Hardy-Weinberg Equation | Detailed

The Hardy-Weinberg principle, also known as the Hardy-Weinberg equilibrium, model, theorem, or law, states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences. These influences include mate choice, mutation, selection, genetic drift, gene flow and meiotic drive.
Consider a population of monoecious diploids, where each organism produces male and female gametes at equal frequency, and has two alleles at each gene locus. Organisms reproduce by random union of gametes (the “gene pool” population model). A locus in this population has two alleles, A and a, that occur with initial frequencies f0(A) = p and f0(a) = q, respectively.[1] The allele frequencies at each generation are obtained by pooling together the alleles from each genotype of the same generation according to the expected contribution from the homozygote and heterozygote genotypes.
The Hardy Weinberg equilibrium comes to be p+q=1
Source:en.wikipedia.o...

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Thanks a ton sir... Now my concept of Hardy Weinbergs equilibrium is crystal clear... Once again thank you sir🖤🖤🌼

[@hussainbiology]

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[@Talia_stamile]

This video helped me pass my biology test. I understand the calculations now. Thank you, sir!

[@hussainbiology]

Glad it helped!

[@kimberlyeverett8447]

Great job. Thank you so much for this explanation on how the Hardy-Weinberg equation is derived, with the use of Punnett Square and corresponding variables and values.

[@hussainbiology]

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[@GRC20121]

I can't see the screen completely for the translation

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[@ramchandrasuthar2848]

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Lotka voltra model

[@bts_rollercosterlifepurplearmy]

Very easy way to understand equilibrium by Huusain Biology...
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[@hussainbiology]

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[@divyanshipal463]

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[@FisheriesTutorials]

Suppose there is a population of cats with the following numbers of genotypes ; BB 50 ; Bb 20 : bb 30. If the population is in Hardy Weinberg equilibrium , then the numbers of the genotypes would be : ( a ) BB 30 : Bb 50 ; bb 20 ( b ) BB 50 ; Bb 10 ; bb 40 ( c ) BB 16 ; Bb 48 ; bb 36 ( d ) BB 36 ; Bb 48 ; bb 16

[@ramchandrasuthar2848]

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[@rahayupertiwi4299]

"If the genotype frequencies in a population deviate from Hardy-Weinberg expectations, it takes only one generation of random mating to bring them into the equilibrium proportions, provided that the above assumptions hold, that allele frequencies are equal in males and females (or else that individuals are hermaphrodites), and that the locus is autosomal. If allele frequencies differ between the sexes, it takes two generations of random mating to attain Hardy-Weinberg equilibrium. Sex-linked loci require multiple generations to attain equilibrium because one sex has two copies of the gene and the other sex has only one."
would u like to help me to explain the text above with an example like this video?

[@medicalworld1111]

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[@hussainbiology]

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[@nasreenfatma6735]

Ur videos are grt..i hav watchd dm lst yr also..nd hav successfully paasd bsc 2nd yr frm c.u..now i m having problm in a question."gene frequencies remain unchanged in each generation but genotype frequency may b changd.explain"..i cant find ny video on ds.plz hlp...😥my exams r knocking d door.

[@hussainbiology]

Aslamualiekum....... First of all thanks for Appreciation....
And for your question , i will let you understand it easily..

[@nasreenfatma6735]

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[@hussainbiology]

Suppose a population has one gene whose alleles are , A dominant and b recessive allele.
Lets say
A dominant allele occurs 4 times.
a recessive allele occurs 6 times.
It can be in the form of
AA genotype.
AA genotype.
aa genotype
aa genotype
aa genotype
Total A = 4
Total a = 6
Or
Aa genotype
Aa genotype
Aa genotype
Aa genotype
aa genotype
Total A = 4
Toatal a = 6
So we can say Frequency of alleles remains same but genotype frequency alters .....
It is like you have two Oranges
And i have two mangoes...
We are only two people ....
2 is the frequency of oranges
2 is the frequency of mangoes.
Lets say we exchange one fruit..
That means You have one orange and one mango now
And same is with me....
But the frequency of oranges and mangoes remained the same therby only changing the variety ( genotype analogy)

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[@janecrus9484]

Pls. Help me with this problem
there are 300 organisms who carry the homozygous dominant gene, 200 organisms who carry the heterozygous gene and 250 organisms who carry the homozygous recessive gene but this data does nit obey the Hardy-Weinberg equilibrium, you should add the same number of species who carry the homozygous dominant gene, organisms who carry the heterozygous gene and organisms who carry the homozygous recessive gene so that the population will be double the original number of individuals. in order to be in the Hardy-Weinberg equilibrium, How much organisms for every genotype you should add and what will be the resulting allele frequencies

[@hussainbiology]

Hi Jane, i have gone through ur question and what i found is that the Heterozygous population number is not correct in the question, bcz when we calculate at Dominant and recessive allele level , like p and q then , the value of p comes out to be .63 and the value of q comes out to be .58 that in total gives us 99 approx.... p + q = 1 ( .63 + .58 = 1 ) Now u will ask how did i calculate these values,,, You easily can see the total Population is 750 , which according to calculation gives us the following frequencies ,
AA ( Homozygous Dominant ) = .40
Aa ( Heterozygous ) = .26 ( which is actually wrong ) there should be .36
aa ( Homozygous recessive ) = .33
but the value of p and q is .63 and .58 which gives us .36 for heterozygous trait ( Aa )
that way population comes out to at equilibrium also.

[@hussainbiology]

I know , it is too tough to explain this problem in words , but if i am wrong correct me , i will accept it and will try to find a way to solve this problem, hope it helps

[@hussainbiology]

I know , it is too tough to explain this problem in words , but if i am wrong correct me , i will accept it and will try to find a way to solve this problem, hope it helps

[@hussainbiology]

see if the values of AA ( p ) comes out to be .40
and value of aa ( q) comes out to be .33
the easily we can calculate the value of pq........ . 63 x .63 ( A x A ) = .39 that is the value of AA or simple p
and .58 x .58 ( a x a ) = .33 that is the value of aa or simply q

[@janecrus9484]

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[@rukhsanamalik1605]

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[@hussainbiology]

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[@ericroutman9969]

Sorry, this video is very misleading . The Hardy-Weinberg model does not care about the dominance relationships among alleles. There is no phenotype involved, so dominance is irrelevant. Second, the fact that p+q = 1 is not specific to HW. The frequencies of two things always sum to 1, if there are only 2 things. Finally, there is no circumstance in which p^2 + 2pq + q^2 does not equal 1, so the derivation of the equilibrium as "1=1" is completely spurious. HW equilibrium refers to the fact that the allele frequencies will never change and the genotype frequencies will stay at P(AA) = p^2, P(Aa) = 2pq, and P(aa) = q^2 . There is no Hardy -Weinberg equation. Take a look at Hardy's original paper - the "Hardy-Weinberg equation" is nowhere to be found. The reason is simple: 1*1 = (p+q)(p+q) = p^2+2pq+q^2 =1. The greatest mathematician in England at the time (Hardy) did not write a paper explaining that 1*1 = 1.

[@joshuachristopherdsouza9021]

p + q = 1: Correct. The equation p + q = 1 is a general principle in population genetics when dealing with two alleles in a population. It represents the fact that the sum of the frequencies of the two alleles (p and q) equals 1. and further Hardy's original paper: While it's true that Hardy did not explicitly present an equation labeled as the "Hardy-Weinberg equation," the principles and concepts described in his paper laid the foundation for what later became known as the Hardy-Weinberg equilibrium

[@ericroutman9969]

@@joshuachristopherdsouza9021 My point is that the fact that p+q =1 and p^2+2pq+q^2 =1 is trivial and does not indicate that a population matches the Hardy Weinberg model. These equations are true whether the population is in HW equilibrium or not. When students are drilled in these equations (as anything other than a math check) they really don't understand the biology of what they are doing. I taught evolution and population genetics to undergraduates and graduate students for 27 years, and the vast majority of these students come to my class thinking that the way to tell if a population is in HW equilibrium is to see if these equations are true - because of the misleading focus on them in lower division classes. For example, Hussein Biology states that "according to HW, p+q=1". (He should say : "because, when there are only two things, the frequency of those two things must sum to 1, p+q=1." This is a general principle of math, not of population genetics) Then he goes on to give his example, where p=.8. Next he calculates p=.2 by subtraction. Then he adds .8 and .2, shows that they equal 1 and states that this "validates the HW equilibrium"!!! No it doesn't. If you do this with any population with p=.8, HW or not, .8+.2 will still equal 1. This video is filled with these kinds of misleading errors.
HW should be taught as a biological model of an idealized population going from one generation to another. The model should explain how we calculate allele frequencies from genotype frequencies in any given generation, how we use probability theory to predict how the allele frequencies in the parents predict the genotype frequencies of the offspring generation, and why that results in equilibrium for the allele frequencies immediately and for the genotype frequencies after one generation. While the equations are true, of course, the focus on them causes all sorts of errors, simply because there is no situation when they are not true (except when a math error has occurred.)

[@muhammadusmanali3596]

20 percent individual show recessive trait ... here why we cannot take q as 0.20 ?? intead of taking square root to find q

[@ritikashaw28]

There are two populations of birds having AA=600, Aa=200 and aa=200. In another population, AA=400, Aa=400 and aa=200. If 20% of the homozygous recessive trait containing birds and 25% homozygous dominant type birds migrate from population I to population II, what will be the new gene frequencies in the population II ?
I NEED SOLUTION OF THIS HELP ME ! PLZ

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[@hussainbiology]

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