# 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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you explain things in a great way. nice and relaxed…I rewatched this one several times (and took not, of course)

Thank you

Nice

thanks a lot for this lesson Totally changed my world of genetics:-)

White ppl have best Genes.

One of the brilliant educational videos and channel i've come across on youtube!

Thanks a lot. It was really helpful.

I LOVE YOUUUU

Why are you talking about mathematics instead of genetics that too in a sloppy way

so is this conditional or unconditional probability?

God damn this is easy after this video

Thanks a lot..u hve cleared a very complex thing easily

Thank you!!!

So, if you have a homozygous dominant parent with AA and heterzygous dominant parent with Aa; What is probability of getting offspring of aa? Would it be 0?