Solving Genetics Problems


>>Hello, welcome to the
Penguin Prof Channel. In today’s episode, I want
to do some genetics problems. So let’s get right to it. Want to make sure everybody’s
comfortable with these terms. Alleles are different
forms of a gene and we inherit one
allele from each parent. What that means is that the
alleles you inherit may say the same thing. If they’re both dominant, we
say you are homozygous dominant for that particular trait. They may also say the same
thing, but both be recessive. We use capital letters
for dominant, lowercase letters for recessive. The other option is that
you might inherit one allele from one parent that’s dominant
and one allele that’s recessive. So, your alleles say
different things and we call that condition heterozygous. Hetero just means other. Okay, a lot of genetics
has to do with probability. And probability is just a
measurement of expectation that an event will occur. Probability values always
range from zero to one. So, zero means absolutely
impossible. One means absolutely certain. So, of course, I convert this
to a penguinized scale and so, all probabilities going to range
in value between zero and one. Probability turns out to be
a little messy when we talk about alleles and this professor
of genetics in Cambridge came up with a way to
sort all this out. So I want to show you
how to use this square and solve some basic
genetics problems with it. The first example I want to look
at is the disease, albinism. Albinism is the lack of melanin, the pigment in our
hair and skin. This lack of pigment is
actually due to an inactive or absent form of tyrosinase. This is the enzyme responsible
for the production of melanin and actually has
copper at the center. Albinism is an autosomal
recessive condition, which means that you
must inherit one copy of the recessive allele
from each parent in order to actually be an albino. So here’s the typical problem. What is the probability
of having an albino child if the parents are both
heterozygous for the condition? So the first thing that
you want to do is write out all of your alleles. So in this case, both the
parents are heterozygous, so they carry the
trait for albinism, but they themselves
are not albino. The question is, if they
both carry the allele, what is the probability
that they’re going to have an offspring that
inherits the recessive allele from each of them, that’s
homozygous recessive and will therefore,
be an albino. How are we going
to figure this out? Well, that’s what that
Punnett square is for. So here we have our
Punnett square and the first thing you want
to do is sort out the parents. Now in this case, the
parents are both heterozygous, so they’re both the same. It doesn’t matter
who you put where. This individual is going to inherit the dominant
allele from both parents. They’re going to be
homozygous dominant. Then you’re going to get — 50% of the offspring
will be heterozygous and then here’s your box with
the homozygous recessive. So what this means is
that the probability of two heterozygous parents
producing an albino child, homozygous recessive,
is one in four. All right, here’s another one. Huntington’s disease
is autosomal dominant. What that means is that only
one dominant allele is needed to actually inherit
this disorder. Huntington’s disease is
a degenerative disease of the nervous system. So what’s the probability of
having a child with the disease if one parent is
homozygous recessive and the other is heterozygous? So the first thing you
want to do, of course, is write out your alleles. So in this case,
this is the parent that actually is
affected, that has or will develop Huntington’s
disease. We’re going to cross
that with a parent that does not carry the
trait, homozygous recessive and we’re going to
see what happens. So we’re going to
make our square here. Here’s the effective parent and
here is the unaffected parent. We’re going to fill out the
boxes and what you’re going to see is that 50% of their
offspring would actually inherit the Huntington’s allele
and 50% would not. So, the probability
of having an offspring with the Huntington’s allele
is two in four or one half. Okay, so hopefully
so far so good. We’re going to start
to talk about unions and intersections here
and Venn diagrams turn out to be really
helpful to do that. It’s a great way to
visualize sets and we’re going to use circles to show
elements of the set. And of course, we’re
talking about relationships, so I thought, “Hey, let’s
talk about relationships.” So we’re going to have
set A and set B in order to illustrate the difference
between unions and intersections and we’re going to get to those
and/or rules for genetics. So set A, if you’re
looking for a relationship, let’s say you’re looking
for a partner that’s smart. Set B, other characteristic
you’re looking for, someone that’s funny. So, let’s look at
the two options. We can have an intersection of
A and B, so an intersection is where the two sets overlap. The union is either A or B.
So what I mean by that is on the left we have the
intersection of individuals with the characteristics smart
and funny where they overlap in the red, versus on the
right we have the union, smart of funny. So what that means is that an
intersection is this and that, it is less likely to occur. So you would say the probability
would be lower than the union, this or that, that is
more likely to occur. And by the way, in
terms of relationships, the more characteristics that
you add and the more sets that you need to overlap,
let’s say you’re looking for smart, funny and can dance. Then you notice that
the intersected regions, that intersection becomes
smaller and smaller. Okay, so probability wise,
when you see that word and, and you’re talking
about an intersection, that means multiplies. So we’re going to do a couple of
examples, like flipping coins. What’s the probability
of flipping two coins and getting heads on both? Okay, so the probability of
getting heads is one half. The probability of
getting heads is one half. So for each coin, the
probability is the same. But this and means multiply, because I am saying I want
this outcome and this outcome. So, one half times one half
gives you the probability of one fourth. All right, here’s another one. What’s the probability
of giving birth to three daughters in a row? Now the gender of any child is
an independent event even though you’ll hear people say
things like, “Oh, you know, girls run in that
family and so forth.” But actually, every child is
an independent event in terms of the probability of what
gender the child is going to be. So the probability of giving
birth to a girl is one half and the probability of giving
birth to a girl is one half, and the probability of giving
birth to a girl is one half. The thing is, I’m saying a
girl and a girl and a girl, so that becomes less
and less likely. And means multiply, so I
multiply the probabilities of each, so the probability
of giving birth to three girls in a row is one eighth. Okay, what about the
probability of getting heads or tails when you flip a coin? So let’s look at that
word or, the union. And I chose this because
of course, it’s something that you intuitively
know already. The probability of
getting heads is a half. The probability of getting
tails is also a half. But that or means you
add them together. So the probability of
getting heads or tails when you flip a coin is one. Okay, what about the
probability of rolling a three or a four on a six-sided die. So you have the probability
of rolling a three on a six-sided die, that
probability is one sixth. You have the probability of a
four, that’s also one sixth. And when I say a
three or a four, so either one will be okay,
I add those two together and I get two sixths
or one third. So, let’s put this
into genetics here. What’s the probability
of having an albino child if both parents were
heterozygous? Yes, we did this already,
but now we’re going to do it without the Punnet square. So now we’re going to look at
both parents being heterozygous. And what is the probability of producing an offspring
that’s homozygous recessive? So this parent has a
one half probability of donating the recessive
allele, so does this parent. In order to get two
recessive alleles, what I’m saying is I
need this and that. So the probability is going
to be one half times one half, that’s going to be one fourth. And hopefully you remember that that is the exact
same result we got with the Punnet square. I feel so good when things
work out the way they should. Okay, so squares do
get ugly pretty fast. And so the value of probability and understanding probability
becomes evident very quickly. This is the dihybrid
cross that I did for my Mendelian Genetics video. So if you want to see
me work this thing, check that video out. You can still, you know, do
it with a dihybrid cross. On an exam, that’s going to take
you a while, but anything bigger than that, you’re really going
to need this probability skill. Look at a trihybrid cross. So we got these parents. Are you really going to do
a square like this, I mean, especially on an exam
in a timed situation? I don’t think so. But understanding probability
makes this really pretty easy to do. Okay, so the P is first. What I want is the offspring
to have two recessive alleles for P. So this parent is
heterozygous, so the probability for them to donate
little P is a half. So is this parent,
same thing, one half. One half times one half gives us
for the offspring, one fourth. For the Q’s, I want an offspring that is heterozygous
for the trait. The only parent with a recessive
allele for Q is this one. The chance of the offspring
inheriting it is one half. The dominant form, the big Q,
has to come from this parent and the probability that it
will donate that big Q is one, because that’s all it has. One half times one is one half. Finally, for the R’s,
we want the offspring to be homozygous dominant for
R, this parent is heterozygous for R, so the probability of it donating the
dominant allele is one half. This parent is homozygous
dominant, so the probability that it will donate
the big R is one. And once again, one half
times one is one half. So finally, if you want to know
what is the probability you’re going to get this offspring, you
have to multiply these together because I’m saying
this combination and this combination
and this combination. And if you do the
math, you’re going to see the probability is one
sixteenth, which is the answer. And see, that wasn’t
so bad after all. So we’ve looked at autosomal
recessive inheritance with Punnet squares
and we’ve looked at autosomal dominant
inheritance. The other kind of problem you’ll
most likely come into contact with is X-linked
recessive trait. So these obviously not
inherited in Mendelian fashion. They are carried only
on the X chromosome. You tend to see interesting
pedigrees with X-linked recessive traits. The most common one that we
talk about is colorblindness, but we also see things
like certain types of hemophilia inherited
in this way. We’re going to use slightly
different nomenclature for this. Because it’s carried on the X,
we actually use the letter X and we use X plus to
mean normal or wild type. And we’re going to do
colorblindness, so I’m going to use this super
script C to show that that is the
allele for colorblind. Now because females have two
X’s and males have only one, the outcome of X-linked
recessive conditions will be different depending
on your gender. So females can be either
homozygous wild time, normal, females can also
be heterozygous. So this is a silent
carrier condition. Females can also be
colorblind, but they have to carry two copies of
the colorblind allele. Males only have one X, so males
are either normal wild type or colorblind. So the difference is that males
cannot carry a sex-linked trait in a heterozygous fashion,
only females can do that. So here’s a typical problem. What is the probability of
having a colorblind child if the mother is a carrier and the father has
normal color vision? Okay, for this I highly
recommend the Punnet square. It’s going to be pretty easy
to solve once you do that. So the mother is a carrier,
so one of her X’s is normal and the other carries the
allele for colorblindness. The father is normal. So you just fill out the square and you’re going to
see what you get. We have one female that
will be totally normal, one female will be a
carrier and for the males, you get one that’s normal,
one that’s colorblind. So the question, the probability of having a colorblind
child is one in four. So there you go,
several examples. I hope that they were helpful. As always, I thank you for visiting the
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