Download 03-131 Genes, Drugs and Disease Problem Set 10 - Key

03-131 Genes, Drugs and Disease
Problem Set 10 - Key
November 27, 2015
1. Two pea plants were bred to each other. The peas from the first generation
(F1) were 50% smooth and 50% wrinkled. What are all of the possible
genotypes of the parent plants? Which genotype gives the observed data?
(This is the correct wording of the problem from the previous problem set).


R
r
r
Rr
rr
r
Rr
rr
The possible genotypes of the parent plants are RR, Rr, or rr.
The only cross that can explain the observed data is between heterozygote dominant and
homozygous recessive: Rr x rr. 50% of the F1 are Rr (smooth) and 50% are wrinkled (rr).
2. A test cross involves crossing an organism that has an unknown genotype with a purebred organism that is
recessive at both alleles. Briefly describe how the test cross could have been used to determine the
genotype of the plants in question 1.
The above is actually a test cross. If the smooth parent pea was RR, the test cross would show 100%
smooth. If the parent smooth pea was Rr, the outcome would be 50:50. The testcross reveals the
presence of recessive alleles.
3. The following pedigree was obtained for several
generations of a family.
i) What pattern(s) of inheritance (e.g.
autosomal/X-linked, dominant/recessive) are
consistent with this pedigree? Briefly justify
your answer.
It is autosomal, as both males and females
Dominant
are affected.
Rr
rr
 It is most likely dominant for the following
reasons:
rr
rr
Rr
Rr
rr
rr
Rr
rr
o The children of “grandfather” also
show the trait.
o The left-hand breeding pair does not
show the trait and none of their
rr
rr
rr
rr
rr
Rr
Rr
rr
Rr
Rr
rr
children show the trait.
 The possible genotypes are shown on the left. Note that the “grandfather” has to be
heterozygous, it is the only way the daughter on the far left can be homozygous (rr) and not show
the phenotype.
ii) What are the possibilities for Susan’s genotype. Briefly justify your answer. Susan is homozygous (rr)
since the inheritance is dominant.

Note: The pedigree can be also be interpreted as autosomal recessive, assuming that the grandfather is
homozygous recessive (rr) and the grandmother is heterozygous (Rr). However, since you should assume that
the phenotype is rare, it would be unlikely that the grandfather is homozygous, so autosomal dominant is the
best answer.
4. The following pedigree was obtained for a family for an
autosomal recessive gene. Could this pattern of
inheritance be consistent with an X-linked gene? Justify
your answer.
Yes, only males are affected. If the mother was a
carrier (XRXr), then the inheritance pattern fits sexlinked recessive, one son inherited the recessive allele
(Xr) from the mother and is affected, the other son
inherited the normal allele (XR)
female gametes
XR
male
gametes
Xr
XR XRXR XRXr
XRY
Y
XRY
XRXr
XrY
XRXr
XrY
or
XRXR
XRY
03-131 Genes, Drugs and Disease
Problem Set 10 - Key
November 27, 2015
5. A homozygous fruit fly with white eyes and wrinkled wings was mated with a homozygous fruit fly with red
eyes and smooth wings. The F1 were all red eyed with smooth wings. The numbers of flies in the F2
generation with different phenotypes were counted and the following were obtained:
Phenotype
Number of Flies
Red eyes, smooth wings
300
White eyes, wrinkled wings 100
Red eyes, wrinkled wings
1
White eyes, smooth wings
1
i) What are the traits and phenotypes in this experiment?
The traits are eye color and wing texture.
The phenotypes are white and red (for eye color), and wrinkled and smooth wings (for wing texture).
ii) Which alleles are dominant?
Red eyes are dominant over white, and smooth wings are dominant over wrinkled.
iii) Use a Punnett square to determine the expected frequency of each possible type of fly after crossing the
F1 flies with each other, assuming the following two conditions:
a) unlinked
If the traits are unlinked, then all possible combinations of genotypes are possible since the
homologous pair of chromosomes bearing the eye color marker and wing texture will be mixed up
during meiosis. The expected ratio is 9/16 red smooth, 3/16 red+wrinkled, 3/16 white+smooth, 1/16
white wrinkled.
RS
Rs
rS
rs
RS
RRSS (red, smooth)
RRSs(red, smooth)
RrSS(red, smooth)
RrSs(red, smooth)
Rs
RRSs (red,smooth)
RRss (red, wrinkle)
RrSs(red, smooth)
Rrss(red, wrinkle)
rS
RrSS(red, smooth)
RrSs(red, smooth)
rrSS(wht, smooth)
rrSs(wht, smooth)
rs
RrSs (red, smooth)
Rrss(red, wrinkle)
rrSs(wht, smooth)
Rrss (wht, wrinkle)
b) linked
If they are linked, then the only possible gametes are RS and rs. ¼ will be homozygous dominant
(RRSS), 2/4 heterozygous (RrSs) and ¼ homozygous recessive (rrss). The only flies that you would
see have red eyes and smooth wings (RRSS or RrSs) or white eyes and wrinkled wings (rrss), in a
ratio of 3:1. You would never see red eyes and wrinkled wings, unless crossing over occurred (which
has clearly happened).
iv) Based on the F2 data and, are these two traits
linked or not? Justify your answer.
r
They are linked because the number of red eyes
and smooth wings or white eyes and wrinkled
wings is in better agreement with the number
predicted by the linked model.
s
Anaphase I
v) Explain how the flies with the rare phenotypes
(last two rows of table) were generated.
Crossing over during meiosis I, see diagram on
the right. This is relatively rare, happening in
the production of ~1/400 gametes.
vi) Can you determine whether these traits are
sex-linked or not based on this data? Why or why
not? No, you would need to know the sex of the
F1 and F2 flies, in addition to their phenotypes.
r
r
S
s
r white eyes
S smooth wings
Meiosis II
R
R
r
r
R Rr r
S
S
s
s
S sS s S
R R r
r
sS
s
R
metaphase I
S
R
R
S
s
R red eyes
s wrinkled wings
Gametes
03-131 Genes, Drugs and Disease
Problem Set 10 - Key
November 27, 2015
6. The inheritance pattern for Retinitis pigmentosa is shown on the
right. What is the most likely mode of inheritance? Briefly justify
your answer.
It is clearly sex-linked since only males are affected.
The question is, on what chromosome?
Y-chromosome, because it is passed from father to son. This
cannot happen with X-linked. In the case of X-linked the mother
provides the X-chromosome to the son.
Note: Since there is only one copy of the Y-chromosome, dominant and recessive do not apply because the
genotype will always be expressed.
7. Plants with red flowers are mated with plants with white flowers. Both the red and white plants are pure
breeding (homozygous). The F1 plants all have pink flowers. What is the mode of inheritance? co-dominant,
RR is red, rr is blue, and Rr is purple (= combination of red and blue, both colors are expressed in the
heterozygote).
8. Rhogam is given to Rh- mothers at childbirth to prevent complications if she has an Rh+ child in the future.
What is rhogam and why does it prevent the Rh- mother from developing antibodies against the Rh factor?
Please use the web and cite your source.
Rhogam are antibodies that recognize the Rh antigen. If they are injected at childbirth, they will bind
to any of the Rh+ red blood cells (from the baby during childbirth) that are circulating in the mother.
Since the Rh antigen is covered by the Rhogam antibody, the mother’s immune system will ignore the
Rh antigen and not make antibodies against the Rh antigen. Consequently, if she has a second Rh + child
it will be fine since the mother has not produced any Rh+ antibodies.
Due to the finite lifetime of antibodies in the blood the mother would need rhogam after every
childbirth.