BIOL 2106
Problem Set #2
Mike Polan
100291925
Oct 14, 2004
1.
When dealing with genes that cause death in infancy, it is known that the only
mode of inheritance possible occurs when two heterozygous recessive carriers mate, and
their offspring is born with the homozygous recessive fatal gene combination. The reason
that this statement can be made is based on the fact that no individuals with the
homozygous recessive gene combination can ever reach an age at which they would be
able to produce their own offspring. Also, due to the nature of recessive traits if a
heterozygous carrier was to mate with a normal individual (homozygous non-carrier), no
combination would yield a homozygous recessive carrier, therefore no offspring would
show the fatal trait.
The use of Punnett squares can exemplify this:
Figure 1. Heterozygous Male
and Female carriers
T
T
TT
t
Tt
t
Tt
tt
Figure 2. Homozygous noncarrier and heterozygous carrier
T
t
Tt
t
Tt
t
tt
tt
In Figure 1, we can see that ¼ of the offspring will express the fatal trait, and the
remaining ¾ will not. More specifically, odds are that ½ of the offspring will be
heterozygous carriers of the fatal trait much like their parents, and ¼ will not carry the
trait at all.
Figure 2 shows us that none of the offspring will be homozygous carriers of the
recessive fatal trait, and there is a 50/50 split between the offspring being born as a
heterozygous recessive carrier, or a normal child lacking the recessive fatal trait.
With recessive traits, it is possible for the fatal gene to skip generations. The
following pedigree chart shows how this can occur.
This furthers the fact that the only chance of expressing the lethal trait comes from the
mating of two heterozygous carriers. There is no other way for the trait to be expressed.
2.
a) Since we are dealing with a rare autosomal recessive condition (non-fatal if
treated), when the offspring of two people display the trait, it is safe to assume that both
parents are heterozygous carriers of the gene since that is the only way for the condition
to arise when the parents do not show the trait.
Therefore the special relationship that led to the illness in two members of the
fourth generation is between the parents from the 3rd generation. Both parents are
heterozygous carriers of the recessive trait. What could very well be an influence upon
both individuals being carriers is the fact that there is a close blood relation between the
parents. This increases the odds of recessive traits being displayed because there is not as
much gene diversification as you would have if the individuals were to mate with people
from the general population who are at less risk of carrying the rare trait.
b) Looking specifically at the unaffected individual II-3, the odds that this person
is a carrier can be figured out by working our way back from generation IV. In part a) we
determined that both parents must be heterozygous carriers. By using that information we
know that the individual III-2 must have inherited the trait from the parents. There are
three possible matches between the parents that could have led to their child being a
heterozygous carrier: Both parents being heterozygous carriers, mother being a
heterozygous carrier and father lacking the trait, or father being a heterozygous carrier
and mother lacking the trait.
Therefore, the odds of the unaffected individual II-3 being a carrier is 2/3, or a 66.6%
chance.
c) Knowing that both III-1 and 2 are heterozygous carriers, we can use a Punnett
square to determine the odds of the unaffected individual IV-3 being a carrier:
T
T
TT
t
Tt
t
Tt
tt
We know that the individual is not a homozygous recessive carrier because the individual
is not expressing the trait. Therefore we are able to scrap the TT result from the table.
This leaves us with a 2/3 chance of the individual carrying the recessive gene without
expressing it.
3.
A historical summary of Brian and Jennifer’s family background relating to
Cystic Fibrosis (CF):
- Brian and his first wife gave birth to a child with CF; therefore Brian must be a
heterozygous carrier of the recessive gene.
- Jennifer’s brother died of CF which means that Jennifer’s parents are both heterozygous
carriers of the recessive trait.
With this information about Jennifer, we are able to determine the odds of Jennifer being
a carrier through the use of a Punnett Square:
Heterozygous male and female carriers
F
f
F
FF
Ff
f
Ff
ff
We know that Jennifer does not have CF, therefore the odds are 2/3 that she carries the
recessive gene, and 1/3 that she is normal.
Keeping in mind that Brian is a Heterozygous carrier, if Jennifer is not a carrier
(1/3 chance), then there is no possibility of Brian and Jennifer’s offspring having CF. If
she does carry the recessive trait, then there is a ¼ chance of the child having CF, ½
chance of the child being a carrier of the recessive trait, and a ¼ chance of the child being
normal.
By taking all odds into consideration, 2/3 chance of Jennifer being a carrier
multiplied by a ¼ chance of the child having CF, we end up with a 1/6 chance of the
child being born with CF (2/3 x ¼ = 1/6).
4.
List of individuals who must be carriers, with explanations:
In all situations where the offspring of a couple expresses a trait, both parents
must be carriers due to the nature of autosomal recessive inheritance. Therefore, I-1 and
2, II-2, 4, and 5 must be carriers of the recessive trait.
Knowing that II-2, 4, and 5 are all carriers, we can conclude that the recessive
gene must be present in the parents. We can state that one of the two parents (either I-3 or
I-4) must have the recessive gene, and the other might have it. There is no possible way
of knowing which of the two carries the gene, but it is possible for both to carry it, or just
one of the two.
In all situations where a parent is a homozygous recessive individual expressing
the rare trait, we know that all offspring will be carriers regardless of the partner’s gene
combination. Therefore we can state that III-3, IV-1, 4, and 5 must be carriers due to a
parent being a homozygous carrier.
All other individuals (II-3, III-1, 4, 5, and 7, IV-2 and 3) might be carriers as there
is not enough information to say with certainty that the person does or does not carry the
trait.