Given Info: 12% red hair (homozygous recessive or q^2) What we need: % for non-red hair (homozygous Dominant or p^2) Step 1. Solve for q^2 q^2= 12% or (0.12) √(q^2) = √(0.12) (radical sign on both sides to get rid of the square) q= 0.35 Step 2. Solve for p Equation= p+q=1 So if solving for p (p= 1-q) p= 1- (0.35) p= 0.65 Step 3. Solve for p^2 (0.65)^2 = 0.42 Step 4. Change it to percent 0.42 x 100 = 42% -Thank you Mr. Anderson for all of your videos!
I watched your videos getting into medical school and now I'm in Medical school and still watch your videos on certain topics. Crazy how some medical school professors can't articulate a concept like the way you can.
hi mr bozeman it is currently 2023 and im taking ap bio in school with my exam in 3 days. I had no idea how to do the hardy weinberg equation because i was trying to solve it as an equation and you've really helped clear it up with me. not only that you've helped me for the majority of the year and im so grateful for that. its amazing that these videos with your concise and simple explanations can still be relevant 9 years later. keep it up!
Thank you professor, you have contributed to my understanding of this complicated equation. For the last problem you have us, q=✓.12=0.346 p= 0.654. Since p2+(2)qp+q= 1, 2pq= 0.452. This is the percentage of the population that are heterozygous for non-red hair. The percentage of the population that is homozygous for non-red hair is 43%
I happen to have an exam including this material tomorrow. I wanted to leisurely review via TH-cam and found this right on my home! What a fun coincidence! Thanks for the upload (:
Thank you so much for helping me understand this! Starting AP Biology tomorrow and natural selection is the first topic we're learning and now i feel a lot more comfortable and ready for tomorrow and the year, thank you!!!
I got everything... but just wanted to know what if we are considering genes having more than 2 alleles like the I gene that codes for the sugar polymer on the surface of the RBC's which has 3 alleles ?? (ABO blood grouping) In that case how does this principle work ? Someone help me with it.. i am just a beginner ....
I cannot be the only one going "exactly what size shoe ARE you b/c those socks are HUGE!!!" And also, this was a beautifully simple way to explain a concept I was not understanding.
it would be good to mention that the p2 and q2 terms are derived from the likelihood of two of those alleles both being present in an individual given their frequency in the gene pool
When would you multiply by 2? My teacher mentioned how if there is, for example, an individual who is recessive, then you multiple that gene by 2 and I'm getting confused with why you would square q instead?
The Hardy-Weinerg law makes it possible to predict the allele frquencies (for a locus in a population) based on the observed genotype frequencies. (True or False)?
You have confused me. I am unsure of your question because I don't know. I would think FALSE because not all locus have the same genes or appearance in certain area.
Can someone help me solve this problem?In a population of 500 individuals, 200 are genotype AA, 125 are genotype Aa, and 175 are genotype aa. What is the frequency of the dominant allele A? What is the frequency of the recessive allele a? Given these allele frequencies, what does Hardy-Weinberg predict will be the frequencies of each of the three genotypes? Is the population in Hardy-Weinberg equilibrium? Fully explain your answer.
Thank you so much for the understanding of concept. I am stuck in this question Given gene A at frequency 0.2 and gene B at frequency 0.6, find the equlibrium frequencies of gametes AB, Ab, aB and ab. Plese explain this question
hey; mr anderson i solved out the answer can u plz tell me whether it's right or wrong the percentage for homozygous non red hair is 43.56.....and i found ur tutors very helpfulll thankssss
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?
q2=.12 for 12% in decimal form therefore q is the square root of .12=.3 p=.7 because 1-.3=.7 so follow the equation p2+2pq+q2=1 2(.3)(.7)=.42 or 42% Hope this helps! :)
My book isn't this awesome, I finally understood this topic, thank you so much!!! (I'm from Mexico and I'm 16 years old so.. my english isn't quite good, sorry)
The question is how many are HOMOZYGOUS for NON-Red Hair...not dominant or recessive. Remember red hair is a recessive trait so although the 2(P)(q) group will carry the trait red hair will not be expressed, but they are HETEROZYGOUS. So, the only HOMOZYGOUS NON-RED HAIR group is P^2 the answer is .427 or 42.7% or rounded off to 43%
yes, good point :). But still, my question is why is that the red hair gene is not expressed in the remaining (most of the) population and thus making it recessive?
What if i told you... a pun is where you have one word mean multiple things or use a homophone for the same effect, not where you swap a word for another word that rhymes. :')
Given Info: 12% red hair (homozygous recessive or q^2)
What we need: % for non-red hair (homozygous Dominant or p^2)
Step 1. Solve for q^2
q^2= 12% or (0.12)
√(q^2) = √(0.12) (radical sign on both sides to get rid of the square)
q= 0.35
Step 2. Solve for p
Equation= p+q=1
So if solving for p (p= 1-q)
p= 1- (0.35)
p= 0.65
Step 3. Solve for p^2
(0.65)^2 = 0.42
Step 4. Change it to percent
0.42 x 100 = 42%
-Thank you Mr. Anderson for all of your videos!
Velcro, the final part of the problem specifies homozygous non-red.
VelcroZGaming g
its actually 43% due to rounding
@@mummiedanser1609 yup its 0.427 so its 0.43 = 43%
@@mummiedanser1609 yeh I got 42.7 so I agree
I watched your videos getting into medical school and now I'm in Medical school and still watch your videos on certain topics. Crazy how some medical school professors can't articulate a concept like the way you can.
hi mr bozeman it is currently 2023 and im taking ap bio in school with my exam in 3 days. I had no idea how to do the hardy weinberg equation because i was trying to solve it as an equation and you've really helped clear it up with me. not only that you've helped me for the majority of the year and im so grateful for that. its amazing that these videos with your concise and simple explanations can still be relevant 9 years later. keep it up!
not ready for that test tomorrow...
never ready
Two years later, how'd you do?
Wil McDowell ayyy
Elizabeth German same...
same...
He's like the Biology version on John Green.
+Christina McKay Wouldn't the biology version of John Green..... Be Hank Green? lol
Trevor Schmahl True XD
Hank is dope🤩but so is Mr Anderson
No, Better !
Hank is bank brah
Thank you professor, you have contributed to my understanding of this complicated equation. For the last problem you have us, q=✓.12=0.346
p= 0.654.
Since p2+(2)qp+q= 1,
2pq= 0.452. This is the percentage of the population that are heterozygous for non-red hair. The percentage of the population that is homozygous for non-red hair is 43%
Thank you again Sir. I should thank you for every time I learn from your lesson. I will begin to be more respectful from here forward.
Mr Anderson, are you my dad?
+Nathaniel Anderson no
+Nathaniel Anderson Sucks for you. He's an intelligent dude.
+Bozeman Science maybe you just have similar fenotypes?
+Nathaniel Anderson haha this made me smile :)
YES
morning of my exam, thank you!! you make this so simple and easy to understand now im not going to lose any marks if it comes up!
I happen to have an exam including this material tomorrow. I wanted to leisurely review via TH-cam and found this right on my home! What a fun coincidence! Thanks for the upload (:
A hardy weinberg question came up in this years ocr biology paper, and this video definitely helped. Thank you Bozeman!
I'm doing hardy Weinberg in school at the moment and thanks so much for uploading this! You are so inspirational to me :)
You explained it so well, it sounds so simple when you explain it!!! Thank you
Cooool analogy with the socks for the masking of the recessive allele by the dominant allele!
you my friend just restored my hopes in learning genetics.
You are the reason I'm passing AP bio
Seriously. We look alike, and have glasses, and have the same last name
He is your dad!! ;))
Thank you so much for helping me understand this! Starting AP Biology tomorrow and natural selection is the first topic we're learning and now i feel a lot more comfortable and ready for tomorrow and the year, thank you!!!
The Greatest Teacher for Me❤ Seriously, you're the best❤🔥
Thanks Sir!!!!! Tried a lot to understand this equation until i found this video. SIMPLE but GREAT!!!
As always, great organization and excellent explanation as to the "whys"
Thank you! This was extremely helpful! It seems so easy now. Not sure why I was having trouble.
I am going to try this equations in different areas to see how well it holds true in other fields.
i was looking for it on bozeman for so many days and finally it is there thank you so much sr
I'm in grad school and this explains it better and more quickly than the textbook.
It is easy to calculate allelic frequencis with homozygous. But, how do you calculate allelic frequencis from heterozygus?
may the science gods bless you, bc u just saved my bio grade
your videos are a blessing for my biolab homework
I'm can almost for certain say that the answer to the question he last gave is approximately 42%.
Thanks! The sock explanation was very creative.
Can someone tell me how to Calculate p , and how it became 86 in the example
I got everything... but just wanted to know what if we are considering genes having more than 2 alleles like the I gene that codes for the sugar polymer on the surface of the RBC's which has 3 alleles ?? (ABO blood grouping) In that case how does this principle work ? Someone help me with it.. i am just a beginner ....
Bozeman Science!! You rock. I wish I was in your class
I cannot be the only one going "exactly what size shoe ARE you b/c those socks are HUGE!!!" And also, this was a beautifully simple way to explain a concept I was not understanding.
it would be good to mention that the p2 and q2 terms are derived from the likelihood of two of those alleles both being present in an individual given their frequency in the gene pool
got 42%.... would that be correct?
+Edris Zalmai ok cool :)
Thanks a lot... Very good explanation! And also thx for talking slowly thats helpfull for german people^^
Thank you so much for your videos!
When would you multiply by 2? My teacher mentioned how if there is, for example, an individual who is recessive, then you multiple that gene by 2 and I'm getting confused with why you would square q instead?
Mr.Anderson, is the answer to the problem 65%?
Thank you for explaining! That was great, esp. the sock analogy! :)
simple and brilliant really appreciate your work
Thank you! You explain very well.
Thank you for a well and easy explanation. 😊
Thank you, this made things so much clearer!
god bless you , explained this much simpler than my teacher
You are Amaaaaaaazing❤❤❤
that was really precise and concise .. Thank you so much
what if I put a black sock in a red sock? is red still recessive
No
Bless your soul this helped me so much!
The Hardy-Weinerg law makes it possible to predict the allele frquencies (for a locus in a population) based on the observed genotype frequencies. (True or False)?
You have confused me. I am unsure of your question because I don't know. I would think FALSE because not all locus have the same genes or appearance in certain area.
how would you know whether the trait is recessive or dominant. For example, how did you know that the red hair trait is recessive.
it told you in the previous question red was recessive
Thank you so much! this was so helpful ... GREAT teacher as always .. :))))))
thanks !! C: i love these videos so much
He never said I hope that is helpful
This Time he hopes its destructive.
Can someone help me solve this problem?In a population of 500 individuals, 200 are genotype AA, 125 are genotype Aa, and 175 are genotype aa. What is the frequency of the dominant allele A? What is the frequency of the recessive allele a? Given these allele frequencies, what does Hardy-Weinberg predict will be the frequencies of each of the three genotypes? Is the population in Hardy-Weinberg equilibrium? Fully explain your answer.
+TheSilvermoondancer
p^2 = 200/500 = 0.4 (AA)
2pq = 125/500 = 0.25 (Aa)
q^2 = 175/500 = 0.35 (aa)
0.4 + 0.125 = 0.525 (A)
And sorry, I don't know if it's in Hardy-Weinberg equilibrium :/
really good video for explaining the basics
thanks
Thank you! The video was really helpful!
You saved my life!
Thank you so mush Mr. Anderson! My teacher has been confusing me so much with this and she recommended me here! I got 65% homozygous for non-red hair!
I got the same answer but I don’t know why everyone is saying 42%.
i love your intro soundtrack
im confused, how did u get .86? please help
cris casicana How did he get the .86
Thank you so much for the understanding of concept. I am stuck in this question
Given gene A at frequency 0.2 and gene B at frequency 0.6, find the equlibrium frequencies of gametes AB, Ab, aB and ab.
Plese explain this question
I love ur videos man
thank you soooooooo much.....keep it up.....it was really very helpful
I love Boseman teaching.
So what is the answer to the last one 43 percent?
hey; mr anderson i solved out the answer can u plz tell me whether it's right or wrong the percentage for homozygous non red hair is 43.56.....and i found ur tutors very helpfulll thankssss
But how do you know if the 12% red hair is Dominant or recessive in the first place?
he was using the red as a recessive trait so we should assume red is recessive
q^2=.12
q=.346
p=1-q=1-.346=.654
2pq=2(.346).654)=.452=45.2%
43% is a little bit too much rounding in my opinion.
Kevin Schmitt this is what i got
that's incorrect. he is asking for "HOMOzygous", so what you're supposed to do is p^2 the p --> (0.654)^2 = 0.427 = 42.7%
@@icecream6741 You're absolutely right.
What video software do you use?
Very awesome video
Mr Andersen, you are God.
Omegasub101 god doesn't exist my brother
Think logical
Thank you! That was very helpful!
thank you. it was very helpful
sir u are crazily awesome
the answer is 44%
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?
What if they give you the heterozygous number?
can someone write out the answer of the problem at the end of the video for me?
q2=.12 for 12% in decimal form
therefore q is the square root of .12=.3
p=.7 because 1-.3=.7
so follow the equation p2+2pq+q2=1
2(.3)(.7)=.42 or 42%
Hope this helps! :)
If you know what an allele is and what genes are, skip to 5:07
Thank you very much, the answer of the last quation 0,44= 44.% :D
My book isn't this awesome, I finally understood this topic, thank you so much!!! (I'm from Mexico and I'm 16 years old so.. my english isn't quite good, sorry)
really like this video bcoz it help me alot
Daddy Anderson saves the day again with his sexy phenotype
The question is how many are HOMOZYGOUS for NON-Red Hair...not dominant or recessive. Remember red hair is a recessive trait so although the 2(P)(q) group will carry the trait red hair will not be expressed, but they are HETEROZYGOUS. So, the only HOMOZYGOUS NON-RED HAIR group is P^2 the answer is .427 or 42.7% or rounded off to 43%
42%?
Dude it's midnight and I have to take four more cornell notes on his videos for AP bio. Kmn
Thank you so much that's so helpful
holy --- now I know some sh*t , thank doc this really help me.
Phenotype whats dominant?
Thank you for saving my biology grade, oh father of my bio grade
So why is the red gene recessive to the non red gene?
Allda Green if in the whole world only 2 % contain red hair then it must be recessive
yes, good point :). But still, my question is why is that the red hair gene is not expressed in the remaining (most of the) population and thus making it recessive?
Nice vid. Thanks
Brazilian here, good class
i'm also brazilian and i must say mr. andersen is an amazing teacher. this class is top notch
Mr Anderson, holy crap, are you the one?
q^2=0.12, q=0.35, 1-0.35=0.65, p=0.65, p^2= 0.4225, 42.25%
"Did you hear that Mr Anderson, that's the sound of inevitability"
+rakesh raja ITS TIME TO STOP THE MATRIX PUNS(lol)
What if i told you... a pun is where you have one word mean multiple things or use a homophone for the same effect, not where you swap a word for another word that rhymes.
:')
+rakesh raja goodness gracious I'm done lol
This guy really put some bass in his voice, when he said that's not how it works.
Mr Anderson can you message me the right answer please. Thank you. I just want to know if I got it right. Thanks a bunch! :)
right awnser is .69
thanks!
Don't listen to him. . . It's 42.25%, or just 42% if you round it. This is the same answer that was repeated further down as well.
oh...I see..okay
why are you lying to him silly boi its obviosly .69 dont lie to throw him off he needs to do good on his test
Thank you !!!