Probability of seeing a genotype given a certain number of chromosomes and alleles
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So let's say I have a triploid species and I have 5 possible alleles for a particular locus. I have found that I can have 35 total possible genotypes (combination with replacement) and I know the allele frequencies for each of the 5 alleles, that I will randomly set as $$F_a = 0.2, F_b = 0.2, F_c = 0.4, F_d=0.1, F_e=0.1$$.
However, I am having a hard time wrapping my mind around how to calculate the number of occurrences of a particular genotype, $ab^2$, in a population of 1000 people.
I understand the allele frequencies, for instance, in the population of 1000 people, given all 1000 observed genotypes, 40% of them had allele $c$. But these are individual values and I have no information on how likely they are to be linked. (I am supposed to assume HW principle holds, but not sure how this helps).
How do I calculate the number of occurrences of the genotype $ab^2$ in the population? I would like an answer that provides good intuition on how to generalize well for any number of ploidy and any number of alleles.
I was told that I should multiply the frequencies together, $0.2*0.2^2$ and then times this by the number of ways I can get $abb$. But I am really confused on which double counts I am supposed to avoid and which double counts I actually want to keep. If I didn't care about double counts, then the number of ways I get $abb$ is $3!$. But if I care about double counts, then I'd divide by $2!$. But we actually want to double count, and that's the whole reason the hardy weinberd equation multiplies the heterozygote probability by 2. Really confused on this part of the problem.
probability
asked Jan 22 at 8:30
JonathanJonathan
388317
$endgroup$
add a comment |
$begingroup$
So let's say I have a triploid species and I have 5 possible alleles for a particular locus. I have found that I can have 35 total possible genotypes (combination with replacement) and I know the allele frequencies for each of the 5 alleles, that I will randomly set as $$F_a = 0.2, F_b = 0.2, F_c = 0.4, F_d=0.1, F_e=0.1$$.
However, I am having a hard time wrapping my mind around how to calculate the number of occurrences of a particular genotype, $ab^2$, in a population of 1000 people.
I understand the allele frequencies, for instance, in the population of 1000 people, given all 1000 observed genotypes, 40% of them had allele $c$. But these are individual values and I have no information on how likely they are to be linked. (I am supposed to assume HW principle holds, but not sure how this helps).
How do I calculate the number of occurrences of the genotype $ab^2$ in the population? I would like an answer that provides good intuition on how to generalize well for any number of ploidy and any number of alleles.
I was told that I should multiply the frequencies together, $0.2*0.2^2$ and then times this by the number of ways I can get $abb$. But I am really confused on which double counts I am supposed to avoid and which double counts I actually want to keep. If I didn't care about double counts, then the number of ways I get $abb$ is $3!$. But if I care about double counts, then I'd divide by $2!$. But we actually want to double count, and that's the whole reason the hardy weinberd equation multiplies the heterozygote probability by 2. Really confused on this part of the problem.
probability
asked Jan 22 at 8:30
JonathanJonathan
388317
$endgroup$
$begingroup$
There are three ways to get $ab^2$, namely $abb$, $bab$, and $bba$, so $frac{3!}{2!} = 3$ is correct.
$endgroup$
– N. F. Taussig
Jan 22 at 10:20
$begingroup$
This is a maths group rather than biology so you can only hope for help in some aspects of the problem. What is the HW principle? You won't find many questions specifically about these topics but you could consider it as a game with 3 5-sided dice and adapt some of the many questions and answers on dice.
$endgroup$
– badjohn
Jan 22 at 11:54
$begingroup$
Your figure of $35$ suggests that you consider $aab$, $aba$, and $baa$ as the same but not $aab$ and $abb$, is that correct? Remember, that even if the frequencies of all alleles were equal, the $35$ cases would not be equally likely. If you want a low tech verification of the combinations, there are only $125$ ordered combinations, you could list all of them to check. Calculating the frequencies of these ordered combinations should be easier and less error prone.
$endgroup$
– badjohn
Jan 22 at 12:00
add a comment |
$begingroup$
So let's say I have a triploid species and I have 5 possible alleles for a particular locus. I have found that I can have 35 total possible genotypes (combination with replacement) and I know the allele frequencies for each of the 5 alleles, that I will randomly set as $$F_a = 0.2, F_b = 0.2, F_c = 0.4, F_d=0.1, F_e=0.1$$.
However, I am having a hard time wrapping my mind around how to calculate the number of occurrences of a particular genotype, $ab^2$, in a population of 1000 people.
I understand the allele frequencies, for instance, in the population of 1000 people, given all 1000 observed genotypes, 40% of them had allele $c$. But these are individual values and I have no information on how likely they are to be linked. (I am supposed to assume HW principle holds, but not sure how this helps).
How do I calculate the number of occurrences of the genotype $ab^2$ in the population? I would like an answer that provides good intuition on how to generalize well for any number of ploidy and any number of alleles.
I was told that I should multiply the frequencies together, $0.2*0.2^2$ and then times this by the number of ways I can get $abb$. But I am really confused on which double counts I am supposed to avoid and which double counts I actually want to keep. If I didn't care about double counts, then the number of ways I get $abb$ is $3!$. But if I care about double counts, then I'd divide by $2!$. But we actually want to double count, and that's the whole reason the hardy weinberd equation multiplies the heterozygote probability by 2. Really confused on this part of the problem.
probability
asked Jan 22 at 8:30
JonathanJonathan
388317
$endgroup$
So let's say I have a triploid species and I have 5 possible alleles for a particular locus. I have found that I can have 35 total possible genotypes (combination with replacement) and I know the allele frequencies for each of the 5 alleles, that I will randomly set as $$F_a = 0.2, F_b = 0.2, F_c = 0.4, F_d=0.1, F_e=0.1$$.
However, I am having a hard time wrapping my mind around how to calculate the number of occurrences of a particular genotype, $ab^2$, in a population of 1000 people.
I understand the allele frequencies, for instance, in the population of 1000 people, given all 1000 observed genotypes, 40% of them had allele $c$. But these are individual values and I have no information on how likely they are to be linked. (I am supposed to assume HW principle holds, but not sure how this helps).
How do I calculate the number of occurrences of the genotype $ab^2$ in the population? I would like an answer that provides good intuition on how to generalize well for any number of ploidy and any number of alleles.
I was told that I should multiply the frequencies together, $0.2*0.2^2$ and then times this by the number of ways I can get $abb$. But I am really confused on which double counts I am supposed to avoid and which double counts I actually want to keep. If I didn't care about double counts, then the number of ways I get $abb$ is $3!$. But if I care about double counts, then I'd divide by $2!$. But we actually want to double count, and that's the whole reason the hardy weinberd equation multiplies the heterozygote probability by 2. Really confused on this part of the problem.
probability
probability
asked Jan 22 at 8:30
JonathanJonathan
388317
asked Jan 22 at 8:30
JonathanJonathan
388317
asked Jan 22 at 8:30
JonathanJonathan
388317
asked Jan 22 at 8:30
JonathanJonathan
388317
asked Jan 22 at 8:30
JonathanJonathan
388317
388317
$begingroup$
There are three ways to get $ab^2$, namely $abb$, $bab$, and $bba$, so $frac{3!}{2!} = 3$ is correct.
$endgroup$
– N. F. Taussig
Jan 22 at 10:20
$begingroup$
This is a maths group rather than biology so you can only hope for help in some aspects of the problem. What is the HW principle? You won't find many questions specifically about these topics but you could consider it as a game with 3 5-sided dice and adapt some of the many questions and answers on dice.
$endgroup$
– badjohn
Jan 22 at 11:54
$begingroup$
Your figure of $35$ suggests that you consider $aab$, $aba$, and $baa$ as the same but not $aab$ and $abb$, is that correct? Remember, that even if the frequencies of all alleles were equal, the $35$ cases would not be equally likely. If you want a low tech verification of the combinations, there are only $125$ ordered combinations, you could list all of them to check. Calculating the frequencies of these ordered combinations should be easier and less error prone.
$endgroup$
– badjohn
Jan 22 at 12:00
add a comment |
$begingroup$
There are three ways to get $ab^2$, namely $abb$, $bab$, and $bba$, so $frac{3!}{2!} = 3$ is correct.
$endgroup$
– N. F. Taussig
Jan 22 at 10:20
$begingroup$
This is a maths group rather than biology so you can only hope for help in some aspects of the problem. What is the HW principle? You won't find many questions specifically about these topics but you could consider it as a game with 3 5-sided dice and adapt some of the many questions and answers on dice.
$endgroup$
– badjohn
Jan 22 at 11:54
$begingroup$
Your figure of $35$ suggests that you consider $aab$, $aba$, and $baa$ as the same but not $aab$ and $abb$, is that correct? Remember, that even if the frequencies of all alleles were equal, the $35$ cases would not be equally likely. If you want a low tech verification of the combinations, there are only $125$ ordered combinations, you could list all of them to check. Calculating the frequencies of these ordered combinations should be easier and less error prone.
$endgroup$
– badjohn
Jan 22 at 12:00
$begingroup$
There are three ways to get $ab^2$, namely $abb$, $bab$, and $bba$, so $frac{3!}{2!} = 3$ is correct.
$endgroup$
– N. F. Taussig
Jan 22 at 10:20
$begingroup$
There are three ways to get $ab^2$, namely $abb$, $bab$, and $bba$, so $frac{3!}{2!} = 3$ is correct.
$endgroup$
– N. F. Taussig
Jan 22 at 10:20
$begingroup$
This is a maths group rather than biology so you can only hope for help in some aspects of the problem. What is the HW principle? You won't find many questions specifically about these topics but you could consider it as a game with 3 5-sided dice and adapt some of the many questions and answers on dice.
$endgroup$
– badjohn
Jan 22 at 11:54
$begingroup$
This is a maths group rather than biology so you can only hope for help in some aspects of the problem. What is the HW principle? You won't find many questions specifically about these topics but you could consider it as a game with 3 5-sided dice and adapt some of the many questions and answers on dice.
$endgroup$
– badjohn
Jan 22 at 11:54
$begingroup$
Your figure of $35$ suggests that you consider $aab$, $aba$, and $baa$ as the same but not $aab$ and $abb$, is that correct? Remember, that even if the frequencies of all alleles were equal, the $35$ cases would not be equally likely. If you want a low tech verification of the combinations, there are only $125$ ordered combinations, you could list all of them to check. Calculating the frequencies of these ordered combinations should be easier and less error prone.
$endgroup$
– badjohn
Jan 22 at 12:00
$begingroup$
Your figure of $35$ suggests that you consider $aab$, $aba$, and $baa$ as the same but not $aab$ and $abb$, is that correct? Remember, that even if the frequencies of all alleles were equal, the $35$ cases would not be equally likely. If you want a low tech verification of the combinations, there are only $125$ ordered combinations, you could list all of them to check. Calculating the frequencies of these ordered combinations should be easier and less error prone.
$endgroup$
– badjohn
Jan 22 at 12:00
add a comment |
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$begingroup$
There are three ways to get $ab^2$, namely $abb$, $bab$, and $bba$, so $frac{3!}{2!} = 3$ is correct.
$endgroup$
– N. F. Taussig
Jan 22 at 10:20
$begingroup$
This is a maths group rather than biology so you can only hope for help in some aspects of the problem. What is the HW principle? You won't find many questions specifically about these topics but you could consider it as a game with 3 5-sided dice and adapt some of the many questions and answers on dice.
$endgroup$
– badjohn
Jan 22 at 11:54
$begingroup$
Your figure of $35$ suggests that you consider $aab$, $aba$, and $baa$ as the same but not $aab$ and $abb$, is that correct? Remember, that even if the frequencies of all alleles were equal, the $35$ cases would not be equally likely. If you want a low tech verification of the combinations, there are only $125$ ordered combinations, you could list all of them to check. Calculating the frequencies of these ordered combinations should be easier and less error prone.
$endgroup$
– badjohn
Jan 22 at 12:00