BIO240 Exam #1 answer key
Fill-in-the-blank (2 pts each – 18 points total)
24. Bacteria reproduce by the process of binary fission .
25. Crossing over occurs during Prophase I of meiosis.
26. In plants, a sporophyte produces spores by the process of meiosis.
27. Nondisjunction is the abnormal segregation of chromosomes during mitosis or meiosis.
28. In which sex determination system is the female the heterogametic sex? ZZ/ZW.
29. A fruit fly with an X:A ratio of 0.5 would be which sex? male
30. A woman has type A- blood, Her first child has O+ and her second child has B-. What is her husband’s
bloodtype? (assume he is the father of both children) B+
31. Only 38 out of 50 individuals with the same genotype display the expected phenotype. This trait exhibits
76% penetrance.
32. Draw the symbol used for an affected pair of female identical twins. (see powerpoint)
Short answer (6 pts each – 54 points total)
33. According to this course’s syllabus, explain the difference between attendance and participation.
The Attendance Policy follows the college guidelines. Attendance is worth a total of 50 points – these
points can be lowered due to missed classes, as well as lateness. A pattern of late arrivals and absence
will result in a grade no higher than a C for the course. Participation is engagement with the day’s
activities – contributions to discussions, group work, and laboratory activities. Participation is also worth
50 points – merely showing up to class will get you zero participation points.
34. In a short paragraph, contrast Meiosis I and Meiosis II.
The same events occur in both stages – Prophase, Metaphase, Anaphase, and Telophase. In Prophase I,
homologous chromosomes pair, and crossing over occurs. In Metaphase I, homologous pairs align at
equatorial plate, and pairs separate during Anaphase I. In Meiosis II, there are no homologous pairs
since the cells at that point are haploid. Chromosomes line up individuals during Metaphase II, and
Anaphase II separates sister chromatids into individual chromosomes (events are Meiosis II are similar
to those of mitosis)
35. Contrast each of the following pairs of concepts:
a. variable expressivity vs. incomplete penetrance
Variable expressivity indicates that a single genotype can yield a range of phenotypes, from mild
to severe. For example, if CB = black, and CW = white, CBCW individuals can exhibit colors from
pale gray to almost black.
Incomplete penetrance indicates than less than 100% of individuals with the same genotype will
express the expected phenotype. For example: even though polydactyly is a dominant trait, some
heterozygous individuals will have the normal number of digits. Their dominant allele, however,
can be expressed in their offspring.
b. maternal effect vs. cytoplasmic inheritance
Genetic maternal effects are autosomal genes in which both parents each contribute an allele.
However, an individual’s genotype does not dictate their own phenotype. Instead, an individual’s
phenotype is determined by their mother’s GENOTYPE (which is responsible for molecules,
enzymes, etc that are present in the egg).
Cytoplasmic inheritance deals with the DNA within mitochondria and chloroplasts. Since both
are only present in the egg (sperm does not contribute either), cytoplasmic inheritance is 100%
maternally derived. A single cell can have hundreds to thousands of these DNA molecules, and
they may be different from one another, leading to heteroplasmic conditions. This varying ratio
of normal to mutant DNA can yield a wide range of phenotypic expressions.
36. In cucumbers, dull fruit (D) is dominant over glossy fruit (d); orange fruit (R) is dominant over cream fruit
(r); bitter cotyledons (B) are dominant over non-bitter cotyledons. The three characters are encoded by
genes located on different pairs of chromosomes. A DdRRbb individual is crossed with a DdrrBb
individual. Give the expected proportions of phenotypes among the progeny of this cross. (show your work
either by Punnett square or branch diagram)
Punnett square
DrB
Drb
drB
drb
DRb
dRb
DDRrBb
Dull,
orange,
bitter
DDRrbb
Dull,
orange,
non-bitter
DdRrBb
Dull,
orange,
bitter
DdRrbb
Dull,
orange,
non-bitter
DdRrBb
Dull,
orange,
bitter
DdRrbb
Dull,
orange,
non-bitter
ddRrBb
glossy,
orange,
bitter
ddRrbb
glossy,
orange,
non-bitter
Branch
¾ D_
¼ dd
½ B_ =
3/8 D_R_B_ dull, orange, bitter
½ bb
3/8 D_R_bb dull, orange, non-bitter
1 R_
=
½ B_ =
1/8 ddR_B_ glossy, orange, bitter
½ bb
1/8 ddR_bb glossy, orange, non-bitter
1 R_
=
37. Some flies have white eyes and others red eyes. The difference in eye color is due to inheritance of one
gene. Two true-breeding flies are mated, with the female white-eyed and the male red-eyed. The F1 are ½
red eyed and female, and ½ white-eyed and male. (There are no white-eyed females, or red-eyed males.)
The F2 includes:
Phenotypes
53 red-eye females
45 white-eyed females
42 red-eyed males
60 white-eyed males
Observed
Number (o)
Expected
Number (e)
d
(o – e)
d2
d2 / e
53
50
3
9
0.18
45
50
-5
25
0.5
42
50
-8
64
1.28
60
200
50
10
100
2.0
3.96
Red-eye
female
White-eye
female
Red-eye male
White-eye
male
Total
2 = _3.96_______
Degrees of freedom (df) = __3____
_0.5___ > P > _0.1___
a. Complete the 2 table.
Based on the information – there are no white eyed females and no red eyed males in F1. Any
differences seen between the sexes such as these indicate that this is an X-linked trait. Let XR = red and
XW = white. The F1 females are XRXW, and the F1 males are XWY. Such a cross will yield the following
expected ratio for the F2 generation: 1 red eyed female:1 white eyed female: 1 red eyed male: 1 white
eyed male.
XW
Y
XR
XRXW
Red female
XRY
Red male
XW
XWXW
Whitefemale
XWY
White male
(For the exam, you would be provided with a copy of the 2 probability table.)
b. Write the null hypothesis (H0).
The is no difference between the observed ratio and the expected ratio of 1 red eyed female: 1 white
eyed female: 1 red eyed male; 1 white eyed male.
NOTE: The null hypothesis is NOT that observed is due to random chance (this is a genetics class,
afterall) – it is that the DIFFERENCE between observed and expected values is due to chance.
c. Briefly interpret of the results of the 2 test. In your answer, be certain that you identify the mode of
inheritance for the white eye color of flies.
The null hypothesis can be accepted since the P value falls between 0.1 and 0.5. This means that
there is a 10%-50% probability differences observed is just due to random chance. White eyes is an X
-linked recessive trait, with red eyes being the dominant form.
38. In guinea pigs, short hair (L) is dominant to long hair (l), and the heterozygous conditions of yellow coat
(CY) and white coat (CW) give cream coat. A short-haired, cream guinea pig is bred to a long-haired, hit
guinea pig, and a long-haired, cream baby guinea pig is produced. When the baby grows up, it is bred back
to the short-haired, cream parent. What phenotypic classes and in what proportions are expected among the
offspring?
Parental phenotypes
Parental genotypes
short-haired cream
LlCYCW
(must be hetero. Ll due to
Baby being ll)
long-haired white
llCWCW
Baby long-haired cream llCYCW
LCY
lCY
lCW
LCW
LlCYCY LlCYCW
Short
Short
yellow
cream
LlCYCW LlCWCW
Short
Short
cream
white
lCY
lCW
llCYCY
long
yellow
llCYCW
long
cream
llCYCW
long
cream
llCWCW
long
white
Branch diagram
½ Ll
½ ll
¼ C YC Y
1/8 short yellow
½ CYCW
1/4 short cream
¼ CWCW
1/8 short white
¼ C YC Y
1/8 long yellow
½ CYCW
¼ long cream
¼ CWCW
1/8 long white
39. A geneticist is examining a culture of fruit flies and discovers a single female with strange spots on her
legs. The new mutation is named melanotic. When a female melanotic fly is crossed with a normal male,
the following progeny are produced: 123 normal females, 125 melanotic females, and 124 normal males.
In subsequent crosses with normal males, melanotic females are frequently obtained, but never any
melanotic males. Provide a possible explanation for the inheritance of the melanotic mutation. (Hint: The
cross produces twice as many female progeny as male progeny.)
Any time that there is a difference in numbers of a particular phenotype between the sexes, this indicates
that the trait is sex-linked. In this instance, the only sex that shows the melanotic phenotype in the F1 is
female. In addition, there is a large difference in AMOUNT of sexes produced in the F1 generation –
roughly a 2 female: 1 male ratio. Since the ratios are thirds (normal female, normal male, melanotic
female) instead of quarters, the presence of a lethal allele is indicated. This implies that the melanotic
allele is an X-linked dominant allele that is lethal in either the homozygous or hemizygous (in males
)condition. Therefore, any living melanotic female MUST be heterozygous.
XM = melanotic
Xm = normal
XMXm (melanotic female) x XmY (normal male)
XM
Xm
Y
XMXm
Melanotic
female
XMY
Melanotic
male LETHAL
Xm
XmXm
Normal
female
XmY
Normal male
40. Shell coiling of the snail Limenaea peregra results from a genetic maternal effect. An autosomal allele for a
right-handed shell (s+), called dextral, is dominant over the left-handed shell (s), called sinistral. A pet snail
called Martha is sinistral and reproduces only as a female (the snails are hermaphroditic). Indicate which of
the following statements are true, and which are false. Explain your reasoning in each case.
a. Martha’s genotype must be ss.
b. Martha’s genotype cannot be s+s+.
c. All of the offspring produced by Martha must be sinistral.
d. At least some of the offspring produced by Martha must be sinistral.
e. Martha’s mother must have been sinistral.
f. All of Martha’s brothers must be sinsitral.
In genetic maternal effect, an individual’s phenotype is determined by their mother’s GENOTYPE.
Since Martha is sinistral, we know that her mother’s GENOTYPE is ss.
a. False – Martha could be s+s OR ss, since we know nothing about her father’s genotype.
b. True – We know that Martha’s mother is ss, so she would have passed an “s” to Martha. She could
be either heterozygous or homozygous ss, depending on the allele from her father.
c. False – if Martha is heterozygous (s+s) all of her offspring would be dextral.
d. False – (same as c.) if Martha is heterozygous, all of her offspring would be dextral.
e. False – Martha’s mother was a sinistral GENOTYPE, but may have had dextral coiling due to HER
mother’s genotype.
f. True – all offspring are determined by mother’s genotype. Since Martha’s mother has an “ss”
genotype, all offspring, regardless of sex, would be sinistral.
41. 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%)