So let’s do some genetics problems based
on Mendel’s Laws. Let’s assume that finger
length is controlled by a single gene. In
humans, the allele for short fingers is dominant
over that for long fingers. If a person with
short fingers, who had one parent with long
fingers, has children with a person having
long fingers, what are the chances of each
child having short fingers? So let’s use
a Punnett square to answer this problem.
OK so a person with short fingers has dominant
allele so we use a capital S to stand for
short. But they had a parent who had long
fingers which is the recessive allele. And
we’ll use a small S to stand for long fingers.
Now, recall that genes come in pairs. When
we talk about homologous chromosomes there
is a gene at each of those two chromosomes
called alleles. So when we talk about a person’s
genotype we have to have two alleles. So the
person with short fingers who has the parent
that had long fingers would have a heterogynous
genotype: a capital S and a small S. And the
person who has long fingers would have only
the recessive, because in order for the recessive
to be expressed, you have to have a recessive
on both of the homologous chromosomes.
So when we do our Punnett square we have to
figure out what would be the alleles that
would be formed during meiosis, which is to
say a haploid cell. So if they have a haploid
cell, they only have one of each of these
letters, so this person would have two possibilities:
they would have a capital S which would stand
for short fingers and a lowercase S which
would stand for short fingers. Notice that
I’ve separated them out so that we represent
them as separate alleles now. But the person
person who is having long fingers is only going
to have one possible allele. All of their
chromosomes in their gametes will have the
long fingered allele or the short S in this
reduced size Punnett square. Now, we have
a heterogynous possibility, and we have a
homozygous possibility. So this child, if
they were conceived, would have short fingers
and this child would have long fingers. So
they would have a fifty percent chance of
having either one.
So let’s look at problem in which there
are two traits in the phenotype. We have the
widow’s peak hairline, which is dominant
over continuous hairline, and short fingers
as we saw in the first problem, which are
dominant over long fingers. So if an individual
is heterozygous for both traits, that is to
say they have unlike alleles for both, they
have children with an individual who is recessive
for both traits, what are the chances of
their child also being recessive for both
traits?
So again, let’s look at the genotypes of
the parents. So the parent, who is heterozygous
for both traits, would have for example, capital
W for the widow’s peak, but would also be
carrying on their other homologous chromosome,
the recessive allele for continuous hairline.
And they also would have short fingers, but
they are carrying the allele for long fingers.
They’re going to have children with a person
who has both recessive traits, which means
they have to have the recessive allele on
both homologs for widow’s peak, and on both
homologs on the size of their fingers.
So what kind of gametes would these individuals
be able to make? This parent is going to have
every possibility of W and S. So, we could
have the two dominant alleles appear in the
same gamete, we could have a dominant and a
recessive, with a capital W and a small S,
we could have a small A and a capital a capital
S, and we could have a small W and a small
S. So, we could write those out like so. But
this individual would only be able to have
one of small W and a small S. So, our Punnett
square does not have to have 16 boxes, but
it can have simply four.
So, in this example we would have a heterozygous
possibility. We would have a possibility that
they have a widow’s peak, but they had long
fingers, or looking at a possibility they
have a continuous hairline and long fingers
and they can have both recessive traits, which
means they have a continuous hairline and
long fingers. So, to answer this problem,
what are the chances of their child also being
recessive for both traits, we could see that
it would be 25% or 1 out of 4. This illustrates
not only Mendel’s law of segregation, where
we see that only one of each letter goes into
gamma, but also, the law of independent assortment,
because the inheritances of the window’s
peak is independent of the inheritance of
the long or short fingers. In the next video
we are going to look at problems that deal
with extensions of Mendel’s laws.
