hi guys welcome to let's get to the marks and what
I'm doing now is I'm going through a few Hardy Weinberg calculations uh this comes up in a level
biology so OCR or AQA these questions are from the AQA syllabus so I'm gonna work through these
okay what do you need to know about Hardy Weinberg well let me get my pen up and run in um so with
the hardy-weinberg the first thing you've got to know is uh this off by a heart is the equation and
if you're taking a level and you can't memorize this
off by heart you might as well give up
now no seriously get learn this the amount of students who don't learn this and then they lose
all their Hardy wine bug marks and it's so basic so P plus Q equals one p squared plus 2pq
plus Q squared equals one memorize that please what does this actually tell us what is
the hardy-weinberg used for well Heidi Weinberg made this equation and the equation basically
States one principle and that is that the frequency so the amount of times that certain
alleles like here in this question we have we have two codominant alleles CG and CB um
and they control for fur color and what animal are we dealing with here squirrels okay so we're
dealing with squares and these two alleles control their fur color now what does p plus Q squared
well p you can do this either way represents and Q represent the alleles so the alleles are
represented by P plus Q what have we got down below we have a second equation and what does
that actually represent well
P Squared is the fact that no squirrel will inherit one allele for
fur color it will inherit two and if it inherits c g c g so homozygous for CG we're using that P
Squared to represent that and then the squirrel could inherit for its fur color cgcb or flip
it the other way around c b c g and that's the homozygous genotype and we will write that here
now you don't have to write these out this is just for me showing you what they represent whenever
you look at the Heidi Weinberg equation we u
se P plus q but it really represents whatever alleles
you're dealing with in the question so here it's c g and CB and that leaves Q squared as the
other homozygous genotype which is CB CB so and these telesphenotypes as well so they might
not give you a genotype they might give you a phenotype and from that you can work out what
one you're dealing with so the phenotype for C this is gray this is brown black and this one
is black Okay so we've got phenotypes as well and the bottom equation t
ells you the genotype
and the phenotype so it depends what they give you in the question why did Hardy Weinberg come up
with this basically what Heidi Weinberg was saying or what this equation or principle is telling
us is that the frequency of the alleles within a population in this case the fur color of the
squirrels will stay the same throughout generation after generation after generation and that the
frequency of the alleles will not change there are a bunch of assumptions that go with
this like
seven or eight and that means that in order for the hardy-weinberg principle to hold true you can
have no migration no immigration um mating has to be random so no one's going away I love a gray
Square all to have babies with no one's saying anything like that there's no selection whatsoever
and as and you have to be geographically isolated um no breeding between populations and all of
these assumptions are basically in real life they don't happen so in real life populations
do
choose who they're going to mate with um there will be immigration and immigration
there will be members of the population leaving and coming in and there will be random
mutation there's never really a population where there isn't random mutation so all of
the principles They Don't Really hold true why do we bother with the Hardy Weinberg then well
it allows us to look at a population and say well if hardy-weinberg principle holds true then we
would expect the frequency to be this and then
when the frequency isn't that and it's a little
less we can say ah why isn't the frequency that oh it's because there's been immigration in this
population or it's because there's been selection in breeding so it allows us to make predictions
and explain changes in the frequency of aliens anyway let's get back to the actual question use
the hardy-weinberg equation to estimate how many squirrels in this population have brown black fur
so they're really asking us how many squirrels are 2pq wh
ich is the heterozygous genotype and
they've given us that two had black fur out of a total of 34. so they've given us Q squared they've
given us c b c b black fur so we're going to find out what Q squared is first that's the only thing
we can do so we're going to do two divided by 34 grab the old scientific calculator and
we're going to go 2 divided by 34. equals 0.0588 okay we can round that up to 0.059 but
there we go left out of that so now we know Q squared so that is Q squared what I
want
to do is square root Q squared and that will give me Q so I'm going to square root the
answer and that gives me q q is equal to 0.24 now I know that Q is 0.24 what must P be well
I can do 1 minus 0.24 and this will tell me that P is 0.76 so I now I know what p is and I
know what Q is I can now work out how oh well I can work out the frequency of the brown black
squirrels I can do 2pq so let's do two times 0.76 which is p times naught 0.24 so if
we go two times 0.76 times 0.24 we get 0
.36 okay it's like 0.365 I'm
just going to keep it at 0.36 what can I now do with that is that my final
answer this is one of the worst hardy-weinberg um questions I've seen because it's only two marks
and there are so many steps actually if we just left it at 0.36 which is what a lot of students
did in the real exam when this question came up um they would only score one Mark so so
that's not the end of the question it said how many squirrels it didn't say frequency
or percentage of squir
rels how many we know the total was 34 so of course we have to
do 34 times 0.36 and we're going to get an answer well my answer is 12.4 how can
you have 0.4 of a squirrel you can put 12 you can put 13 you can put anything in
between okay so that's the estimated number of squirrels that have black fur awesome I
hope that helped we're going to go through a couple more of these so you might want
to grab a pen and paper pause the video when the question comes up have a go at
it and so forth ri
ght let me clear this okay another question just move my camera
slightly so I can see what we're doing maybe I shouldn't write in such um let's write in Black
that would be a bit more sensible wouldn't it it's going to let me change my color yeah
of course okay so first things first Hardy Weinberg three marks that's definitely worth
writing out the equation for I mean some of you won't bother writing the equation out in the
exam because you'll think oh my gosh it's only two marks I don't ha
ve time to write the equation
out you have always got to write the equation out that took five seconds of your three minutes that
you've got on this question so let's go through um we can do the same thing if we want if if I
want to take it the long route let's take it the long route and just show you um in the actual
thing once you've got used to doing this you won't have to do this every time okay I've got
three genotypes w r w r w r s and SWS so let's just assign these at the bottom here
we've got WR
w r we've got the heterozygous which is wrws and we've got the well there's no we've got the
other homozygous there's no recessive here because this is another co-dominant um this is
another codominant one so super duper duper what is p and Q then well p is w r and Q is w s it's
the worst R in oh gosh it's only getting worse wrws right what is the question actually asking us
let's get down to it 51 are resistant to warfarin are these two phenot genotypes are resistant
while t
he phenotype is resistant to warfarin so they've told us the phenotype resistant
to warfarin they've given us the genotype of the resistance to war for instance so
we know that these are the guys that are resistant to warfarin these are the ones that have
resistance they're they're saying 51 is resistant so I need to change 51 into decimals we can't use
percentages in the hardy-weinberg equation because it's all out of one so let's change it there 0.51
51 have those genotypes they're resist
ant and the question is asking us estimate the percentage
of rats which are heterozygous so they want that bit so it's definitely going to be less than 51
because the total resistant rats are 51 how many are heterozygous so it's got to be less than 51.
well now we know that's 0.51 this has to be 0.49 so if that is 0.49 we can now work out what
Q is and then we can do work out what p is and then we can come back here and do 2pq
and get our answer so let's get to it 0.49 that probably will sc
ore you one out of three
marks the examiner seeing that on the paper you're then going to want to
square root 0.49 to give you Q and that will give you 0.7 you're then going
to look for p plus Q equals one so 1 minus 0.7 obviously in an exam you're just doing
this in your head there's I'm just trying to show you so p is 0.3 lovely jubbly now
we can come down and do 2pq so we go two times 0.3 times 0.7 and we've worked out 2pq and that equals this is 2pq equals 0.42
now we need to change th
at back into a percentage just times 100 that should be
standard and there we go we have scored four marks three marks four we've got a Bonus one
did we right you'd get a mark for 0.7 you get them up for calculating 0.3 and obviously you get the
important Mark for The Final Answer you'd also get a Mark here so you can or here so the maximum
three marks but for all of this working you can get up to two marks without the correct answer
okay now let's go over here give two assumptions that mus
t be made when using the hardy-weinberg
equation those assumptions are going to be uh so no random mutations I mean in what
population would you have no random mutations uh no immigration there you go there's two uh we
can continue and say um no selection in breeding oh what's going on yeah I can continue and
say a whole bunch of others that go here there's like seven altogether so I can say no
selection in breeding no immigration etc etc hi guys looking at this next question again I've
cu
t out bits of the question we don't need this was unusual because this one was worth four marks
but when you actually do it it's not very hard so let's uh start with how you need to start every
single one and you'll want to pause the video and have a go at this and then I'll go through
it and hopefully by now you're smashing these out right humans have genetic resistance
to infection a recessive allele gives increased resistance to infection by
the malarial parasite in a population hi guys
back to another hardy-weinberg equation so
let's get into this one some humans have a genetic resistance to infection a recessive allele gives
increased resistance to infection by the malarial parasite in a population the proportion of babies
born who are homozygous for the salil is 0.01 okay use the Hardy Weinberg equation to calculate
the expected proportion of heterozygotes they love to ask this so again they want us to come out
with the proportion so they want like naught point they wan
t a decimal What proportion
are heterozygous well they've just given us one of the homozygous now was it recessive
was it dominant and does it matter really probably not so a recessive allele gives
increased resistance in a population the proportion of babies born who are homozygous
further resistant alial so the recessive one that's going to be over here so we now know we
have not they are homozygous is 0.01 so they've told us there it's homozygous I always use this
for homozygous recessi
ve I don't think that was necessarily that important for you to know because
they're working out the heterozygous so we now take that back to Q so we do the square root of
0.01 remember p and Q are they single allele they are not the genotype which is homozygous the
heterozygous genotype so and they're not the phenotypes okay so let's square root 0.01 and we
get 0.1 so this means that Q is 0.1 we then use p plus Q equals one to go one minus 0.1 equals
0.9 we know that p is equal to 0.9 why
am I writing all this working out because it says
four marks I'm like wow normally they're two marks but we're going to pick up loads of
marked here just for showing our working so all we've got to do now is work
out the heterozygous proportion okay let's whack this in 0.1
times 0.9 times 2. and that equals 0.18 and that is so many marks it is unbelievable
so what Mark where would we get these marks well if you worked out that that Q squared plus
0.01 they'll award you a mark If you then s
quare rooted 0.01 to get 0.1 they'll award you a
mark If you then use the Hardy Weinberg Top Line of the equation to work out that P is 0.9 you get
another Mark and then for the final answer of 0.18 um we get the final Mark so 0.18 awesome
Heidi Weinberg principal predict this is basically saying what is the hardy-weinberg
principle well the hardy-weinberg principle um predicts the frequency now you can say
proportion as in 0.1 0.5 or you can call it a frequency but I'm going to use the wor
d frequency
predicts the frequency of alleles in a population and it says that that frequency uh will
remain constant so it basically says the frequency of alleles in a population will
remain constant so it's really a two marker what's the third Mark for do you think
it's for saying one of the seven things that are needed for the Heidi Weinberg equation
to be reliable and for it to work we need no immigration okay um no selection basically
the hardy-weinberg principle knee selection I don'
t know what knee selection is but yeah
so basically the hardy-weinberg principle um you need to be yeah oh gosh a highly Weinberg principle restart for the
50th time a Hardy Weinberg principal relies on an unnatural situation where nothing that
normally would change the frequency of alleles is there so there's no mutations there's
no immigration immigration no random uh no selective mating and that type of thing so um
there are no real populations out there in the earth that the Hardy Weinb
erg principle holds
true for um in all populations you're gonna get some of these that will counter the Hardy Weinberg
principle but it doesn't mean it's not very useful as we shall see from the rest of this question
so the table shows the frequency of some alleles in the population of cats in three cities so
we've got Athens Paris and London over here the frequency of the allele is shown here or
the proportion depending on what word you prefer so we have the proportions in each city of the
different alleles okay going across in rows in each City okay white cats are deaf right um
would the Heidi Weinberg principle hold true for white cats so look at this do we think that
the pop the frequency of the white cat allele the frequency that gives them white fur will
that hold constant and that is that like let's just take London London the frequency is 0.004
are we expecting that to go up or down or will it be hold true with the hardy-weinberg and in 50
years will it still be 0.04
well if they're deaf that may well affect their survival chances and
so and also they deafness will probably not be selected for when it comes to mating it's going
to mess up their mating rituals and things like that so actually we would say no it wouldn't
hold true because white cats would be less likely to mate less likely to survive um and
so forth so being deaf is a disadvantageous allele basically a disadvantageous characteristic
so we need to put all of that down there let's look at
C I'm not going to bother typing all
that the videos gone on too long anyway what is the evidence from the table that non-aguti and
blotch are alleles of different genes so what this means is they're saying uh what they're
asking here is nonaguti and blotched yikes so if we just look there what's the evidence
that non-agutian blotched are not of the same gene are you not Big T Little T which would be two
different aliens of the same gene well if we look at the data for Paris the frequency o
f non-aguti
is 0.71 and the frequency of blotched is 0.78 again Hardy Weinberg tells us that P plus
Q equals one if I add those two alleles together 0.71 and 0.78 we get a number bigger
than one so um for instance 0.71 plus 0.78 is bigger than one let's not try to write like
that again okay so that means they're not the same a little they're completely different
genes uh if we look here this would look like evidence because this adds up to just
under one so you might say oh but we've got t
his data from Paris and from London where we
can see that if you add the number of alleles together the frequencies in that population it's
greater than one so yeah let's get on to the hardy-weinberg question next hair length in cats
is determined by a single Gene with two alleles the allele for long hair H is recessive
the allele for short hair Capital H is dominant let's get down to it with the
hardy-weinberg so we've got P plus Q equals one p squared plus 2pq plus Q squared equals one so
we're going to go through this now so the allele for long hair little H is recessive Big H is
dominant use the information in the table and the Hardy one big equation to estimate the percentage
of cats that are heterozygous for hair length in which city in London okay so we're looking
at London we're looking at heterozygous for hair length they've told us that the allele for long
hair is recessive so when I come to here and I go 0.33 that's my Q squared although you might use P
that's my
Q squared 0.33 they want us to find out heterozygous so it's the old 2pq again so we just
take that up to here we square root 0.33 to find Q just Square rooting that right now and we get 0.57 Q is 0.57 we know then that P must be that was a bad attempt at muting um P must be 0.43 so what is 2p cubed we go oh I've left
no room we go 2 times 0.57 times 0.43 0.49 and what did they ask for percentage
or frequency estimate the percentage so our answer for 2 p q is 0.49 we need to multiply by
a hu
ndred to put in a percentage it's 49 percent of the cats are heterozygous for hair length super
I hope that's helpful I will do another video where I go through the rest of the AQA questions
on Heidi Weinberg there's about 16 questions that have been released in the last 20 years on the
hardy-weinberg so um I will go through the rest of them that's another eight I'll go through
them a bit quicker hopefully all right take care I hope this is helpful uh you can use this
to work through in pra
ctice let's get to the marks
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