Allele frequencies of the next generation

Click For Summary
SUMMARY

The discussion centers on calculating allele frequencies using the Hardy-Weinberg equation based on a gene pool of 75 light-colored beans (F) and 25 dark-colored beans (f). The total number of beans is 100, leading to an initial frequency of F (p) at 0.75 and f (q) at 0.25. The confusion arose from a misinterpretation of the question regarding the next generation's allele frequencies, which remain unchanged under Hardy-Weinberg equilibrium. Ultimately, the correct frequencies for the next generation are confirmed as 0.75 for F and 0.25 for f.

PREREQUISITES
  • Understanding of the Hardy-Weinberg principle
  • Basic knowledge of allele frequency calculations
  • Familiarity with dominant and recessive traits in genetics
  • Ability to interpret genetic problems involving diploid organisms
NEXT STEPS
  • Study the Hardy-Weinberg equilibrium in detail
  • Learn about dominant and recessive inheritance patterns
  • Explore real-world applications of allele frequency calculations
  • Investigate factors that can disrupt Hardy-Weinberg equilibrium
USEFUL FOR

Students studying genetics, educators teaching genetic principles, and anyone interested in understanding population genetics and allele frequency calculations.

Adriane
Messages
9
Reaction score
0

Homework Statement


Starting with 75 light-colored beans and 25 dark-colored beans, the 100 beans represent my gene pool. Light beans = F and dark beans = f.
p = (total F alleles) / (total number of beans used)

And now I'm stuck where it's asking me to "practice determining the initial allele frequencies (p and q) for both F and F alleles, but change the top number to genotype ff x 2."

Homework Equations


Hardy-Weinberg equation = p^2 + 2pq + q^2 = 1

The Attempt at a Solution


I got this far:
I have 75 light beans and 25 dark beans. Light beans = F and dark beans = f.
Number of offspring with genotype FF (2 light beans) = 75 x 2 = 150

Number of offspring with genotype Ff (one dark one light) = 25 x 1 = 25

Total F alleles = 175

I thought to find the allele frequencies of p and q in the next generation I had to square them and then do 2pq and add it, but nothing I do adds up to 1.
 
Physics news on Phys.org
Adriane said:

Homework Statement


Starting with 75 light-colored beans and 25 dark-colored beans, the 100 beans represent my gene pool. Light beans = F and dark beans = f.
p = (total F alleles) / (total number of beans used)

And now I'm stuck where it's asking me to "practice determining the initial allele frequencies (p and q) for both F and F alleles, but change the top number to genotype ff x 2."

Homework Equations


Hardy-Weinberg equation = p^2 + 2pq + q^2 = 1

The Attempt at a Solution


I got this far:
I have 75 light beans and 25 dark beans. Light beans = F and dark beans = f.
Number of offspring with genotype FF (2 light beans) = 75 x 2 = 150

Number of offspring with genotype Ff (one dark one light) = 25 x 1 = 25

Total F alleles = 175

I thought to find the allele frequencies of p and q in the next generation I had to square them and then do 2pq and add it, but nothing I do adds up to 1.

It was not very easy to make out what the question was (it would have been better to quote from your book or problem sheet) but it seems the question is nothing about the next generation but about the one you've got. You've quoted Hardy-Weinberg, which assumes genetic equilibrium so the next generation will have just the same frequencies of everything as this. I would bet it says that the dark beans are ff, doesn't make too much sense otherwise. Small letters conventionally represent recessive genes. So you re told the frequency of ff, not too difficult to work out the frequency of f.
 
epenguin said:
It was not very easy to make out what the question was (it would have been better to quote from your book or problem sheet) but it seems the question is nothing about the next generation but about the one you've got. You've quoted Hardy-Weinberg, which assumes genetic equilibrium so the next generation will have just the same frequencies of everything as this. I would bet it says that the dark beans are ff, doesn't make too much sense otherwise. Small letters conventionally represent recessive genes. So you re told the frequency of ff, not too difficult to work out the frequency of f.
Thank you so much, turns out I was being played by my teacher. He wanted to see if we would come to this conclusion, and I over thought the whole thing. Thanks again.
 
So what is the answer?
 
The answer is just what you said, that in Hardy-Weinberg equilibrium the allele frequency would stay the same through each generation. So basically all that work for a simple answer.
 
So the allele frequencies in next generation remain the same as for this generation (most probably, for large enough populations etc.) but the question was what are they?
 
It was .75 for F and .25 for f.
 
Adriane said:
It was .75 for F and .25 for f.

I don't think so.
 
Last edited:
epenguin said:
I don't think so.
Then what do you think it was?
 
  • #10
Adriane said:
Then what do you think it was?

The rule is I shouldn't answer for you. But you could not have a more classic first-lesson example of inheritance in a diploid organism with dominant/recessive genes; does a phenotype 3:1 ratio remind you of anything?
 
Last edited:

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
6K
Replies
12
Views
4K
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
16K
  • · Replies 2 ·
Replies
2
Views
690
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 13 ·
Replies
13
Views
4K