Hardy-weinberg equation :D
Moderators: honeev, Leonid, amiradm, BioTeam
Hardy-weinberg equation :D
hey!
i have 2 questions.
1. In a certain population, the dominant phenotype of a certain trait occurs 91% of the time. What is the frequency of the dominant allele?
i got 0.7 or 70%...
it's either that or 0.91
i'm not sure.
which one do u think it is?
2. The allele for the hair pattern called widows peak is dominant over the allele for no widow’s peak. In a population of 1000 individuals, 510 show the dominant phenotype. How many individuals would you expect of each of the possible three genotypes for this trait?
Okay so i got
homozygous dominant 90
heterozygous 420
homozygous recessive 490
it's that.. or...
510, 410, 80 (respectively)
can someone please check my answers... and check which ones are right?
thank you so much! i appreciate it greatly
Here are some of hte equations
p + q = 1
p^2 + 2pq + q^2
i have 2 questions.
1. In a certain population, the dominant phenotype of a certain trait occurs 91% of the time. What is the frequency of the dominant allele?
i got 0.7 or 70%...
it's either that or 0.91
i'm not sure.
which one do u think it is?
2. The allele for the hair pattern called widows peak is dominant over the allele for no widow’s peak. In a population of 1000 individuals, 510 show the dominant phenotype. How many individuals would you expect of each of the possible three genotypes for this trait?
Okay so i got
homozygous dominant 90
heterozygous 420
homozygous recessive 490
it's that.. or...
510, 410, 80 (respectively)
can someone please check my answers... and check which ones are right?
thank you so much! i appreciate it greatly
Here are some of hte equations
p + q = 1
p^2 + 2pq + q^2
lol yeah
one more thing,
if
AA - 6
Aa - 9
aa - 0
---------
15
then how do u find p and q.
because... since the recessive is 0.. there shoudl still be a q. but if u find the square root of 0 it'll be 0. but there should still be q in the heterozygotes. so q can't be = to 0... the recessive allele should never completely be removed bc of the heterozygotes... can someone clarify i'm confused
would u do 9/15 = 0.6 / 2 = 0.3/ 0.4 = 0.75?.. ahah i.. idk
one more thing,
if
AA - 6
Aa - 9
aa - 0
---------
15
then how do u find p and q.
because... since the recessive is 0.. there shoudl still be a q. but if u find the square root of 0 it'll be 0. but there should still be q in the heterozygotes. so q can't be = to 0... the recessive allele should never completely be removed bc of the heterozygotes... can someone clarify i'm confused
would u do 9/15 = 0.6 / 2 = 0.3/ 0.4 = 0.75?.. ahah i.. idk
Who is online
Users browsing this forum: No registered users and 3 guests