Hardy-Weinberg equilibrium applied to population genetics problem

[@karlatrujillolcegsolana3702]

Thank you! I have been trying to figure this out for days. You do an amazing job simplifying this. You rock.

[@GeneticsLessons]

You can find many more examples for practice if you search my channel for "Hardy-Weinberg".

[@Andr295]

Thank you so much! This video really explains the principle very well to solve much more complex genetics problems. Thanks again.

[@scratchoriginalsdh]

Okay, that one completely confused me. Can someone just give me a 20 page essay to write please?!

[@GeneticsLessons]

Here is a play list of hardy-Weinberg problems: kzbin.info/aero/PLzBiAPKi8vOPcxFbw6UvVSL-PqLtju3b8

[@alanabeaumont2650]

I do not understand why getting the root of p^2 does not give me the value of p. This should make sense?

[@GeneticsLessons]

If you know the frequency of p^2 - but usually you are not given this information. Only frequency of dominant phenotype which is made of homozygous P^2 and 2pq heterozygous.

[@mfundo1pumla]

Well explained, thank you.

[@GeneticsLessons]

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[@mfundo1pumla]

@@GeneticsLessons I passed and graduated 5yrs ago, thanks to you GeneticsLessons, without your videos witch break everything down for students, I wouldn't have made it.

[@johnamani6546]

Thanks,l can now score all related questions about this topic

[@GeneticsLessons]

Great 👍

[@ramjiyadav2118]

very well explained sir ...

[@GeneticsLessons]

Thanks for watching.

[@tousifpk9496]

For 2nd generation?

[@GeneticsLessons]

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[@harshagarhwal7266]

Hi sir from where these values of p an q come 0.55 and 9.45 respectively

[@CsprsSassyHrly]

This may be too late to answer cause I see you asked two years ago, but in case someone else has the same question, I figured I'd answer. He's taking the .15 from the p-squared and adding it to half of the 2pq frequency (.8/2 = .4), .4 plus .15 is .55 = p... then subtracting .55 from 1 to get q. The reason for this is because the 2pq is a heterozygote diploid, so half of that genotype frequency is A and the other half is B. Hope that helps.

[@jacksonmulwa9406]

0.45

[@GeneticsLessons]

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[@domination06]

why not just square root the p^2 value and the q^2 and see if they add up to 1 or not... in which case they do not... meaning they aren't in equilibrium . This is a long ass way for a simple square root lol

[@GeneticsLessons]

In this video I want to show how to solve this problem on the most basic level for deep understanding Genotypes ratios and difference between OBSERVED and EXPECTED ratios, your method is also applicable but gives the answer YES or NOT without explanation in detail of how to calculate Expected Ratio and compare with Observed ratio and what is Hardy-Weinberg equilibrium.

[@bleuemoone8710]

If you get tin the habits of doing the square root, you might square root anyways to solve for p or q which is wrong cause you need to take into account the heterozygous

[@TrinanjanDutta]

ur calculations are wrong..square root of p gives 038 not .55!!..plz check...

[@GeneticsLessons]

Watch video one more time. My calculations are correct! You messing OBSERVED RATIO and EXPECTED. Your mistake is that you assume that this gene pool in Hardy-Weiberg equilibrium. Ha-Ha - who told you so? This is what we have to find! If you add A alleles in AA genotype and A alleles in AB genotype you will get frequency of allele A 55%. and 100%-55% means frequency of the allele B 45%. Ask yourself what is A? what it's frequency in this genepool? .55 or 55% Right? Then AA would be .055 x 055=0.3025 or 30.25% EXPECTED RATIO! Now back to your mistake: This is simple math - according to your assumptions if frequency allele A 38% then frequency of the allele B would be 62% ! And this is absolutely not so cause you don't have to be a mathematician to see that the frequency BB genotype is much smaller then AA genotype (3 times smaller) and A and B alleles in AB genotype are equal - so frequency of A allele can not be less then allele B ! Your method can be correct ONLY if this gene pool in Hardy-Weinberg equilibrium - and this is something we don't know until we do the calculations as I showed them. If this gene pool would be in Hardy-Weinberg equilibrium then you can find the frequency of the allele A in the genepool by taking square root of AA genotyope frequency. Hope this does not blow your mind :-)

[@TrinanjanDutta]

i knw this is not in hardy weinberg equlibrium...i find ur method is very much tedious...while it can be calculated very simply..thats all..buddy....no need to strain this ne further...:)

[@shivangirairai5235]

these r option 1) 0.00005 2)0.993 3)0.000000025 4)0.971

[@GeneticsLessons]

Sorry If you need an answer you have to write the problem not from your head "something like that" but exactly as stated in the problem without grammatical and stylistic and punctuation mistakes - I'm not a mind reader.

[@shivangirairai5235]

+Nikolay's Genetics Lessons i already sent u question and then there options

[@GeneticsLessons]

Your question does not contain any numbers!

[@vimbaichingwadza8324]

THOSE DECIMALS ARE WRONG EYY

[@GeneticsLessons]

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