BIO240
Exam #2 Answer Key
Professor Jill Carroll
35. Explain how concordance studies in twin pairs are used to determine if a trait has a genetic component, and
give an example.
Concordance in twin pairs measures the number in which both members of a twin pair share a trait. For
example, concordance of 34%for heart disease means that out of 100 twin pairs in which one twin has
heart disease, 34% of the time, their twin will also have heart disease. Monozygotic twins are identical,
so any trait that is SOLELY genetic will have a concordance of 100%. If there is higher concordance in
monozygotic twins vs dizygotic twins (who only share about 50% of alleles), the trait must have a genetic
component.
Any number of hypothetical examples could be used:
Obesity
Monozygotic concordance = 67%
Dizygotic concordance = 46%
This indicates that genetics must play a role in obesity, but is not the only factor.
36. Joe is color blind. Both his mother and father have normal vision, but his mother’s father (Joe’s maternal
grandfather) is color blind. All Joe’s other grandparents have normal color vision (all grandparents are still
living). Joe has three older sisters – Patty, Betsy, and Lora – all with normal color vision. Joe’s oldest
sister, Patty, is married to a man with normal color vision; they have two children, a 9-year-old color blind
boy and a 4-year-old girl with normal vision.
a. Using standard symbols and labels, draw a pedigree of Joe’s family
(NOTE: you could have formatted this slightly differently – listed Joe’s mother before the father,
simplified Joe’s two sisters with a single circle with a “2”, etc)
b. What is the most likely mode of inheritance for color blindness in Joe’s family?
X-linked recessive. Only males have the trait, and they inherit the trait from their mothers, who
are carriers. It is unlikely to be an autosomal recessive trait, because we would not expect two
unrelated males marrying into the pedigree (Joe’s father and Patty’s husband) to both be carriers
for a relatively rare trait.
c. If Joe marries a woman who has no family history of color blindness, what is the probability that
their first child will be a color blind boy?
0% since Joe cannot pass his color-blind X to his son, and there is no indication that his wife is a
carrier.
d. If Patty and her husband have another child, what is the probability that the child will be a color
blind boy?
XN = normal vision
Xn = colorblind
Patty must be a carrier (XNXn) since she has an affected son and her husband is XNY. There is a
25% chance that there next child will be a color blind boy.
XN
Y
XN
Xn
XNXN
Normal
girl
XNY
Normal
boy
XNXn
Normal
girl
XnY
Colorblind
boy
(NOTE – the question did NOT ask what the probability of a son being colorblind is – which
would have been 50%)
37. You are doing lab work with a new species of beetle. You have isolated lines that breed true for either blue
shells and long antenna, or green shells and short antenna. Crossing these lines yields F1 progeny with
blue shells and long antenna. Crossing F1 progeny with beetles that have green shells and short antenna
yield the following progeny:
blue shell, long antenna
green shell, short antenna
blue shell, short antenna
green shell, long antenna
37
43
82
78
a. Which shell color and antenna length alleles are dominant?
Crossing true breeding (which means homozygous) blue shell and long antenna with a green shell
and short antenna yields 100% F1 with blue shells and long antenna (as stated in the problem).
Therefore, blue shell is dominant over green, and long antenna is dominant over short.
B = blue b = green
L = long
l = short
b. What would the expected numbers of the different phenotypes be for the F2 generation if the two genes
assorted independently?
If the two genes assort independently, the F1 individual would be able to produce 4 different gametes
with equal probability. The individual that has a green shell and short antenna is capable of producing
only one type of gamete – always giving recessive alleles for both traits.
25% blue long
25% blue short
25% green long
25% green short
bl
BL
Bl
bL
bl
BbLl
Blue
Long
Bbll
Blue
short
bbLl
green
Long
bbll
green
short
Since there was a total of 240 offspring, you’d expect 60 of each phenotype.
c. If the genes are linked, calculate the recombination frequency between them. Is the F1 individual is cis or
trans configuration?
The progeny numbers do not exhibit roughly equal amounts of all 4 different phenotypes, which does
indicate that the two genes are linked. The most common phenotypes have one dominant trait and one
recessive trait (blue short = 82 and green long = 78). This means that the F1 heterozygote must
be in the trans configuration, since it gave one dominant allele and one recessive allele together.
Recombinant offspring (blue long and green short ) were created when crossing over rearranged the
alleles.
Recombination frequency = (37 + 43)/(82 + 78 + 37 + 43) = 80/240 = 33.3%
38. An individual has the following genotype. Gene loci (A) and (B) are 15 cM apart. Indicate all the possible
gametes that this individual can produce, and the proportions of expected progeny genotypes if a testcross
is performed on this individual.
A
b
a
B
nonrecombinant gametes would be A_______b, and a_______B and recombinant gametes (due to
crossing over rearranging them) would be A_______B and a_______b
The recombination rate between these two genes is 15% - so recombinant gametes would be expected to
be 7.5% each, which then leaves the nonrecombinant gametes with a probability of 42.5% each.
A testcross would be to cross this double heterozygote with an individual that is recessive for both traits.
AaBb x aabb
Offsring genotypes
A_______b
_________
a
b
42.5%
a_______B
_________
a
b
42.5%
A_______B
_________
a
b
7.5%
a_______b
_________
a
b
7.5%
39. Define the following: transformation, transduction, and conjugation.
Transformation is the ability of bacteria to take up foreign DNA from its environment, and incorporate
into its own genome.
Transduction is a virus introducing bacterial DNA into a bacterium. During the lytic cycle, fragmented
bacterial DNA can enter a viral protein coat. When the viral particle goes to “infect” another cell, it has
injected bacterial, not viral genetic information.
Conjugation is a direct transfer of genetic material from one bacterium to another via a conjugation
bridge – a cytoplasmic connection in which DNA directly goes from the cytoplasm of one bacterium into
the cytoplasm of another bacterium.
After any of these, crossing over can occur within homologous regions, and recombination of alleles
occurs.
40. Contrast the lytic and lysogenic cycles, and how this leads to either generalized or specialized
transduction.
Lytic cycle – a viral particle injects its genetic material into host cell. The host cell’s DNA becomes
fragmented. Viral genetic information is replicated, protein coats are made, new viral particles are
assembled. The host cell then bursts/lyses – releasing viral particles.
Lysogenic cycle – similar to lytic cycle, a viral particle injects its genetic material into host cell. However,
the host cell’s DNA does NOT get degraded. Viral DNA(a prophage) incorporates into host cell, being
replicated every time the cell divides. At some point, the prophage may splice back out, and cell will
enter the lytic cycle.
Generalized transduction occurs when bacterial DNA enters a protein coat – ALL bacterial genes have
an equal probability of this occurring. This occurs with the lytic cycle, since the bacterial genome is
fragmented, any fragment has the potential of entering a viral protein coat.
Specialized transduction occurs in the lysogenic cycle. When the prophage splices back out of bacterial
DNA, it may carry a bacterial gene/s with it. Since the prophage can insert at only particular points of
the bacterial DNA molecule, bacterial genes closest to this region have a greater chance of being carried
out by the prophage compared to those that are further away from the insertion locations.
41. A female rat that is heterozygous for an autosomal reciprocal translocation has 36 eggs that were generated
from the following 9 meioses: 3 by alternate segregation, 3 by adjacent-1 segregation, and 3 by adjacent-2
segregation. She is mated to a chromosomally normal male. What is the probability that her surviving
offspring will inherit a chromosome bearing a translocation?
All gametes produced by adjacent-1 segregation and adjacent-2 segregation are inviable due to
containing unbalanced genetic information. In meiosis I of alternate segregation, one cell gets both
normal chromosomes while the second cell gets both chromosomes that contain the translocation. Since
these are the only two options that will yield actual offspring, the probability that her translocation will
be inherited is 50%.
42. Two nonhomologous chromosomes have the following segments:
A •B C D E F G
R •S T U V W X
Draw chromosomes that would result from the following chromosome rearrangements.
a. displaced inverted duplication of EFG
G F E A•B C D E F G
(NOTE: this is just one example – there are several ways to show this, as long as the duplicated
sequence is backwards and not directly after the original sequence)
b. reciprocal translocation of CD and W
A •B W E F G
R •S T U V C D X
c. Robertsonian translocation
XWVUTS•BCDEFG
A • R (may or may not have a centromere – tends to get ‘lost’ during cell division)
43. In children that express recessive genetic diseases, such as cystic fibrosis, it has been occasionally found
that only one parent is heterozygous for the disease allele. In these cases, it has also been confirmed that the
parents are the true biological parents of the affected child. Offer a genetic explanation for this observation.
One parent is heterozygous (Aa) while the second parent is homozygous ‘normal’ (AA).
(NOTE – there is more than one correct answer for this question):
ONE explanation to why they had an affected child (aa) is uniparental disomy. Both chromosomes (each
of which was ‘a’) came from the heterozygous parent. This usually orginates as a trisomy – two
chromosomes from the heterozgote, and one chromosome from the other parent. One of the three
chromosomes gets ‘lost’ in later cell divisions. In this case, the child would have started as ‘Aaa’ and
then lost the ‘A’, leaving the ‘aa’ affected genotype.
A SECOND explanation is pseudominance. The child should have been ‘Aa’ but due to a microdeletion,
the ‘A’ allele was lost. This leaves only one copy of the gene. Since the copy that is left is the recessive
disease allele, that is the only phenotype that can be expressed.