Genetics Practice Problems
Transcript Genetics Practice Problems
Pseudohypertropic muscular dystrophy is a disorder
that causes gradual deterioration of the
muscles. It is seen only in boys born to
apparently normal parents and usually results in
death in the early teens. (a) Is
pseudohypertrophic muscular dystrophy caused
by a dominant or recessive allele? (b) Is its
inheritance sex-linked or autosomal? (c) How do
you know? Explain why this disorder is always
seen in boys and never girls.
(a)Recessive allele because…
(a) if it were dominant then the heterozygous female
would die before she was sexually mature, and thus
could not pass on the trait
(b) Sex-linked trait because…
• it only occurs in males.
(c) It is never seen in girls because…
a) for a girl to be homozygous recessive (and express
the trait) XmXm, her father would have to be
hemizygous recessive XmY…. and her mother would
have to be heterozygous or homozygous recessive
(XmXM or XmXm).
b) It is impossible for a homozygous recessive male or a
homozygous recessive female to have children because
they will die before being old enough to have
• A phenotypic wild-type fruit fly (heterozygous
for gray body color and red eyes) was mated
with a double mutant black fruit fly with purple
• The offspring were as follows: wild-type, 721;
black-purple, 751; gray-purple, 49; black-red,
(a) What is the recombination frequency between
these genes for body color and eye color?
(b) Are these genes on the same chromosome? If
so, how far apart are they?
• b+ = gray body
• b = black body
• pr+= red eyes
• pr = purple eyes
Total flies = 1566
(a) The percent recombination is therefore 6%
• Red-green color blindness is caused by a
sex-linked recessive allele. A color-blind
man marries a woman with normal vision
whose father was color-blind. (a) What
is the probability that they will have a
Mother must be a carrier
since her father was
(Answer) Probability of having
a color blind daughter is
(1/2 x 1/2) or 1/4 or 25%
(0.25) This is because both
sex and colorblindness are
involved in the solution.
Determine the sequence of genes along
a chromosome based on the following
• A-B, 8 map units
• A-C, 28 map units
• A-D, 25 map units
• B-C , 20 map units
• B-D, 33 map units.
#3. Solution Steps for solving
Choose the greatest map distance (in this case 33) and
place the genes involved at opposite ends of a line
representing a portion of the chromosome in question
Now choose a gene combination with either B or D in it.
For example: A-D = 25 map units
Since A-B is only 8 map units, A must be
in the middle
Now choose on that has A in it :
You already have A-D and A-B, so all that is left is A-C, but where
does it go…which side?
Since the distance from A to B is about 8 and the distance from C
to B is less than the distance from A to B the C-gene must be to
the left of the B-gene.
• A man with group A blood marries a woman
with group B blood. Their child has group O
(a) What are the genotypes of these
(b) What other genotypes and in what
frequencies, would you expect in offspring
from this marriage?
Father = AO
Mother = BO
First Child = OO
b) See punnett square
• The genotype of F1 individuals in a
tetrahybrid cross is AaBbCcDd.
• Assuming independent assortment (no
linkage) of these four genes, what are the
probabilities that F2 offspring would have
the following genotypes?
AaBbCcDd x AaBbCcDd
The pedigree below traces the inheritance of a vary rare biochemical
disorder in humans. Affected individuals are indicated by filled-in circles
A) Is the allele for this disorder dominant or recessive?
B) What genotypes are possible for the individuals marked 1, 2, and 3.
a) The allele is most likely
1) #2 individual with the trait marries a
woman with the trait and 50% of their
offspring are normal. If the trait
were recessive one would expect
the following 100% affected
Bb ( if 2's genotype were bb he would
not have the disease and if BB all his
children would have the condition.)
bb (all normal individuals are
• Disorders Matching
• Self quizzes by Chapter (Campbell)
How much should I study?
“…you should plan on studying between 2 and 3 hours for every hour you spend in
lecture. Thus, for a 3-hour lecture course, you should be spending 6 to 9 hours per week
outside of class studying your notes, doing assigned readings, preparing for exams, and
so forth. For a schedule with 12 credit hours, that means 24 to 36 additional hours of
out-of-class effort to adequately understand the material.”
–”Study Skills” Jerry A. Waldvogel, Clemson University
This class is 7.5 hrs a week….which means 15 to 23 hrs per week [about 3 hrs per day]