@OlyGian sorry but that's not true. if you want to find the percentage in the population which carries the allele, you have to add 2pq (those which have it in combination with another allele) and q squared (those which have it exclusively).
p^2 + 2pq + q^2 AND you can use the equation p + q = 1..That's how I was always taught.
It doesn't honostly matter in the HWE if it's set to 1 or not, you're finding allele frequency of a number that's usually already been given to you...So technically there is no "right" way
Ok, thanks. Sorry if its different to the equation you use, this is just how i was taught it and as far as i know it doesnt make any difference - it still works. Anyone who can use the HWE will probably be able to factorise and expand brackets anyway so theres no real need for that bit. What youre saying is basically the same as arguing that 2+2 = 4 is wrong just cos i didnt write it equal to 2^2 as well.
Theres one on electrophilic addition, and i did make a really long one about physical chem for CHM4 (AQA) but when i tried to upload it it said that it was too long... i might see if i can split it up into shorter sections and try uploading it again.
Thank you so much for doing this. I have an adjunct for my recitation portion and she isn't very good at explaining things. She just zooms through like its a race.
I know what its like when people teach stuff really quickly, you think youre just beginning to understand it and then you have to move on to the next topic! Thats why i spend so much time going through stuff myself.
Right, of course! A dominant allele will 'overpower' the recessive one, making the victim only carry one. When one carries two recessive alleles, they will suffer from CF. ( CF ís autosomal, right? )
I think this actually makes sense to me. Thank you very much! I will make sure to subscribe to you, hope you will make more educative videos.
You also just saved my test, which I have tomorrow ;-).
Umm... im not exactly sure what youre asking but q^2 is the frequency of the homozygous recessive organisms (which is 1/2000) and youre given the value of this in the question.
2pq is the frequency of heterozygotes which is found by doing 2 x frequency of the dominant allele x frequency of the recessive allele.
I hope thats what you were asking, if its not then just say and i'll try to answer whatever it was that you wanted to know. :)
Well, Cystic Firbrose is caused by a recessive allele. I thought that this would come down to 2pq ( as q=rec ) and q² ( idem ). So, that would mean that both 2pq and q² carry CF? I could very well be mistaking, I just don't understand.
Oh right, ok. I think i understand what you mean now.
Yeah, CF is caused by a recessive allele so to actually develop the disease you need to have 2 recessive alleles, so only the q^2 people will suffer from CF.
Anyone who is heterozygous (2pq) will carry one defective recessive allele and one normal dominant allele so they wont actually suffer from the disease but they will be carriers.
This is SO great!!! I can see clearly now!! excelent :)
carolalila 9 months ago
Thanks for the video, I subbed! Sub me back!
lambergino 11 months ago
ohhh thank you
mandylla 1 year ago
thanks... i realy understanded it ... so thank u again soo much
bobfrasoa 1 year ago
Wait... What Weinberg are we talking about here?
Did Steven Weinberg do biology!? I thought he was a theoretical physicist!
Am I confusing him with someone else?
Thymonico 2 years ago
@Thymonico your an idiot
jackarse1 1 year ago
I discovered, probably 9 months ago, that his name is Wilhelm Weinberg. Sorry... :P
Thymonico 1 year ago
I wish we were allowed to use calculators on the AP Bio exam, so I'm not caught up doing the square root of a non-square number
kaysterG 2 years ago
ok sorry:P
get it now!
thnx! =D
oscarhoogland 2 years ago
thanks! get the theory now, but... if you wanna know the % of carriers, don't you have to add the numbers of 2pq and q^2? Or am I talking rubbish?
oscarhoogland 2 years ago
Nope, you're talking rubbish, sorry. :-P
2pq is the percentage of carriers, q^2 is the percentage of homozygous recessive organisms.
OlyGian 2 years ago
@OlyGian sorry but that's not true. if you want to find the percentage in the population which carries the allele, you have to add 2pq (those which have it in combination with another allele) and q squared (those which have it exclusively).
idster7 2 years ago
Not necessarily, Stranger
The equation technically can be
p^2 + 2pq + q^2 AND you can use the equation p + q = 1..That's how I was always taught.
It doesn't honostly matter in the HWE if it's set to 1 or not, you're finding allele frequency of a number that's usually already been given to you...So technically there is no "right" way
lampboy08 3 years ago
That is not the Hardy-Weinberg Equation
The HWE is (p+q)^2 = p^2 + 2pq + q^2 = 1
Genotype frequencies sum to 1
Allelic frequencies sum to 1
Stanger35thGT 3 years ago
Ok, thanks. Sorry if its different to the equation you use, this is just how i was taught it and as far as i know it doesnt make any difference - it still works. Anyone who can use the HWE will probably be able to factorise and expand brackets anyway so theres no real need for that bit. What youre saying is basically the same as arguing that 2+2 = 4 is wrong just cos i didnt write it equal to 2^2 as well.
OlyGian 3 years ago
nice i had a similar problem today in my genetics book.....
joseerock 3 years ago
That was awesome, Thank You!!! Loved the humor too!!
P.S. Anything on General Chem?
nike7754 3 years ago
Theres one on electrophilic addition, and i did make a really long one about physical chem for CHM4 (AQA) but when i tried to upload it it said that it was too long... i might see if i can split it up into shorter sections and try uploading it again.
Oh, and thanks for subscribing! :)
OlyGian 3 years ago
Thank you so much for doing this. I have an adjunct for my recitation portion and she isn't very good at explaining things. She just zooms through like its a race.
You made it understandable! God Bless you!
sansaviera 3 years ago
Youre welcome. Im really glad it helped you. :)
I know what its like when people teach stuff really quickly, you think youre just beginning to understand it and then you have to move on to the next topic! Thats why i spend so much time going through stuff myself.
OlyGian 3 years ago
Thanks, that really helped :) btw, what's the title of the song? ^_^
OnyxTenshi 3 years ago
Its called 'I'm a Realist' by The Cribs. :)
OlyGian 3 years ago
thanks a lot ;)
OnyxTenshi 3 years ago
Right, of course! A dominant allele will 'overpower' the recessive one, making the victim only carry one. When one carries two recessive alleles, they will suffer from CF. ( CF ís autosomal, right? )
I think this actually makes sense to me. Thank you very much! I will make sure to subscribe to you, hope you will make more educative videos.
You also just saved my test, which I have tomorrow ;-).
~Carr
fr4gm4st3rcarr 3 years ago
I've got a question:
2×p×q = 0.043 ( heterozygot ), which comes down to 4.3%.
But how about q²( homozygot )? It's also recessive right? Or do I get things wrong now? Hope you'll answer :).
Thank you, sure was helpful.
Carr.
fr4gm4st3rcarr 3 years ago
Umm... im not exactly sure what youre asking but q^2 is the frequency of the homozygous recessive organisms (which is 1/2000) and youre given the value of this in the question.
2pq is the frequency of heterozygotes which is found by doing 2 x frequency of the dominant allele x frequency of the recessive allele.
I hope thats what you were asking, if its not then just say and i'll try to answer whatever it was that you wanted to know. :)
OlyGian 3 years ago
Well, Cystic Firbrose is caused by a recessive allele. I thought that this would come down to 2pq ( as q=rec ) and q² ( idem ). So, that would mean that both 2pq and q² carry CF? I could very well be mistaking, I just don't understand.
Thank you for your explaination.
~Carr
fr4gm4st3rcarr 3 years ago
Oh right, ok. I think i understand what you mean now.
Yeah, CF is caused by a recessive allele so to actually develop the disease you need to have 2 recessive alleles, so only the q^2 people will suffer from CF.
Anyone who is heterozygous (2pq) will carry one defective recessive allele and one normal dominant allele so they wont actually suffer from the disease but they will be carriers.
Hope that helps. :)
OlyGian 3 years ago
Just wanted to say thank you very much, this really helped a lot. Any chance you can send like 2 more examples lol :)
sharkman14 3 years ago
Yeah, i can make a whole video of examples if you want. Ive got loads of free time now my exams have finished. :)
OlyGian 3 years ago
Awesome, I can understand this! I wish I could say the same about my lectures...
kneedownbrown 3 years ago
Thanks agian
ManchesterUni2009 3 years ago
Youre welcome. Ive just made a chi squared video too - its uploading right now...
OlyGian 3 years ago