Biology 115
Name___________________
Exam I
Discussion Section________
Spring, 2003
Circle the most correct answer: (80 pts)
- In
watermelons, bitter fruit (B) is dominant over sweet fruit (b),
and yellow spots (S) are dominant over no spots(s). The genes for these two characteristics
assort independently. A homozygous
plant that has bitter fruit and yellow spots is crossed to a homozygous
plant that has sweet fruit and no spots.
If an F1 progeny plant from this cross is backcrossed to
the sweet, nonspotted parent, what phenotypic ratio is expected in
the offspring?
a.
3:1 d. 1:1
b.
1:1:1:1 e. 2:1:1
c.
9:3:3:1
- A
woman has blood type A and is also type M. She has a child with blood type AB MN. Which of the following men might
be the father of this child?
a.
George O MN
b. Henry AB M
c. Claude A N
d. Bill B MN
e. Tom O N
- The
type of inheritance shown in the pedigree below is most likely: (This is a real pedigree of a family in
Sao Paulo, Brazil, where a condition called ectodactyly was segregating.)
- autosomal
dominant
- autosomal
recessive
- sex-linked
(X-linked) recessive
- sex-linked
(X-linked) dominant
- Y-linked
- A panel
of cell lines was created from mouse-human somatic cell fusions. Each hybrid line was examined for the
presence of human chromosomes and for the production of enzyme X.
On the
basis of these results, which chromosome has the gene that codes for
enzyme X?
a.
2
b.
4
c.
6
d.
8
e.
none of the above is correct
- Huntington
disease (HD) is a rare autosomal dominant condition in humans that is
usually not expressed until an individual reaches middle age. HD is a neurological disorder that ultimately
is lethal and practically all affected individuals are heterozygous for
the condition. A woman, who
eventually expresses the symptoms of HD, has a son. He marries, has a daughter, and is
killed in an auto accident at the age of 26. His daughter eventually marries and has a son. What is the probability that this son
carries the HD allele? (Assume all
those who marry into this family are normal.)
- 1/2
- 0
- 1/4
- 1/8
- 1/16
- In
summer squash, a cross of a white fruit, disk-shaped fruit plant with a
yellow fruit, disk-shaped fruit plant resulted in offspring of 3/8 white,
disk-shaped fruit plants, 3/8 yellow, disk-shaped fruit plants, 1/8
yellow, sphere-shaped fruit plants, and 1/8 white, sphere-shaped fruit
plants. The dominant alleles for
white fruit and disk-shaped fruit can be represented by W and D,
respectively while the recessive alleles for yellow fruit and
sphere-shaped fruit can be represented by w and d,
respectively. What are the
genotypes of the parents in this cross?
- Ww
dd x Ww Dd
- Ww
Dd x ww Dd
- Ww
Dd x ww dd
- WW
Dd x ww Dd
- Ww
DD x ww Dd
- In
humans, red-green colorblindness is caused by a sex-linked recessive
allele. A woman has normal vision
but her father is colorblind. She marries
a man with normal vision but whose own father was colorblind. What is the probability that their
first child will be colorblind?
- 0
- 1/8
- 1/4
- 1/2
- 1/16
- In Drosophila,
a recessive autosomal allele, ebony (e), produces a black body
color when homozygous and at an independently assorting autosomal locus,
the recessive allele, black (b), also produces a black body color
when homozygous. Flies that are ee
b+_, e+_ bb, or ee bb
are phenotypically identical and have black bodies. Flies of e+_b+_
genotype have the wild-type grey body color. Ebony flies of eeb+b+
genotype are crossed to black e+e+bb
flies. The F1 flies are
then allowed to intercross with each other to produce an F2. What will be the expected phenotypic
ratio in the F2?
- 13:3
- 9:3:3:1
- 9:4:3
- 12:3:1
- 9:7
- Two
genes, a and b, are located 6 map units apart on the genetic
map of Drosophila. If one
is able to look at 100 meiotic events in Drosophila, in how many of
these 100 meiotic events should one expect to find a crossover event
(chiasma) occurring between a and b?
- 6
- 3
- 12
- 0
- at
least 50
- The
trait in the pedigree is for a relatively common genetic condition. Based upon the most likely mode of
inheritance of this trait, what is the probability that a child of the first
cousin marriage (III-2 and III-3) will have this trait?
- 1/4
- 1/2
- 1/9
- 1/12
- 1/6
- A
plant of AB/ab genotype is testcrossed by crossing it to an ab/ab
plant. If the two loci are 10 map
units apart, what percentage of the progeny are expected to be AB/ab?
- 10%
- 20%
- 45%
- 5%
- 90%
- You
have been given a virgin Drosophila female that has bristles on her
thorax that are much shorter than the normal wild-type. She is mated with a wild-type male
(with long bristles) and the following F1 progeny are
obtained: 1/3 short-bristled
females, 1/3 long-bristled females, and 1/3 long-bristled males. A cross of the F1
long-bristled females with their brothers gave only long-bristled F2
flies of both sexes. A cross of F1
short-bristled females with their brothers gave 1/3 short-bristled
females, 1/3 long-bristled females, and 1/3 long-bristled males. The best hypothesis to account for
these results is:
- A
mutant sex-linked allele is present that causes short bristles when
heterozygous with the wild-type allele but is lethal when hemizygous or
homozygous.
- Two
different gene pairs that affect bristle length are segregating, one is
autosomal and the other is sex-linked.
- Two
autosomals gene pairs are segregating that affect bristle length. One gene locus has a dominant allele
that suppresses the activity of the normal allele at the second locus.
- One
autosomal gene locus is involved.
It contains an allele that causes short bristles in females when
heterozygous and short bristles in males when homozygous.
- Short
bristles are caused by a mutant allele carried on the Y chromosome.
- An
organism is diploid with a XXY sex chromosome complement. It could be a:
- male
fruit fly or male human
- male
fruit fly or female human
- female
fruit fly or male human
- female
fruit fly or female human
- intersex
fruit fly or male human
- In
humans the diploid number of chromosomes is 46. How many chromosomes (as opposed to chromatids) are there in
metaphase II of meiosis in humans?
- 2
- 4
- 46
- 92
- 23
- Genes A,
B, and C are independently assorting and control production
of a black pigment through the production of intermediate molecules in a
certain snail species. Assume that
A, B, and C act in a pathway as follows:
A B C
colorless à colorless àcolorless
à
black pigment
The
alternative alleles that give abnormal functioning of these genes are
designated a, b, and c, respectively. A black AABBCC male is crossed to a
colorless aabbcc female to give a black F1. The F1 are crossed with each
other. If colorless and black are the only
phenotypes observed, what proportion of the F2 are colorless?
a.
1/64
b.
37/64
c.
9/64
d.
27/64
e.
3/16
- In Drosophila
the recessive alleles for brown (bw) and scarlet eyes (st)
represent two independently assorting autosomal genes. These genes interact so that the bwbw
stst genotype gives a white eye color. If a bwbw st+st+ fly is
crossed to a bw+bw+ stst fly, what
proportion of the F2 will have white eyes?
- 13/16
- 7/16
- 1/4
- 1/16
- 3/4
- Indicate
whether each of the following statements is true (T) or false (F). (10 pts.)
_____ a. Homologous chromosomes separate from each other during meiosis I.
_____ b. A two-strand double crossover between two genes will lead to two
recombinant chromatids.
_____ c. One would use two degress of freedom when testing the
acceptability of data fitting a 13:3 ratio using the chi-square test.
_____d. The genetic distance between two genes located far apart on a
chromosome will be more accurate than the genetic distance between two genes
located close together on the same chromosome because double crossovers do not
occur between genes that are close to each other.
_____ e. Segregation of the two alleles in a heterozygous individual
usually occurs during anaphase II of meiosis?
(Assume there is no crossing over between the gene and the centromere.)
- You
are told that in a family with two children, one of them is a girl. What
is the probability that the other is also a girl? (This is actually a
difficult question and the correct answer is not the obvious answer. Do
this one last.) (2 pts)
- In Drosophila,
cherub wings (ch), black body (b), and cinnabar eyes (cn)
result from recessive alleles that are all located on the same autosomal
chromosome. A homozygous wild-type fly is crossed with a cherub, black,
and cinnabar fly, and the resulting F1 females testcrossed with
cherub, black, and cinnabar males. The following progeny were produced
from this testcross:
ch b+ cn 105
ch+ b+
cn+ 750
ch+ b cn 40
ch+ b+ cn 4
ch b cn 753
ch b+ cn+ 41
ch+ b cn+ 102
ch b cn+ 5
_________
Total 1800
a.
Determine the linear order of the genes on the chromosome
(which gene is
in the middle). (2 pts)
b.
Calculate the map distances between the three loci. (4 pts)
c.
Determine the coefficient of coincidence and the interference
for the three loci. (2 pts)