemisttree]

to answer the latter question:
q = frequency of recessive alleles expressed as a percent or decimal. in this case 0.23 therefore, what I mean by the algebraic (p = 1 - q) is exactly that: (p = 1 - 0.23) and yes fallenrose is correct, .77 or 77%

Oaks, I believe the question was directed to Fallen. Moonbear wants to help Fallen and wants to know where to begin...

 

[fallenrose]

thanks for all the help i understand it now