Biology 1AL - Review, Lab Exam

Uploaded by UCBerkeley on 15.10.2012

>>INSTRUCTOR: The sooner we start, the more we can cover. As mentioned I'm just going
to go through problems in the exam reader I don't have my laptop here, I'll tell you
which ones they are. I've chosen them specifically based on three things, one if there was e
mail request I went through what people requested. Two, I try to group them by topics. And then
three, tried to pick ones that that would illustrate the slight differences between
them. So let's just run through the first one, which is question 43, from spring 2012.
Okay. So, if you have your exam reader, please pull
it out. If you don't, hopefully the person next to you has one. And I would like you
to read through the problem, I'll give you 2 minutes to read through the problem.
It's No.ハ43, spring 2012, which is the first one in the exam reader, No.ハ43.
Okay. Let's go through this. I'm hoping this is the best way to do this in terms ofハ
remember last time I said, look the traits, look at the alternative form, look at them
independent of one another and then make the probabilities as if they're independent if
they don't match the data and then they're linked and then you to figure out how they're
linked. (Reading).
So those are the letters weハshould be using. A... (Reading).
So we can use DT or AA or CA, but that's the basis of the letters chosen based on the newtons,
a female fly with normal wings, curly antenna, and you... (Reading) and that is true breeding.
Okay. So, if it was true breeding that means for
many generations it breeds the same phenotype. Which means it' homozygous for that trait,
and or if the trait is sex linked in the males it's hemizygous, so it says from here, it
says we see differences between females and males with respect to thorax but not for these
two it tells you that in the problem. So because we see a difference here, we know that thorax
is X linked, so, X linked just means that trait is physically on the X chromosome it
has nothing to do with genetic linkage, because there's no second trait to be talking about
genetic linkage with. To have genetic linkage, so it means it's physically on the X chromosome
we can talk about three linked, if we have a second locus, then we can potentially talk
about genetic linkage. So not for antenna and wings so let's think about this, let's
figure out what this would be and then for the F1s let's look tat dataハ not the F is
but for the data. Normal wings to angled wings we see 300ハ these numbers correct?
30750ハ I usual will I do about 1,000ハ but it's for reason it'sハ a thousand 50,
is that right? 300 and 750ハ let's just do this. So it'sハ
750ハ a 1,050. And then for this it's a 1,000 and 50. I'm
not quite sure why I chose 1,050. It's a one to one ratio, what gives you a one to one
ratio? Hetero with homozygous, so we know the female is heterozygous, because a males
from a try breeding population which mates is homozygous.
So what does that make the mutation? Dominant or recessive? Recessive.
So the mutation is recessive. So the females genotype is like this...
[Instructor writing on board] And the males genotype is like this...
[Instructor writing on board] Now, I look at another trait so I don't know
if it'sハ I know it is not an X chromosome because it was told to us that wings are not
an X chromosome, let's look at the next locus. Antenna verses curly, so normal antenna 300,
750ハ one to one ratio. What gives us a one to one ratio? Hetero with homozygous recessive
is so we know the male for antenna is homozygous, the female is heterozygous is the dominant
or recessive? Dominant. [Instructor writing on board]
And then the male is like this... [Instructor writing on board]
Now, I'm just going to say they're genetic linked, if they don't match the predictions
and then go back and make them linked. We know that thorax was sex linked, so the male
is dark thorax, true breeding, and for the thorax we see normal a one to one ratio, also,
is that correct? I believe so, right?
Just help me out with the data. Normal thorax, 525, just do dark.
Yes? It's a antenna which is Hetero with hemizygous, so this makes this the hetero, right.
So that which is the mutation is? Recessive. So let's determine if these two traits are
genetically linked or not. So we can look at these independent of one another. We said
if we do this branching diagram that one half of them should be AA + and one half should
be AA in terms of phenotype and one half should be CA +, and one half should be CA and one
half should be CA + and oneハhalf should be CA, remember it's one to one ratio and
so now I'm looking at them, so I should be expecting a one fourth, a one fourth, a one
fourth, and one fourth. For this...
So, someone help me out and do the counting. So this would be... Normal antenna. Normalハ
sorry, normalハ normal wings, normal antenna, normal wings, normal antenna, 150, 300, is
that the numbers? Okay. So I'm coming up with 300.
Let's just try the next one, which is normal antennaハ sorry normal wing, curly antenna.
Normal wing, curly antenna. 750. Let's do this... Angled wings, normal antenna,
angle wings, normal antenna, 750, and 300. Does that look like it's unlinked?
No. They're linked.
And which of these classes of offspring represent a parental type of chromosome? This one here?
This one, right? Because this is the largest number.
And those represent the recombinants, so this means when I go back and take my genotype
here, this is saying they're unlinked but we know they're linked but what we don't know
if it's like that... Or like this... [Instructor writing on board]
Because in words they're both the same. But what we see is we should have the wild type
mutant coupled on one of the chromosome and we should have mutant wild type on the other.ハIf
that's the case and then it ha has to be like that answer and we're done.
That's the genotypeハ so again, I looked at them independent of one another.
Iハ if I'm looked a them it's Hetero with homozygous I did with the next one, looked
at the two of them. Independent and then I said what would be my predictions, and doesn't
match predictions so I have to go back and make them linked and realize that the largest
category is due the fact that it's a parental chromosome. I think the key thing here is
in words only, this individual is heterozygous with the traits genetically linked, but the
predictions are different. This is the wild type alleles on the same chromosomes this
is with different chromosomes, these are still homologous pair, this is one member of the
pair this is the other member of the pair. Let's do another problem, let's do 44 now.
Which is going to hope be an extension of this.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Sure.
>>STUDENT: [Indiscernible]. >>INSTRUCTOR: Which case it was, case one
or case two? That's what your question is, right?
Okay. So, from the data, the largest number should
be due to the fact that they have a parental type of chromosome, so from the data we know
that the offspring are due to the fact that they got a parental type chromosome that's
the combination and this is the combination which chromosome.
So what we're seeing here is you canハ so let's just make the Punnett square for case
1, so if it was case 1 this is where notation system is so useful it's A + A, CA +, like
this... [Instructor writing on board]
So it a parental would be AA +, CA +, a parental would be AA CA, and then the recombinant would
be this coming down if you want I could color code this where this is that yellow chromosome
this is the white chromosome, a cross over would look like this...
[Instructor writing on board] We'll ignore this set of chromatins but if
that was the case, then we get A + like this... Picking up that yellow, the white here, like
this... Picking up thatハ sorry, the yellow.
Picking up this CA +. And to keep it color-coded I'll write these
now in yellow. Where that's one chromosome, there, that's the other chromosome there,
there's one recombinant, there's the other recombinant.
Okay. That's case 1.
If it had been case 2, notice case 2 would be would be the most common, this would be
the most common, I would just change the letters for case 2 now. Not in yellow so we don't
have ミany confusion. Put in orange.
Case 2 would be... [Instructor writing on board]
This would be the parental that would be the parental that would be the recombinant and
that would be the recombinant, but what we saw from the data was that the largest numbers
were due to the fact that they had a wild type allele, and a mutant.
So we go to here, which is largest number or wild type and mutant.
It's this one... Okay.
That help? >>STUDENT: Yeah.
>>INSTRUCTOR: So now let's do 44. So next page flip it over, 44.
Yes? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: How do we know it is not C.? What is C.?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Oh, you mean choice C? We know
that the mutants were dominant. So the CAハ so we know that the mutation was recessive.
And we know that this mutation was dominant so it has to match the case from the very
first part, do you want me to go over that first part again?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Okay.
All right. So back here, when we've got the offspring, we knew that this one was the true
breeding so it has had to be homozygous something, this had to be the heterozygous since we saw
normal wings that meant this was a recessive mutation, since we saw the curly antenna we
know this had to be dominant mutation since we saw this we knew the thorax has to be a
recessive mutation that's where we figure out the choice of upper and lower case and
the choice of letters. So let's do 44. So please read through 44, I'll put it on the
board. I tried the overhead last time I don't think
it worked as good as just rewriting things on the board.
So read through it while I'm writing it out. It takes me some time.
[Instructor writing on board] And I may pick a student to come up and discuss
this. And the only reason for that is you heard
it from me multiple times so maybe hearing it from a student from a different voice will
make a difference. So be potential, be prepared for a shout out.
So one's left, people are leaving now. Get them! You're mean.
Normal abdomen. Blue [Indiscernible]. Fat abdomen.
Normal bristles. Fat abdomen, blue.
125, 125, 375 and 375, so notice for the last problem we didn't have to do the math that
it was 1,050. Would you like me to give you 2 minutes to try it on your own? Go for it.
I'll twiddle my thumb. Who's got the answer?
Few hands, excellent that was about a minute and a half.
All right. Letユs go ahead and try to analyze the problem.
So let's look at abdomen, normal to fat the ratio is what?
3 to 1, which implies it's aハ what? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: Hetero with a heterozygous, so is fat dominant or recessive? Dominant.
So we know now it's like that, I don't know if it's sex linked yet.
Haven't figured that out but I know the phenotype is Hetero with hetero. Could it be sex linked?
No because the male can't be heterozygous. So let's look at the bristle morphology, we
have normal 500 to blue, what gives us the one to one ratio, Hetero with homozygous recessive.
Now I have a heteroハwith homo recessive. But it doesn't say anything about which of
these were true bleed breeding, right? So I don't know which of these is dominant recessive
yet from this information, because it's just not enough information there.
But it says later on I took a female fly with a blue bristles made it to a dad, the dad
had normal bristles and it got 3/4 blue bristle what is gives me a 3 to one ratio? Heteroハwith
hetero and which one is dominant, so one is 3/4 so from that blue bristles is dominant,
[Instructor writing on board]. So, now, here, we had said we had a one to
one ratio, which is a heteroハwith the homozygous recessive so I know blue bristles is dominant.
This is homozygous recessive. This is blue bristles, the homozygous male,
the female. Weハ I'll go over it again. Later on another
cross gave us some information where we saw 3/4 at blue bristles and 1/4 had normal bristles.
But from that we can tell which was dominant and which was recessive, okay so sometimes
a cross doesn't tell us all of the information we need. A substance cross has to be made.
So from this we know it is like, this let's just figure this out if this was unlinked,
so we had a heteroハwith hetero. Is that correct? Yes. So 3/4 of those should
be FA, the dominant fat abdomen, 1/4 should be the normal abdomen, and for the second
locus we said it was a one to one ratio is that right? One to one ratio.
So 1/2 should beハ should be wild type, 1/2 should be mutant blue bristles 1/2 should
be wild type, 1/2 should have blue bristles so if they're unlinked I expect 3/8, 3/8,
1/8, and 1/8. So, help me out with the data. Fat abdomen,
no more bristles how many? 375, fat abdomen blue bristles? 375.
Normal abdomen? No more bristles? 125. And 125.
Exactly what we would predict so are the traits unlinked or linked?
Unlinked. So we're done. Whatever answer matches them
being unlinked would be the answer. Okay.
This is where people usually start to like, okay, I'm starting to get this, that moment.
Any questions about this? Yes?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: You have to talk much louder.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Is the tree that I just drew
from one of the parents? The diagram that I've done is based upon what we saw in the
offspring. Again, looking at the traits independent,
we said 3/4 of the offspring should have the fat abdomen, 1/4 should have the normal abdomen.
And then when we look at the second trait we said 1/2 should have the normal bristles
one half should have the normal bristle and the same is true down here when I look at
them independent of one another. For the offspring.
And then when I look at the offspring I see it matches that prediction so I'm going to
say they're unlinked. Okay.
So let's do another problemハ question? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: So, the question is if it's normal does it get a plus? First I'm not sure
what you mean by normal, do you mean wild type?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: So the question is, if it's
aハ if it's says it's formal or wild type, how do you indicate that? Always.
Without exception. Add a plus.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Usually, if it's normal it equals
wild type, not necessarily 100% of the time, but usually, okay.
All right. So, let's do another problem. I think have
marked here, let's see, 44. Okay. Question 39 from fall 2011.
This one is going to be much more difficult, but you're going to see the same concept being
applied. Slightly different though.
Okay. I'll give with you a few minutes to read it
as I go through it. [Instructor writing on board]
Okay. This is a little bit more difficult to do
if I gave you the data and then asked you to determine the genotype from the data. You
could do that, but this tells you sort of what's going on, and from that you to generate
the data. Okay. It's all the further that's going to go down.
So let's quick think about this. I have a true breeding population, it has
white fur. So, I'm going to refer to this mutant as the
W mutation, that means somewhere on one of the chromosome is the genetic locus for W.
Okay. For population 1, to keep track of it, I'm
going to write that as W 1. Over W 1. Mutant, mutant true breeding.
Okay. For every other genetic locus in that individual,
whether it involves fur color, eye color, number of toes, every other locus its homozygous
wild type. This is a single mutant population. Homozygous, wild type for everything else.
No other mutant alleles. I'm going to say that again. Because that
kept coming up over and over again during office hours it is wild type at every other
locus, whether or not that locus has anything to do with fur color or anything else in that
fly. For a population 2, I know there's a locus
that corresponds to this t white mutation. That is the absence of brown fur.
And it says it was sex linked. So on the male, I know the male is like this...
[Instructor writing on board] And the female is like this... [Instructor
writing on board] The X chromosome, but it tells you in the
problem, these are two different genetic loci. So that means, there's the W 1 locus, I'll
do it for the female, there's the homologous pair.
And yellow will be the one from the other parent.
There's the W 1 locus. Defective, there's the W 1 locus, defective.
Homologous homozygous mutant but I know there'sハtwo loci involved so what is the genotype of the
second locus here? Mutant or wild type?
What I just spent about 3 minutes going over, what is it for every other locus?
Wild type. So what is it for the W 2 locus?
Wild type. And every other locus that I could possibly
imagine. Now, put it on the same chromosome it maybe
physically linked but not genetically or maybe genetically unlinked but I'm just trying to
show you to you see how these are two loci, so when I write the notation for the W 1 locus
I'm going to write this to put the 1 before the 2 so I can keep a little bit more track
of it. [Instructor writing on board]
X W 1 +, W 2 mutant. This one that they're unlinked because we'll make predictions that
they're unlinked and for the W 2 locus, what is the male?
Wild type or mutant? Wild type is this wild type for every other
locus except for this... Doesn't matter if it has to do with eye color
or anything else its wild type. That's what the statement there are two genetic
loci involved tells you, and we'll just start with them being unlinked.
Okay. Now, that we know what it should look like,
let's make our Punnett square and see what the predictions are.
The male can make that type of gamete and that; the female can make one type of gamete.
Okay. I'm sorry I can't pull that board down anymore.
So X W 1 W 2 + and the Y and X W 1 +, W 2, and that's the only one it can make.
So fill in your Punnett squares I'm going to give you a moment to do that.
Continue to use our notation system. [Instructor writing on board]
Okay. Let's go through this.
We said the male can make 2 types of gametes can give this chromosome or that. So we indicated
that or that, each is 1/2. We said that the female can make 4 gametes
but they're the same of if it was unlinked. So what is the phenotype of the female offspring?
Brown fur or white fur? Brown. Excellent. Because they have wild type information
here, which is dominant, and they have wild type information for this, so they are brown.
In terms of color. And what's the phenotype of the males?
White. Because they're mutant here.
It doesn't matter that they're plus here, they lack one of the necessary steps.
So these are white. Okay.
Question? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: I can't hear you and don't know whose talking, yes talk loud.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: So, yes, so I'll say it for
probably the fourth or fifth time now, it's mutant at the one locus, what it is at every
other locus? Wild type. Every locus, every locus, whether or not it whether or not it's
with fur color. The only time we have double mutants we tell you they're double mutants
because mutations are rare event so the answer to have rare event is so low, unless we tell
you it's happened, ignore it. So the key thing here, and I'm going to say it, because apparently
can't stress it enough, what is it for every other locus other than the mutant?
Wild type. Okay. Great. I think we don't need to do that
more; we'll do another problem like this and you'll see again the same thing. So did that
take care of that? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: Yes? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: Okay. So the question is: For the W 2 locus, why didn't we write pluses
like this... [Instructor writing on board]
Because if that's the case then that's not mutant so it couldn't have come from the mutant
population that was mutant for the W 2 locus. So we said this came from a true breeding
population, that was mutant, it's called population 2, so we made it mutant for population 2 homozygous.
That's why it doesn't wild type it was mutant for that locus, but wild type for every other
locus which happens to be the W 1 and any others. See how this is working?
So, you see the value of the mutation, the notation, because that's really the difference
so that's how we keep track of things. Question? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: It would not affect the answer if any way in fact, could there be crossing
over here? No.
But if there's crossing over it's the same, all four of them are the same.
Right? Now, it could potentially have another problem,
where they could be linked, but see I haven't given you any specific data, but I couldハ
if I had made them linked, then if they were linked, we'll just do that it's not on the
problem, but if they're linked, then it would be W 1, W 2 + X W 1, + W 2, let's say we take
the female here. Let's say we take that female that's the only possibly linkage that can
exist it has to be a wild type mutant, mutant wild type because of this...
Right. This is a parental chromosome and that is
a parental chromosome so if I took this F1, let's just do that problem.
This is a made up problem on the spot. So let's take the F1 female.
So you guys do thisハ I'm making the problem upright now. F1 female and cross it with a
male homozygous mutant for both loci. True breeding population both mutations.
[Instructor writing on board] This is the only other question I can do here
in terms of permutations. Okay, the genotype here should be X W 1, W
2 over Y, homozygous, well, in this case its not homozygous it's hemizygous, but mutant,
mutant. Right a. And the genotype here is like this...
The female is this genotype. And I don't know yet if they're linked or not. But we're going
to make some data, and from the data you'll be able to tell many if they're linked or
not. Let's say, when I make my Punnett square, that's one parental type, if it's linked,
the other parental type is XW 1, W 2 plus, the other recombinant is X W 1 + W 2 +, and
the last recombinant ask X W 1, W 2. And males can only give recessive information,
or no information. So first off we'd see no difference between males and females, so I'm
going to have to look at both I'm going to look at one. And say I foundハthe following
data... 250, 250ハ whoops.
250, 250. Would you say they're genetically linked or unlinked? Linked. Look at the phenotypes
what's the phenotype of this? White.
Because it's got lack of information for the W 2, so these are white.
[Instructor writing on board] What's a phenotype of these? White.
Thank you. What's the phenotype of these? Brown.
Excellent, thank you. And what's the phenotype here? White.
So, if I saw a 3 to one ratio with 3/4 of it being white I would say they were genetically
unlinked. Notice if they had been linked, let's just
make a number doesn't matter it won't change the answer, say they're map distance is 40,
this would be .2, .2, .3, .3, if that was the case, then not 250 would be brown, but
a number other than 250, 200. So that's how you would know they're linked.
Okay. In fact that's every permutation that can
answer that type of question, right? What we did there where two loci? We figure that
out. We figured out predictions whether they were
linked or unlinked. And we've done both, right?
So that's a all the different types of questionsハ so if you go back and look at the exam reader
and you see any question like that you will see they're variance there of because that's
the only options we have, linked or unlinked. Any questions about this one?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: The original problem. Way back
over here? Okay. >>STUDENT: [Indiscernible]
>>INSTRUCTOR: It's an assumption. Mutations are single.
Everything is wild type, everything. Okay.
So, the assumption is, if you see mutation it's single mutant, if it's true breeding
you know it's homozygous or hemizygous, wild type or everything else.
All right, should we try one other problem? Yeah. Question?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Do you want to do 40 on that?
Okay. They did anotherハ well, actually you know
what, we did how you would do the problem right here, but you want to do a 40 specifically,
we'll do 40 specifically, so that's fine. What I did was I did instead a general problem,
which allowed you to see any problem I can ask.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: So what we did was we made our
predictions and I had said that if they were unlinked it would have been 1/4, 1/4, 1/4,
1/4 and if they were unlinked we would have seen in the data, the data will tell us the
answer. If in the data I had seen 3 to 1 ratio of
white to brown that would have told me they were unlinked.
If I had seen a ratio other than 3 to 1, such as 2 to 8.8 to .2 that would have told me
they were linked. Yeah? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: So now let's do the actual question, but realize this was the actual explanation
for every question now let's do the question. Okay.
Um the scientist... (Reading). And they mated with a brown furred male.
So, let's just rewrite this, so there's no confusion.
W 1, W 2, I put the plus on top. So there's noハ [Instructor writing on board] and it
was made into a brown furred male, which means it's W + 1, W 2 plus.
Why so let's go this Punnett square here, so one becomes + +, so what's the phenotype
of all of these? Brown.
Because they get down the information from mom. And what the phenotype of the dad's?
White, white, brown, white. Okay.
So the only difference is that in this case, I have to change the notation, because in
my sample problem here, I had chosen the male to be homozygousハ enzyme sorry, hemizygous
recessive in this case it's hemizygous dominant. That's why all of the females would be brown
brown brown brown, because this information is dominant, dominant. Okay.
So does that make sense? Does that make sense in.
So it was the exact same problem, so it was good that we did that because you saw the
other possible mutation. So let's doハ question? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: For the chart that I have here. Okay.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: You would only get the 3 to
1 ratio if it was unlinked. If it was linked it would be anything other than a 3 to 1 ratio.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Um, so, you're not going to
see any option for it to having been unlinked so it's not even a possibility because that
would never match any of the data, so A through E has no data that would match a link problem,
it would only match it if it was unlinked. Okay.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Oh all of the females have brown
furs, no, inside they do, but if it's linked a number other than 1/4 would be brown. And
it says, sorryハ a number other than 1/4 would be brown so it wouldn't be all of the
males have white, right. So see how this is working?
Okay. Let's move on. Okay. I'm hoping you're starting to seeハ
let's do 44 now. One page over.
Okay. You can see that to explain this with all
of the questions, I mean we're 50 minutes into the lecture.
So 44, go ahead and look at that, I'm going to erase these boards and put it here.
[Instructor writing on board] Okay. So here's the question basically, how
much DNA does a daughter share with her mother in.
So, mom gives half the DNA, dad gives half the DNA to the daughter. Right.
Because the daughter's a diploid. So daughter shares with mom half the DNA.
Just like you share half of your DNA with mom because it came from mom and the other
half your dad. So we've got the first part.
That's 50%. Now, let's do brothers with brothers.
Okay. So, brothers get 100% of their DNA from mom,
because they're haploid. So there's no dad. Sad but there's no dad.
All the DNA comes from mom. Let's just look at this.
So, if the brother had gotten the A allele, so clearly somewhere on a given chromosome
in the mom that's diploid there's going to be a difference.
So if a brother got the A allele, what's the chanceハ so let's say, A prime, what's the
chance the other brother gets A prime? So one brother got the A prime, what's the
chance that the otherハ brotherハ when he got gametes from mom got the A prime? Half.
That's it. So it's 50 ever. So brothers share with brothers 50%.
Any questions about that? Now let's do sisters with sisters.
So half they're DNA from mom, half from dad. We've already said that for maternal genes
they should share half. So, they'll share 25% of their genes, maternal
genes with each other. How much do the paternal genes do they share?
100%. Because dad can only give one type of gamete, he's haploid, so 50% of the DNA so
it's 1/2 x 100%. So that's now 50%, so sisters share 75% of their DNA.
All of they're paternal DNA is shared equally. Because dad's haploid.
Mom is diploid, so they share half the DNA from mom, but mom only contributed half the
DNA. Okay.
All right. So, let's see here. I think that's the end
of the genetic-type questions let's do a sequencing question, we're getting a lot of these.
So the next page from that one a question 46 it's sequencing one.
I'm hoping as you do some of these questions, you start to see the similarities.
So let's quick do this. I'm doing this question first, because it
ties together a lot of concepts that people seem to be missing.
So, hopefully I can show you the concepts again.
[Instructor writing on board] Okay.
We do two types of sequencing here. Two types.
Group 1 uses a radioactive primer for visualization, so let's first do group 1.
And we're doing the dideoxyA lane. Our primer is radioactive I'm going to make it orange.
And we have a million templates like that. Remember that.
Large number. And then what we should get here at this first
position is a subset we'll get the dideoxyA. So a subset will be one longer than the primer,
but all of the rest of the subset, whatever the ratio is debts the DA, any questions a
about that in. This subset is determined upon the dideoxyA
to deoxyAs so over here they'll get DT, DT, they'll all get this...
Over here, get DT, all get that. DC they all get that.
Realize it's one by one this direction, right? DC as it stands, TC and but then a subset
will get the dideoxyA. And this got the DA, but a subset get the
dideoxyA. DA, DA. And then a DT.
So all of these will have DT. But because the primer is use to visualize
it, I will see a band at the first position relative to the primer, I will see a band
at this position relative to the primer, we'll see a band at this position relative to the
primer and I will see a large band to this primer, it's a primer that allows me to visualize.
So, what I would see there is data that should look like something like lane 1, which is
3 bands at the top, a band down here. Okay.
One position, whatever this happens to be, one more, and one more.
So that's with the primer visualize. [Instructor writing on board]
Let's now do the dideoxy that's used to visualize it. So it's the dideoxy that visualizes, to
pretend they're just white. You know, if I do this on the same board I'm
going to regret it. So I'm just going to redraw it. Everything's the same. [Instructor writing
on board] Sometimes I regret trying to save time, so
I'm just going to redraw it. Okay here's where the primer so, in this case,
the primer is not radioactive. But if the dideoxy that's Fluorescent, so
I should get dideoxy here, where would I get everywhere else. DA, is that clear? Dud anyone
else not see that? Everyone sees it and then we get DT, DT, DC, DC, DC, a subset gets the
dideoxyA gets the DA, DT, blah, blah, gets the DA, subset, gets a subset, of dideoxyT,
gets the DA, DA, DT. Will you see this band?
No. Because you have no way to visualize it. So
in this case, there's results would look like the first position. This position... And that...
But the upper band would be missing, relative to this, the upper bands missing because the
primer allowed you to visualize here, the Fluorescently labeled molecule here allowed
you to visualize it. Do you see the difference? No.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Oh, sorry.
Okay, it's not going to changeハ okay. So, do you guys see how this is the dideoxyハthat
as allows you to visualize it. This is on the gel, when you're on the gel, would you
seeハ would there be a band that's there? There's a band of DNA but you can't see it.
It's invisible. There's no way to visualize it.
Why? Because there's nothing thatハallows you visualize it there's no Fluorescently
labeled molecule there. Over here, you can visualize it, because the
primer is radioactive and that's what exposes the film.
Okay. Do I need to go over this again or can we
move on? Move on? Question?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: So the question was, why do
you have termination here effectively? Because they subset will get the dideoxyA whatever
the percentage is, and all of the others got the DA. Okay.
And then all of the others goハa DT because there was no dideoxyT, all got the DT on and
a subset got this... So in that tube the subset is longer than with primer and this then this
long and this long. But these will expose the film when you run them on the gel because
they will be at that band, those are bands and it's radioactive so it will expose the
film right there. When you run this gel, there's a band there, but it won't florescence because
it has no Fluorescent molecules there so as a result you can't so that it's there.
Okay. Let's do a sequencing problem.
Yes, question? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: So, the question is why are the there no Fluorescently molecules here,
do you see any red molecule here? No. That's the reason why there isn't.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Why isn't there? Let's do numbers.
Okay. So let's do a million of these. So there's a million templates.
I'm convinced you don't want me to draw a million. So 2, 4, 6, 8, 10.
Let's make the ratio 10%. How many will stop here?
What percentage? 10%.
So 10% of a million is 100,000. So that means all of the rest of them 900,000
get the d DA, is that correct? And then, 900,000 get this, 900,000 get thatハ
so now they're all 900,000 like that, right? What percent will get the dideoxy this time?
10%. Percentage doesn't change so there's 90,000 molecules like this... [Instructor
writing on board] which mean's there's so how many would be like, this? So 81,000 and
that would leaveハ this is 271, so it would 629,000 molecules like this...
[Instructor writing on board] Based upon those percentages, but not a single
one is visible because they don't have any Fluorescent label in it.
Does that make sense? Let's do another problem like this.
I'm just going to pick a sequencing gel. Yes?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: So the answers were for this
one it looks like that, which I think it was lane 1. And for this one, it's like [Indiscernible]
which I think is lane 4. >>STUDENT: [Indiscernible]
>>INSTRUCTOR: One is running three prime to five prime, okay.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Um, I can talk to you individually
like they're not running in reverse orders so I'll meet you can individually on that.
They're not running reverse orders though, I guarantee you that. So let's do a problem
application. Okay.
Should we stand up and stretch? Yeah stand up. Stretch. Just a minute. Get
some blood moving and then I'm going pick a problem. And I'll have you guys do it.
We have the exam on Thursday night so let's just doハ this is page 14, let me figure
out what exam it is, I think it's the very first one. It is a spring 2012, question No.ハ41.
This now is allows me to see if you can apply this.
Okay. [Instructor writing on board]
Okay. First off there seems to be some confusion.
The DNA migrates to what pole? Positive or minus, through smaller pieces migrate faster
or slower? Fast sore the bottom of the gel is drawn here, unless indicated otherwise
that's the smallest end, so previously the question was, why is this down here? Because
this was the smallest piece so, this was a small piece, this was a bigger piece so the
gel with in that direction. Same is true here, so actually, by the dideoxyC lanes, so I see
a band, at the very top, so let's think about what would yield that result? The primer is
radioactive, so there is where the primer goes.
I probably wrote this wrong, did I? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: CGT? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: ACG, thank you. As you're writing this and talking it's kind
of hard not to Mays make mistakes but there we go. This is where the primer should anneal,
did I do it wrong? ACG CGT so here's where my primer anneals.
That's radioactive so let's just do this... Everything ends up at the full length.
So I got a DA essay, a DT. A DC, DA, DA, DT, DG, DG, DG, DC, DA.
>>STUDENT: [Indiscernible] >>INSTRUCTOR: TGGCCハ
>>STUDENT: Other way, go left. >>INSTRUCTOR: This is why it's typed for you
on the exam. Okay. TGGハ you saying my mistake is in the template?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: Okay.
All right. So, do we see any termination? No. So what would yield that? You forgot the
dideoxyC. So there's no dideoxyC.
So if there's no dideoxyC, will we ever terminate? So, so you get full length. Okay.
Does that make sense? So the absence of dideoxyC means it never
stops it goes to the end. But because the primer is radioactive you would see that band.
If you had used Fluorescently labeled dideoxyC would you see a band? No, because there's
no dideoxyハso there's no way to florescence, does this make sense? Let's do the dideoxyA
lane. Okay.
DideoxyA lane we see big band here, and here. Now, the intensity of the band, the darkness
is reflective of the number of molecules. Again, asked in office hours, why is the one
band gray, and the others are black? Because there's less black that's why it looks gray.
So it's a numbers game, okay so, the bottom band has a lot more in it than this band.
This band has more than that band. There's no band up here. So the question's asking,
why don't you see a band up here? So let's just do the normal reaction.
I should have had my T bind, my G bind, my C bind, I should have had for a subset there
is the dideoxyA lane, so I should have had a subset that went dideoxyA to stop, others
got the DA. So I'll circle that in yellow so you know
that's a stop. So we see that band in the fact that there's
a lot of them. Then in the others, again we still have the
primer, always in these. That wasn't orange.
So we still have all of the primers in those molecules so we have a million primers at
least to template the million molecules. And we get DA, DT, DC, a subset get the dideoxyA,
we see a band that's fairly intense, some got the DA, and they've gotten a DA there,
DT, DC, and then we go on, okay. And then we got another subset set in the
dideoxyA. They stop. And then we should have gotten
DA on out to this dideoxyA. We should have.
But we don't see that band. Why don't we see that band?
There's two few molecules to visualize it, that means the ratio of dideoxyA to deoxyA
is higher than it should have been. So as a result, if you remember back here, I did
a number oF10%, that's higher than it should have been. So, this here, would have been
a high ratio of dideoxyA to deoxyA. Does that make sense? This why this band doesn't appear
there's too few molecules to develop the film. Okay.
Let's do another problem. Let's switch gears.
And let's do the touch a ploid question. Anybody remember where that's at?
Anybody remember where the [Indiscernible] is.
The Zenu(?), 4 N? I had it marked but my post it fell off.
Someone see that question? Where is it at?
So I've chose closed my thing, which is the same exam?
Spring 2012, what question No. 11. Okay. Let's quick do this.
So probably the last one we'll have time for, we'll see.
4 Nハ oh this is a different question than what I was going do, but that's okay.
Let eats not do this one because this isn't the one I was looking for, becauseハ let's
do the number of possibly permutations. Summer 2011, is that the one? Summer 2011,
17, that the reason I want you to doハthis one is allows you to see applying principles
of meiosis, but I think this one is more challenging so I would rather do the more challenging
one. Females make gametes that are 2 N.
[Instructor writing on board] Males make gametes that are 1 N.
So, by way of review, let's look another humans first that are 2 N.
Okay. So in terms of review my humans, 1 N = 23,
2 N = 46, so that means I have 23 chromosomes, No.ハ1, No.ハ2, No.ハ3.
When I make gametes, how many types of gametes can I make relative to chromosome No.ハ1?
2. How ant chromosome No.ハ2? 2. How about chromosome No.ハ3? 2. So it becomes
2, that's a possible ways for the number of unique chromosomes x 23.
Anyone not see that? That's for a human.
There are 23. So now let's do this... To help you it says 1 N = 12.
N = 12. So that means 4 N = 48 that means there are
12 chromosomes in a given set. And there are 4 sets.
Does that make sense? 4 N, N = 12. Let's pretend somewhere along
these chromosomes there's a difference. So I'm going to call it A prime, A double prime,
A triple prime and A four prime. Just to keep track of the four different A type chromosomes
this is B, C, D E F G H etc. How many different types of gametes that you can make that are
diploid here? So I'll give you a minute the figure out that. How many types of diploid
gametes can you make? Write the possible combinations. You got 4?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: You got 4, right?
(No sound). Okay.
Let's quick do this. I heard answers 1, 4, 6, 16, for just this locus, I can get A prime,
and with it A double prime. I can get A prime and A triple prime. I can get A prime and
A four prime I can get A double prime, A triple prime I can get A double prime, A four prime
and lastly A triple prime and A four prime. 6 possible ways to get diploid gametes for
that chromosome type A. Right?
These are diploid. I cannot get A prime, A prime because when
we pair up, remember, they pull to different poles.
So there are 6 possible types for this chromosome, A, how many for B?
6. How many for C?
6. There are 12 of those.
So it becomes 6, raise to the 12th power. All right.
Yes? >>STUDENT: [Indiscernible]
>>INSTRUCTOR: Because there are 12 types of chromosomes the A chromosomes which is No.
1, the B which is No.ハ2, C No.ハ3, etc., etc., I did this first because I wanted to
see how you see application there, are two times for chromosome type No.ハ1, two for
type 2. So we raise 2 the possibly ways to the types of chromosome. So this was something
I think everyone understood this was a little bit extension. Now, how many types of gametes
and so the way this workings is you get an egg + a sperm + a sperm, so how many different
types of gametes can you make for this individual, four and gametes are haploid. How many types
for a given chromosome No. A? 4.
You can get A prime, or a A double prime or A triple prime or A four prime. So for mom
it was 6 to the 12. For dad 1, it's 4 to the 12. For dad 2, it's 4 to the 12.
Mom, dad 1, dad 2. Number of unique gametes, number of unique gametes. I think the difficult
part here was getting this fact that there's 6.
So people have asked me in office hours, how many can you get if it's 6 N or 8 N, it's
large number and I would never do that to you. There is a formula I once figured out
but it's complicated formula, I would never do that. But I think this is relatively doable
to just write those out. Okay. All right.
So, one last problem, pick it because those were the ones I wanted to go over.
What's that? 10 spring 2012 is the winner.
Okay. And I have office hours on Wednesday, I believe
it's 10 to 2 four hours it's in bSpace, that's where I'm going to look to see where they
are. So we have this large genome like this, 3 x 10 to the 9th base pairs here's gene 1,
here's gene 2, it says I have primer pairs, locus 1, should yield aquatic of thousand
base pairs if it's done forward and reverse and this should yield a product of 2,000,
if it's done properly we have forward and reverse primers. It says only the four primer
for locus one was used and reverse primer for 2. So just going to use yellow to indicate
one of the primers. Doesn't matter where I put it. There's that
and let's say the other primer is here. I can switch the strands it's not going to change
the answer. So, we have the primers anneal.
We extend it. We allow it to extend for one minute, which is 60 seconds, it goes one thousand
nucleotides per second, so how far should this get?
60,000. Nowhere near the other primer to anneal.
Even if the other primer was in this direction, nowhere near, so when I denature this, I'm
going to get back this... Template and this... But this one here I'm going to call P 1, the
product 1, looks like this... Can my primer anneal to that?
No. Because it's the same polarity. I'm going
to switch this primer here, to be orange. Will that primer anneal?
No. Because the sequences are complementary you never got anywhere near it. So I will
still have this original template there which is why every round I will increase by, you
know, one or two, this molecule, and I increase by 1, this molecule as well.
And this one will be about 60,000 as well. But because I don't run the reaction long
enough to get all of the way over, if I had then I could get am amplification bus I don't
do it that long, every round this anneal will primer back to this and I will get a product
like that, and every round I will get a product like this, so these non units increase by
2, so I increase by 2 every round but I'm not exponential increase, so how many rounds
did we do? 30 rounds so I would have approximately 60
molecules that are 6,000 long. There is no exponential increase because the
primer can't anneal here, we can't get forward and reverse primers to anneal that's the reason
you have forward and reverse primers so the reverse primer had been done correctly, would
have annealed somewhere in here, to start lying us to amply fie it because we get both
strands then of our target. Okay.
So, one of these answers should hopefully be matching that.
Okay. All right. Ah, do you want to do one more or not?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: For those who want it, I'm doing
one more, for those who don't you're free to leave. What's the question?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: I have the to finish this because
webcast will finish in about 2 minutes. It may take that long to find the problem.
All right. Does someone else have a problem you want me to do?
>>STUDENT: [Indiscernible] >>INSTRUCTOR: 45-second midterm.
Sorry. Okay.
It's a PCR one so the gel tells us the size and the amount of DNA.
So from the PCR product, we have 400ハnanograms of DNA.
[Instructor writing on board] Because there's the PCR product it's the same
intensity as the lower band, does that make sense? And the size of it is 2,000 base pairs
long. So, if that DNA was 2,000 base pairs long,
the molecule weight per mol of base pair is 600ハgrams per mol so if I have a mol it
would be 2 x 10 to the 3, 6 x 10 to the 2, 12 x 10 to the 5, 1.10 x 2 [Indiscernible]
to the grams, if I had a mol, but I don't, I have 400ハnanograms so as a result I have
to take the amount of DNA I do have, which is 400 xハ I
like to put things in scientific notation. If I had a mol it would be 1.2 x 10 to the
6ハgrams per mol. I want the number of molecules. So I multiply
by Avogadro's number, 6 x 10 to the 23. This becomes 24, over 1.2, which is 20. This
isハ ハ13, that's 23, that'sハ ハ10, sorry, that's 10.
So 2.0 x 10, 11. With math being very quick, so double-check my math. But that's the way
you would do it that's the weight, in the band, that's the molecular weight for 4,000ハ
2,000 base pairs long, and that gives you the number of mols x Avogadro's number. That
gives you number of molecules, so good luck on Thursday night.