Lecture 8 Recording
We'll do some test questions. I will refresh a bit your
member you on the
Chapter 2 as well.
And on Wednesday we'll start in chapter 4, right?
I believe...
We're right in October 10th, right, on Friday. Yeah.
Alright, so in the previous lecture we discussed the centiplegic inheritance.
And again, set the cosmic inheritance is the inheritance that
is the result of inheriting the gene expression.
expression pattern.
DNA from organelles.
from mitochondria and chloroplast.
So we have a number of mitochondrial chloropor from in cells, or hundreds, right?
and all of them may be slightly different.
And as a reflection, you may have a plant that has different branches.
with different genetics. So that can be a prize to specific type of flower plus. Specifically here is the one with the pigments, give you the green color, one
It has no pigment white. Four branches that have a mix, right, that are able to give you this so-called variegated.
type of inheritance where you can have
two different disabilities of having pigmental not having
statement.
Plants, right?
So there are a number of
genes that are on mitochondria that
So mostly mitochondria, you also see now with energy, right? So energy production generation
and they're getting different type of diseases associated with mutations in these genes. As you recall, the nuclear genomes are
Linear right so trainees repairs of chromosomes versus mitochondrial genomes are circular in
and do not contain.
as many and in general they're fairly small as you see it's 16,000
And nevertheless it has plant m genes. So the pedigree of splintecongrel diseases, so inheritance is fairly simple.
So you recall that mitochondrains
mothers, right? So that means if you have the affected female
It will be
the gene, the augmented gene will be passed to everyone.
Male
And now when we have affected males and females, it became very simple. Males don't pass it.
but females do.
right? So males here are affected as to see, right? So they have a disease, but they cannot pass it.
they don't pass anything in the title class grades.
So simply there is no microphone.
firm cells.
But females do.
So that basically describes the cytoplasmic inheritance that
females can cast the disease. So first of all, risk of proclocrossis, right? It's going to keep you different.
This is the use of female to cause a disease.
which may have passed the disease, right? So, but may also not, yeah.
It is different because that linkage is based on the inheritance through the nuclear genome.
So in the nucleus you have the autosomes, so we have 22 pairs of autosomes and we have a pair of sex chromosomes X and Y. So if this is a nuclear inheritance you can have two possibilities, we will refresh them today.
This one, I didn't hear this come from the, not from the nucleus, but from the organelles, the structures inside of the cytoplasm.
And because when the gametes are formed, you form as a male, a female gametes, they're different by the material they bring. Those cells
Permint, holland green, nuclease,
So they have equal contribution to inheritance of nuclear genes.
a few
female whole womb brings all cytoplasm and male does not. So the sperm cell does not have the cytoplasm so it cannot bring mitochondria.
So therefore it cannot affect the progeny, right?
And therefore if you do reciprocal truss, if you take diseased female and
The process was a normal male. This inheritance will be passed to all progeny, males and females regardless.
If you take diseased male and normal female, the diseased male cannot pass the cytoplasmic inheritance. They have the disease because in all somatic cells it present, but when the gametes are formed, the gametes are formed
Some mature sperm cell does not have cytoplasm and therefore it cannot pass the disease. So the first rule, erycoproclocrossis, will give you different results. Second rule, females pass the disease.
and the males do not pass it straight.
Okay, so that was fairly simple. We'll do some
some of the tests, right? So then we'll try to refresh the chapter
to our knowledge.
So
straightforward question so you have two dihybrid individuals right do you understand what is the definition of
Monohybrid and dihybrid and anyone telling me
What is monohybrid individual and what is
Yes, please.
Two heterozygous characters absolutely spot on so hybrid
Just a definition means you already have two different alleles, right? So hybrid means have to reside, right?
It's not neither recessive nor dominant, but has to result.
And if it's a monohybrid...
would basically mean that it's only one character.
the house that's at the house.
So you put half, for example, A capital A small and then B small, B small.
I'm presenting you information about two characters, right? What actually you will call this organism is a monohybrid.
because the second character has two similar alleles.
Right? A dihybrid means that this individual has both characters, so A, B, C,
encoding one character phenotype, let's say, shape of seeds, right? And B, encoding the color of seeds.
are represented here as both alleles, dominant and recessive. So they have hybrids and they have two characters, dihybrid.
So the dihybrid can be either this, or it can have, for example, two other genes.
C small, C small, and D capital, D capital. It still will be dihybrid.
because c's and d's are represented by the same length.
letters, right, so small or capital, right?
So hybrid always means two different oles at present.
And monodiatry means how many of these characters, or gene sets, are present in the hybrid form.
Anyone has a problem with this?
Everything's clear?
All right, so we have here two dihybrid individuals across
Very classical F1 cross
And if a and b are independent, it's a gradient.
What is the probability of the probability
is that this homozygous recessive at both A and B. So
What is a homozygous recessive? Is it A small, A small, right?
be small, be small. Okay? So...
what would be the chance of getting this individual?
Yeah.
1 16th. So how did you calculate that?
That's perfect. Okay.
Yes, yeah, so homozygous recessive is the rarest form, right?
The calculations could be done by well if you know that
They are A, capital A small, B capital B small.
the you can calculate it for one allele because they independently segregate it and that means one force right the chance between
Two, the same individuals that have recessive alleles, right? So, dominant allele.
And then the same would be for second gene 1 fourth. So 1 fourth time 1 fourth gives you 1 sixteenth. So that was
Simple. Alright, so we have snapdragon plants. It may be tall or short.
Red or white.
right? Flowers. And we have the parents with phenotypic characteristics.
presented here, right? So by those letters, they're crossed, right?
And here you have the
numbers, right? So these are characteristics. Okay?
So
The question is, what is the most accurate description of the genotypes?
The parents in cross one, so you look at the cross one, both dihybrids. So that's first.
B. One parent in cross two.
is it dihybrid? So that basically means the second is not, right?
C, both parents in cross 3.
One more.
We agree that mono hybrids means that they have only one allele that is
in heterozygous state and second is not right. D
One parent in cross two is homozygous at both genes and homozygous at both genes could be isobose capitals
Both are small or one cap, both one small. So it could be
A capital A capital, B capital B capital, or in this case, it's L, R, and P, W, right? Oh, vice versa.
And then Feynol says, well, nothing of this four is correct.
No, no, sorry. All of the above are correct, right? So that would be number six, I guess, right? So all of the above are correct. So choose which one is correct, A, B, C, D, or E.
And please, if you have the answer, yes.
I don't know, it presents to you five cases, right?
So either A or B or C or D or E are correct answers.
Thank you.
So one of the fives that are listed here, ABCDE, is correct.
So no, this question is not a trick, right? So the first four give you individual answers, right?
So one of them could be correct.
And the fifth gives you an answer that all of them are correct.
So the number 5 would be here true only if all 4 are correct.
Or it can be that either A or B or C or D are correct.
and all the others are wrong.
right yes
It's an E. It is correct answer, so well, guide us through A.
first.
Absolutely, so that's the easiest answer. It's clear, 9 to 3 to 3 to 1, and we learned that in dihybrid cross, we are getting this only when we are crossing both heterozygous parents.
and heterozygous by both alleles are dihybrids. Therefore, the parents in cross one are both dihybrids.
So I'm hybrids. Anyone wants to do B?
We already have the answer is E, so all of them are all four are correct, but guidance through B for example.
Yes, please.
Not even monohybrid, right? Yeah, absolutely. Brilliant. So you spotted that the ratios are 1 to 1 to 1 to 1.
And that only can happen if it's a test cross, right? And the test cross is when you take someone that is receptive to the bugs.
Right? So, well that doesn't mean that the second one is actually dihybrid.
Right? Does it or does it not?
So one of them is not, but the one of them is, right?
Of course, if you have this one-to-one-to-one-to-one, the second must be dihybrid, right?
you don't even have to go through C and D anymore because if two are correct, you have to give one answer that means that E is correct, right?
So you can of course try all of them, right? So to be certain. But as soon as you're absolutely sure on two. So that means that the answer cannot be one or the second. So...
it has to be more than 2. And the only choice that is given is E, where all of them are correct. But let's do it anyway. So C,
Both parents in cross-street are Mona.
hybrids.
So, monohybridz means that at least one gene is heterozygous, right?
So if you notice here there are zeros
Right? So what do zeros represent?
they represent the power, right?
So which is
and they represent
And then we have here three to one segregation for
white flower as a color, right?
All right.
and
We have 3 to 1 segregations for the short versus
Paul, so there are three shorts
for heights versus one Paul, right? So there is one heterozygous for
for height, right?
and we don't have any
Anything in this category, so we have red flowers, they are not existing, right?
Suggesting that in the cross we didn't have at least four colors one of them did not have
the heterozygosity. So one
only one was the hybrid image 8.
So
Next one is D, one parenine cross 2 is homozygous parenin.
both genes, right?
And that was easy because we agreed that that was a test cross. So that's where almost a gothicity of both genes comes. So the test plant is recessive.
homozygous at both genes. So all of these above, correct?
You're good with this?
Any questions?
Yeah.
So
In the third cross we have SW, right?
Right like this.
is a distance SW, right? So cross to SW.
Okay.
So based on this
we see that the largest number signifies here that both are dominant. So therefore,
Short and the red are dominant.
short and red are dominant. So as soon as we see white,
The only way is
They're recessive.
Right?
White cannot be otherwise white right as a what threat would dominate
So that already tells me the size of them is dihybrid.
So the stability of them, each of them is monohybrid, right?
So both parents in cross three are monohybrid.
So
If we assume that only one
was monohybrid, we would have 50-50.
Right? So, because this...
This is half and then
This is another house. But we don't have 50-50, we have 3-1.
Right? And three to one is only possible
if both of them are heterozygous.
Tall.
So tall is recessive, right? So it means that both of them are
monohybrid.
You can have 15 genes and if only one of them is a hybrid, it has two, so it's a monohybrid.
Right?
question. So anything else?
You good?
A true breeding fly. So you remember true breeding means it doesn't segregate, right?
with curly wins
And wall type body, so you
Currently, right?
Wild type buddy.
Wild type
Likely is done not necessarily but likely is gone. So
do something like this.
Bye.
is crossed to a turbulent fly with wild type winds.
black color buddy right so wild type wings so
Well, assume that this is a wild type point, so, don't worry.
Black body, right? Black body color.
So both of them are true breathing, right?
Introborated means that they don't segregate, so we can be absolutely certain that that's what their genotype is.
F1 flight
are all wild type.
Of course it makes sense, right? So because these will dominate these and these guys will dominate these. So therefore,
The progeny
is this, right?
Okay.
The F1 is crossed to a fly with curly winds and black body color.
So this guy is called, is crossed to curly winds.
and black body.
What is it?
Wow, it's a test cross, right?
and are producing the ratio of one quarter, one quarter, one quarter, and one quarter.
Okay.
In the progeny, so there are only two possibilities for you. In the progeny, the wild type flies and black body, so A,
asks you wild type flies and black bodies.
Right? It doesn't say what the allele is, but at least one is capital because it's wild-tied, right?
Curly winds flies, so wild-tapped flies and blank body.
You're well-tapped twice.
and leg body, curly wind flying.
are differentials. So the second one is curly wind flies.
We don't know what this is, right? So it doesn't say about the body. So it gives us two choices. Wild type flies and black body. So this is a wild type, right? We don't know the second one, but it doesn't matter. And the black body. So black body can only be if both of them are.
recessive, right? And then it says curly wind flies. Well, we just said that curly winds is a mutant, so we know this genotype, right? And we don't know anything about the body.
And the B
gives you the wild type flies and black body wild type flies
And black body, right? So again, we don't know this.
and currently waiting flies.
We don't know this so it tells us that this is
Parentals.
And this is recombinant.
Right? So which one is true? A or B?
These are parents.
So parental alleles should look like them.
All right.
because recombinant alleles would have
this
Pair it with this.
or these pairing with this.
What do we have here? Capital C.
Right? Present?
present. Okay? We don't know this.
Right? Doesn't matter. Now we have...
small B's right
Do we have such a parent?
We don't.
That means that the answer is B, very confident.
Right?
So again, we did the cross between the two
true breeding, right? So you see, it doesn't have any information about genetics, but it tells us
that they are true breeding, right? So we can define what they are.
typically. Yes.
We don't, I can switch, it will not change anything, right? So what you have to define for yourself, right? So it doesn't matter...
Typically, wild type is dominant, right? So it doesn't tell you otherwise, assume this, right? But if we assume otherwise, it will not change our answer, right? So all we care that
We produced a heterozygous in F1.
Right? And then we test cross it. Okay?
So you don't even have to calculate anything. So when you have a description
It says wild type flies and black body, right? So that already is not a test anymore, right?
And Curlywind Flies is a test anymore, right? Like we don't know because it's...
Second is it not in the...
is not now, right?
But because it gives you wildcard flies, right, so as capital C,
and
Black body, right?
You don't have such animals in your crop.
You have it here, but not here. This one is heterozygous. This one is wild-type, right? That's what tells us.
F1 flights are all wild type, right? So you cross the wild type to fully recess it, right?
black body. So that means it's recombinant, right?
near phenotype that appear. So therefore
The B answer is correct.
You're good with that? Any questions?
Okay.
This one is fairly simple.
Raise your hand when you have an answer.
It's a C, which is more common in males than females.
So how did you did you say?
Simple, right? So you could go through all others that we discussed today.
Reciprocal crosses give different results, right? That's what we agreed on cytoplasmic inheritance, correct? Males cannot pass on the trait. Of course they cannot because it's in cytoplasm and sperm doesn't have
the mitochondria, right?
If a female has a trait, all of her offspring will also have the trait. Absolutely, because females get the cytoplasm if it's there.
All the individuals, all organisms, males, females will have it, right? So that means that C is the one that is not true, right? Or you could...
do it alternatively as gentlemen said because it's not sex linked right so it tells you that the cytoplasmic so therefore it is
the one that you have to dig, right?
Okay.
You have a cross?
And the question is, is the progeny
in the progeny, what is the probability of an individual heterozygous at all genes, right? Heterozygous two different letters present. So you have to calculate individually for each.
Character right characteristics and then multiply them together right so well when you have the answer
raise your hand
Yes, please.
So, normal day about.
Okay, so there is no ratio that is correct here.
Let's try, I don't know, I haven't done this. Let's calculate. So...
heterozygous for the first one.
Is it health? 1 and 2?
I'm asking hello. How? Right? 1 and 2. So for the second cross.
heterozygous is also one in two
So it's 1 in 4 already, right? So cross between these two.
What is the possibility here?
one, so it's still one in four, then cross between d, d,
and DD, so what is its ability?
So now we have one in here.
here
Hetirozavus.
One half, one and two.
So when you cross
you
You get this, you get this.
You get this, and you get this. What is the possibility of heterozygous?
1 and 2. So now we are 1 sixteenths, right? So...
Next one.
Health, so 132. And next one.
Half. 164. So the right answer is B.
So you calculate individually for each, right?
So...
if you did.
The first one just repeating right so this is the one allele
two allium, right? One, two.
Half would be caterositis.
Christ and regardless so when you cross
like this
It still does the same.
Right? So when you cross like this.
It's still half, right?
So when you cross
like this.
than all of them.
So you want right this one on
All of them are done.
It's one, so you multiply with one, right?
Okay.
So we have a rare disease.
As long as you hear the rare disease or anything doesn't say it's a common or frequent whatnot
You always assume that the second individual that is not affected
don't have anyone recessive, Aleem, right?
So that means that
This guy is affected, right? This is healthy. So that means that this one is heterozygous.
All right.
And this is coming from outside is homozygous, right?
So we can say that this individual is A, A recessive, right? Small a, small a.
This individual is capital A, capital A.
Right?
So
It's not sex lean, right? So because
Tell me why is it not sex-league? Yeah?
I know why, because someone at the top before me has this next slide set up with notes.
And apparently when it disappears, sorry about that. You should have told me earlier.
I'll let you through the list.
Thank you. Alright, so tell me why is it not sex linked?
Yeah.
Both males and females, right? So, and if you look at in this, so this disease, like most sex links are associated with X chromosome, right? And if it's a recessive disease, if it's a recessive, that means that this lady
God, the genes from both parents, but the male is not affected here, right? So it means that this X chromosome is normal.
So if this was dominant, right, then the male passes X chromosome to daughter, then the daughter
So it is not sex. So it's uh, uh, all the song, right?
So again, we agreed that this one is heterozygous, A capital A small, right? So, God, the allele.
from the solder. This one is normal, right? So, homozyg, right?
So that means that these are individuals
have a chance, what are the chances to get the recessive allele?
Yeah.
one and two 50-50 right so one parent is a capital a capital and One is a capital a small right so 50 percent so here
We have a diseased individual, right? And we look at that.
that.
supposed to have
And allele disease, the allele, correct?
Otherwise you cannot get the disease.
So it means that A capital A small, A capital A small, right?
Now we have two unaffected individuals here, right? So what is the probability of individual A?
of carrying the allele.
Yeah.
Two thirds, right? So the reason as you recall why we are talking about three here, not one quarter, two quarters, whatnot,
because we already eliminated
the one that is recessive.
Right, so we have a
a carrier likely, but not a disease. So if you have this
Cross.
between two recesses, so a disease.
here
1 quarter, so we know it is not a disease right so that what's required it remains
is two a's and one aa, right?
So three choice
and only these guys can carry
It's always too serious.
So we have two thirds and we have
a B individual at one third.
Right?
So what is the probability that A and B will have an effect?
So now we calculated probability for each partner.
But we know that affected child is
one quarter, right? So therefore to know what was the probability for an affected child
We do.
this multiplication, right?
And we get
No, one third, sorry, it's one and two.
So this guy was one and two, right?
So, in order to give an answer about whether the child is affected or not,
We need to have a probability for excessive allele to be carried by each.
partner, right? So we calculated partner B to have 50% health.
Right, so how did we get this?
Here's parents.
in here.
This must be a carrier because God recessive allele from the father, right?
And this one is because it's rare so we should not have any allele, right? So the probability of
allele here is one and two, right? So these two parents have two serves of one and two, and then the probability of them
getting this recessive with one and four, so therefore we have
1.12, so the B is correct answer.
Right?
So the probability of an affected child
Also 1 in 12, right?
So it's either a recessive, small a small a, or both capital a's.
So
12th total 12th or 12th right so we deduct one 12th that the probability to have an affected child so all other children will have
the probability of not having the disease and there's 11 12s. Right. So.
Yeah, if 12 is affected, then all others are not affected.
carriers, they may be not, but these probability is 11 twelfths.
So that's a probability for you to calculate for heterozygous child, right? So I give you the calculations and answer right away.
simple right so you remember a couple lectures ago I asked you to do the homework to calculate
This is a slightly different example, right? But the idea is the same. So if we have this rare disease and we have to calculate the probability of heterozygous child
Okay, we'll look at these affected individuals. So it has small a, small a recessive, right? That means that the partner, it's a rare disease, the capital A's. We get the same scenario. So we have
the carrier right so plus non-effectively right therefore annual this three
have equal probability of being Isaac
Patrozegas or homozygous, right?
So here we agree that these two are carriers, right? So we have a disease and then we have these two individuals
that have a probability of two serts being heterozygous and one sert being
homosects, right? So now we have to calculate probability
for heterozygous, and for this we have to pair each of this probability, right?
with one of these two, right? Say this probability was
Only with this one why because
These two cannot give you a recessive for heterozygous, right? So they both have dominance, right? So they cannot give you the probabilities. Therefore, your equation will have three parts. We'll have multiplication, so probability between this probability
and one of the two, right? And then this probability with the
So we have here
Two and so two-thirds right in here
from building for E
have probability for this heterozygous, right?
and we multiply it by half because heterozygous you get in monohybrid cross.
one in two, not one in four. So you remember that the heterozygous monohybrid cross
You have
1 force, 2 force, or half probability for heterodytron.
and
one force for ability
We always calculated before these guys because it's one-fourth
disease, right? We could calculate the same for one-fourth no disease as well, right? So, but if you ask for heterozygous,
That means 2 fours of half.
That's why in this equation, each of three in the end have half.
So again, 2 thirds probability by 1 half probability
But by one half that's what you get to basically form on a hybrid cross.
then you multiply two-thirds by...
these probabilities, so by again half and then finally one third times
One health here and one health.
So together, sorry, it gives you five twelfths.
If you calculate these independently and add them up, it will give you 512.
Any questions?
It's a bit more difficult but still the same logic applies, right? So you calculated probability of either A or B to carry an allele, right?
And then because any of them of these two alleles can pair with any of these two, right? So you have to have four variants, right? So these probability
by one of two and this probability by one and two but because
These two do not have any chance of getting the heterozygous. We eliminate
right? And we always in the end multiply by 1 in half.
because that is a probability of getting heterozygous in the cross.
Do you think you will be able to do it in the exam?
I only had one half
Shaking great, so I okay too
So
Ask me questions. So where is the difficulty? What is
What part is unclear in my calculation?
Let's start from, I'm not sure you understand this completely.
So let's start with this from the top.
Is it clear to you why this double capital and this double small a's
Okay.
So that concept is clear. It's a rare disease, so it means that the second partner is not affected and doesn't have the recessive allele.
So now we have a daughter
Obviously, the daughter carries one allele from each parent, so the daughter is heterozygous.
So again, rare disease and the partner has no
Recessive allele, so it's double capital, right?
So when you cross this
double capital and recesses.
So the chances of getting the recessive allele is one and two.
The chances are 1 and 2.
This is potential tearing and this is potential tearing. So there are four.
potential parents, only in this too, which is health, you can get recessive allele.
So therefore
This individual B that is participating in the cross
has a chance either to have a recessive allele or not.
50-50, right? So just imagine that
This individual produces just two whole cells.
right, and 50-50 chance either carry this or this combination.
So in this cross we have a disease, so two small a's.
The only possibility of getting that if both of them are.
Hatteras Agus
Right?
So that means that the
The individual A
has a chance of carrying the allele in two-thirds
out of three possible. Why? Because if you cross two heterozygotes,
And you don't have recessive, which is not, we know that this guy is not, doesn't have a disease, right? So that leaves you only with three units. One,
would be A A and 2 A. So therefore the chances of this individual to carry
after zygosil is 2 thirds and the chances are that it does not are 1 third.
third.
Do you have any problems so far with calculations like this?
No?
Perfect. So now these two individuals meet. We don't know which one of the sperm cells would carry this.
or that, right? So heterozygous or homozygous. And we don't know which one of the own will be participating. Therefore we should
assume that this perm can lead either of these two all cells and this perm can lead either of these two all cells.
So therefore, we have to in order to give all different probabilities
we have to calculate first the chances of this sperm meeting this all
two-thirds by one half. And then we need to know what would be the outcome of independent assortment. And as we agreed,
So it will be 1 and 2 to get the heterozygous. That's why we multiply.
So then you do the same for this verb to be this O, right? So again,
by one and two and then this perm to be this all right so and again one and two and we did not calculate this perm and this all because
because it does not have any recessive. And so now we have two, all the three independent events. So because they are independent, we sum them up.
Right? And that would give you five twelves.
Better now?
Thank you.
All right, now the one.
individuals with white or green colored seeds, right, were crossed to individuals with striped or plain colored pots.
Okay, so we have here two characters, right?
So we have crosses here, so three different crosses. You notice probably right away that we have here nine
to 3 to 3 to 1. If you always look at 9, right? So that means that these letters represent dominant phenotype, right? So therefore white,
dominates green, right, and strike, dominate plain.
Okay, so the question is here, what is the most accurate description of phenotypes?
Right, so white and plain.
Dominant? No.
because the stripe is dominant. So white and stripe dominant, so the B is the right answer. So green and plain are dominant, no they are not.
Green Striper Dominant, none or not, none of the bugs.
So the right answer is wide and strict. So it's one of the easiest questions you probably can get, right? So again, always look.
and one of the crosses likely will give you 9 to 3 to 3 to 1. And the 9 will always tell you which ones are dominant.
So it means that opposites are recessive, right?
Okay.
So we have the same, right? So cross, so again, we have wide and striped dominant.
right so and just when you
when you presented by letters, so we assumed that G capital dominated.
rate than P capital domination.
So what is the best description of parental genotype in the cross?
2
Right?
So what's your answer? Well, look at the ratios that you see in the cross 2.
Okay.
So which answer? Yes, please.
here
So, D, can you speak up again?
Why? Why?
Right?
One of them mind-mono hybrid, yeah?
You're right, so D, it's a D. So basically
When you said that one of them on a hybrid, that means that in this
The only two that give you monohybrid in both parents are these two possibilities.
So because this is dihybrid, right? This is dihybrid, right?
and out of these two
If this was
One of them was test cross. It would give you one-to-one.
right for the stripe versus plane. So therefore the only one that can give you three to one ratio in this cross is the D, right?
so we cross two.
Any problems with this one?
Okay, so again the zeros signify that we
Have a monohybrid cross so on the wrong of that
is given the possibilities of
So one of them has to be recessive.
and the one that is not a cesare, the one that gives you the stripe or
plane and the stripe of plane are ratio is three to one and this is possible only if both of them have the rositis so it cannot be between after and homozygous so
That's why the answer is D.
Okay, so that's pretty much it for the test.
should not have any sense of
Unexpected, let's say, in terms of
There's different calculations of probability and test. We calculated it all. So in this remaining time, I will a little bit go through the sex linkage just
just in case right so
Just to refresh your memory.
Okay.
So how did we do, how did we learn that there are inheritance associated with sex linkages? So you remember one in the autosomal, right? So there is a proper process.
always give you the same results because it doesn't matter which parent carries the dominant allele.
But in the sex linkage, it does matter why because
parents are not equal in terms of their caring acts chromosome. Y chromosome has very few genes so there are very few
cases where you pass the
disease Y-Y chromosome. But X chromosomes are very common. So therefore, if you cross the male that is affected, right?
So it has only one X chromosome. So this X chromosome can be passed to the daughter, right? So because the male
passes Y chromosomes to sons, right? So that means that
In terms of inheritance of disease, sons keep the generation, so they only can get
the recessive alleles from their mother, right? So sons have X and Y, right? So that means they get Y from father and X from mother, so therefore they can only get a disease if the mother passes them a recessive allele, right? So we recall this different process that gave different phenotypes.
between
and also this drosophila depending on the color of the animal.
that was associated with X chromosome being present on email and email. I will not go through the details but basically
showing you here the difference. So when you cross red female
How's the color in both X chromosomes white white male
versus when you cross white female with red male you get two different outcomes.
In one cross you get half and half.
right so but all of them being red and another eight cross you will get the white white
male and red female.
because white as a recessive can only be observed
in the mail right away because in
female can be dominated by.
right so different crosses give you reciprocal process give you a different
results.
So you may have some test cross about the color, right? So you recall the color in cats.
on X chromosome, X linked, right? So if you are male, right, so because you only have X chromosome, the male can be either orange or black, so they can carry a little bit more
defines orange color or black color. If it's a female, right, so it can be either black, orange, or calico.
It means that if you get X from black and X from orange, right?
And because in somatic cells one of the X chromosomes in the female
get suppressed, right? So it means that you can have fur where cells randomly express black
orange, so you have this mix of colors on the fur as you see here.
Right?
By now you're probably familiar with these symbols, right?
Um...
You are familiar with a pedigree, right? So we've done it many times
So again remember if we have a see the disease so that if it's a rare disease we always assume that the partner is not affected and doesn't have rare disease.
assess it a little so it's important, right? Do remember these so you may get them in your test. So what signifies a cross, what shows twins, right? So affected individuals.
heterozygotes. We actually didn't see that, right? So because in most of the tasks you're supposed to figure out yourself.
But this was a pedigree that was combined and
shown by a doctor so doctor if they know that's how they show the heterozymes right so if it's a sex lean then the carrier has a dot again in
In our test process, you will not know this, right?
So we have a definition for propositors, so the individual that we investigate, and that we're trying to deduce what was the
phenotype of parents here, so we have a disease
We would have to figure out was it was it sex linked was it
Was it autosomal, was it dominant or recessive?
right so all this
steps right so if you have disease
likely their homozygous recessive, right? So the only possibility for them would be if the appearance harm.
If it's a rare disease, that means that they're
dear
Yeah.
mates, right, so they are not affected and so on and so forth. So you can calculate the probabilities all the way up.
So we've done this before.
What else?
Yeah, so again, you should be able to calculate all these probabilities based on all the
examples we have done. Just seeing that there is anything new here, right?
Okay, so you recall this, right?
what are we dealing with here?
Is it sex-laved? Is it...
Autosomal dominant recessive what it is
you
So
One, two, three kids, right?
One of them is not affected, two are affected.
Can be sex linked
So if it's sex linked you should have more or less equal
more or less biased toward one sense, specifically male. So we have this more or less equal representation. Could it be Y chromosome?
Now right so we have the effective female right?
So like Litz-Ava-Somol, if it was dominant, all three are supposed to have it, right?
but yes, what do you think?
So it's a dominant disease.
would you expect all of them to be infected?
It's a trick question.
How many alleles do you have to have affected in order to have a disease if it's dominant?
One, right?
Is the chance the second a little normal?
Right this individual is not effective doesn't have to leave so therefore it will provide the normal allele
And this guy can provide a normal allele, at least one. So there is a chance for this individual to have
disease.
Again.
because they switched to a new lecture, so sorry.
And I don't know how to inaccurate this next slide step.
Okay, so this guy is affected right, but we don't know what it has
Thank you.
Capital or let's assume it's a recessive right so let's start with simple case So if this individual is a recessive right that means too small is right so this is highly likely
not having a disease, but two kids out of three do have a disease. So in theory it is possible that it is a carrier.
That's not a rare disease anymore, right?
So now we have a disease individual here if the individual is recessive, right?
Again, this individual is supposed to be normal.
right so it doesn't have to have a disease if it's rare
Some kids are affected.
So, highly likely this individual has a
the carrier, right? So, and the same in this cross.
If this was dominant we can assume that
one allele A-capital and one allele A-small, right? And these both are A-smalls, right?
So that means that this guy
maybe a capital a small right and this
has to have both of them as models. Same in here, right?
So therefore this in theory can be a dominant
trade with only one allele affected.
Right?
And so that's what we have here, right?
So when you look at this, always remember that if it's a disease that
basically seen in every generation, right? So, randomly male or female
This individual can still have a normal allele, right?
So therefore, there is still possibility that some of the children are not affected. Right?
Any problems with this? Okay.
Yeah.
Yeah, so always assume that the disease is rare. So most of the disease is predominantly rare. So they should not be. In theory, you can have something like that when the allele is so common, right? So that it can be a recessive that is present
Because here we have three marriages and in two of them the partner would have to carry the allele.
easier and higher probability that it's not
a very common recessive, likely a dominant trait.
So the trait is dominant, all infected individuals must be homozygous recessive, and that's what we've got. And infected individuals must have at least one gene.
dominant holding up. And it's a rare disease, so highly unlikely they have both.
dominant audience.
Right? And that's what I was telling you. There is a possibility that
This is a recessive allele, right?
So it just means that there are a lot of carriers.
And so this is possible if the disease is not rare. So that means that you have two types of the same pedigree, right?
that gives you two probabilities. This probability is way more common, right? So because most of the diseases are
So if you're in your test, it tells you that it's a rare genetic disease, your answer would be here and your logic would lead you to a given outcome of the disease.
If it tells you that this is common,
then your logic would be that you have a recessive, right?
then this would be normal, right? Sorry, this would have to have a recessive allele because you cannot get
disease and then every time you have a disease appearing in next generation shows you that you were mating with heterozygous with a carrion, right?
So again, the test on your exam will tell you in these specific cases where you have a rare
or very common disease. Okay, any questions?