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### Course: AP®︎/College Biology > Unit 5

Lesson 3: Non-Mendelian genetics- Variations on Mendel's laws (overview)
- Thomas Hunt Morgan and fruit flies
- The chromosomal basis of inheritance
- X-linked inheritance
- Genetic linkage & mapping
- Pedigree for determining probability of exhibiting sex linked recessive trait
- Pedigrees review
- Extranuclear inheritance 1
- Inheritance of mitochondrial and chloroplast DNA
- Non-Mendelian genetics

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# Pedigree for determining probability of exhibiting sex linked recessive trait

Pedigree charts can be used to determine the probability of offspring exhibiting an X-linked recessive trait, such as color blindness. By scrutinizing genotypes and utilizing a Punnett square, the probability of color blindness in offspring can be calculated. Created by Sal Khan.

## Want to join the conversation?

- At the end of the video, Sal says there is 50% chance that the next child of Tom and Barbara will be colorblind. Does it matter that the couple already have 3 children and one of them is color blind? Will that mene that the next child (according to Punnet square) will definitely be colorblind?(11 votes)
- no.. it isnt a set thing. the punnet square shows the probability of the offspring getting those specific genes(22 votes)

- is there any trick to solve pedigree problems in MCAT?(9 votes)
- IF the parents do not show the trait, but some of their children have that trait, that means the trait is recessive. Does that help?(7 votes)

- At the end of the video, Sal says there is 50% chance that the next child of Tom and Barbara will be colorblind. Does it matter that the couple already have 3 children and one of them is color blind(5 votes)
- Lets look at this a different way, when you flip a coin it is 50/50 for you to get heads or tails. If you have flipped it 3 times before and got heads each time it is still a 50/50 chance on the 4th flip. There is no direct causal effect on the next flip from the previous flips.(8 votes)

- How does one decide which parent to start off with? For example, in the video, Sal uses the mother's parents as a starting point. But in real-life examples, how do you decide? Or is it always started with the mother's side?(6 votes)
- It starts with whichever parent has the pedigree. Hopefully, the parents aren't siblings, and therefore you can trace whichever parent is traced in the pedigree you're given.(5 votes)

- Is there a pedigree for codominance?(6 votes)
- Yes, try making one with B for black coloring, W for white coloring. BB would be black, WW would be white, and BW would be gray.(4 votes)

- Let's say the mother is a carrier XB-Xb and the father is healthy XB-Y. They'd have 1 healthy son, 1 affected son, 1 carrier daughter and 1 healthy daughter. If they were to have another daughter, is she more likely to be a carrier or not?(5 votes)
- There is a 50/50 chance of her being a carrier or not(4 votes)

- Sorry I might be confused, but are Tom and Barbara siblings?(4 votes)
- Nope, only Barbara and Mary are Bill and Bonnie's kids. Otherwise there would be a line going from Bill and Bonnie to Tom as well.(6 votes)

- If a recessive trait shows up in all the generations, how can we figure out that the trait is recessive?(6 votes)
- how is it that Xc-Y is a carrier and not X+Xc? pls explain. Thanks(4 votes)
- is their a way to tell if they are different(3 votes)

## Video transcript

- [Instructor] We are
told the pedigree chart represents the inheritance
of color blindness through three generations. And we see this here. The standard convention is a square is male, circle is female. If it's colored in, that means
that they exhibit the trait, in this case it's color blindness. So Bill exhibits color blindness. His phenotype is color blind, while Bonnie does not
exhibit color blindness. Color blindness is an
X-linked recessive trait. If Barbara is expecting another child, so this is Barbara right here, what is the probability
that it will be colorblind? So pause this video and see if you can figure that out on your own. All right, now let's work
through this together. So they're asking us about
their next child here. What is the probability that
it is going to be colorblind? And to help us with that, we can try to figure out the
genotypes of Tom and Barbara. So Tom is pretty straightforward. He is male, we know that
'cause there's a square there. So X, he has an X chromosome
and he has a Y chromosome. And color blindness is an
X-linked recessive trait. And so let me just make
clear what's going on. So I'll do lowercase C for
colorblind, colorblind. And I could do a capital
C for the dominant trait, which is not colorblind, but since they look so similar, I'll just use a plus for not colorblind, not color, not colorblind. And so Tom, his phenotype,
he is colorblind, and he only has one X chromosome, what the colorblind trait is linked to. And so that must have the
recessive allele right over there. So this is Tom's genotype. But what about Barbara? Well, we know Barbara's going to have two X chromosomes because
Barbara is female. And we know that both of
them can't be lowercase C because then Barbara would
exhibit color blindness, but how can we figure
out her actual genotype? Well, we could look at her parents. So Bill over here is going to have the same genotype as Tom, at least with respect to color blindness. He is male, so he has an X
chromosome and a Y chromosome. And because he exhibits color blindness, that X chromosome must have the recessive colorblind allele associated with it. Now, Bonnie, we do not know. She will be XX, will
have two X chromosomes. Like Barbara, we know that both of these can't have the recessive allele because then Bonnie would be filled in, she would exhibit color blindness. But we don't know whether she is a carrier or whether she isn't. But let's just think about where Barbara got her chromosomes from. One of her X chromosomes
comes from her father. And the other one comes from her mother. So if she got this X
chromosome from her father, her father only has one
X chromosome to give, the one that has the colorblind allele. So if this is from her father, it must have the colorblind allele here. And we know that the one from her mother does not the colorblind allele because if it was like this, then Barbara would be
colorblind, and she isn't. So we know that this must be a plus here. It is the dominant non-colorblind allele. And so now we know both of their genotypes and we can use those to then figure out the possible outcomes for their offspring. So for example, Tom can contribute a X chromosome that has
a colorblind allele, or a Y chromosome. And Barbara, right over here, can contribute an X chromosome that has the colorblind allele, or an X chromosome that has
the non-colorblind allele. Barbara is a carrier. And so let me just draw a
little Punnett square here. And so we have four possible
outcomes for their children and they're all equally likely. So you can get the X
chromosome from Barbara that has the colorblind allele and the X chromosome from Tom that has the colorblind allele. You could have the X chromosome from Barbara with the colorblind allele, and the Y chromosome from Tom. You could have the
non-colorblind X chromosome that does not have the
colorblind allele on it, and get the colorblind
X chromosome from Tom. Or you could have the
non-colorblind X chromosome and the Y chromosome from the father. So there's four equal scenarios. And so in how many of these scenarios is the offspring colorblind? Well, here we have a colorblind female. She has two of the recessive alleles, so that female will be colorblind. This is a female carrier, but they will not show the
phenotype of being colorblind. This over here is a colorblind male, has only one X chromosome and it has the colorblind allele on it. And this is a non-colorblind male. So out of four equal outcomes, two of them have the
offspring being colorblind. So two out of four, that
would be a 50% probability that the offspring will be colorblind.