Biology 321 QUIZ#1
Spring 2011 25 pts.
NAME______________________________
1. (3 pts.) The hormone gibberellin is a major determinant of plant height and dwarfism can be
caused by a failure to produce or to respond properly to this compound. Many different gene
products are required for the biosynthesis of gibberellin and for the proper cellular response to its
presence. Given this information do you think that dwarfism in plants can be inherited as a
single gene trait? Circle: YES Then support your answer with a 1-2 sentence explanation.
No credit if no explanation.
Student Answers: If a mutation in one gene inhibits the entire pathway needed for the synthesis
of gibberellin, then it can be a single-gene trait. (One mutation causes a phenotypic change).
If one gene product in the sequence is disrupted it could cause dwarfism. For example, the
biosynthesis of gibberellin may function properly, but the cellular response may be incorrect.
Therefore that single gene for cellular response could cause dwarfism.
2. (5 pts.) American elms (Ulmus americana) lined many city streets before Dutch elm disease
killed most of them in the mid 20th century. It has been accepted for many years that American
elms are tetraploids (4n=56), but there have been persistent rumors of elm trees that were
triploids or diploids. In the 1990s it was shown that a resistant strain of elms found in
Washington DC was 3n = 42. More recently diploid elm trees (2n=28) have been found in the
wild growing alongside the tetraploids.
Circle True or False for each statement. Answer false any part of the statement is false. If
there are two statements, the first statement is true and you are to decide if the second statement
is T or F.
F The information given above implies that the elm strains differ with respect to the number of
different chromosomes (double-stranded DNA molecules) per genome copy.
F Each product of meiosis I in the tetraploid strain would have 14 chromosomes and 28
chromatids
T Each product of meiosis II in the diploid strain would have 14 chromosomes and 14
double-stranded DNA molecules.
T
A commercially valuable fungal-resistant strain called Jefferson has been derived from the
3n strain. Asexual propagation of this strain via mitosis does not present a problem even though
the uneven number of homologs means that some chromosomes will not have a pairing partner.
T Although there is no direct evidence, a wild diploid strain may have may have been the
original source of the disease resistance seen in the triploid stock. This speculation makes some
sense since an accidental cross of a 4n elm with a 2n elm would produce a triploid offspring.
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3. (9 pts) You are exploring the inheritance of two traits in summer squash. Different parental
plants (all of unknown ancestry) were used in each cross.
Trait
fruit color
fruit morphology (shape)
Alternative phenotypes
green or yellow
disk or sphere
Number of progeny in each class
Parental phenotypes
1
yellow disk
y
y
d
c c m m
X
green
sphere
green
disk
0
0
d
yellow
sphere
yellow
disk
500
500
y
green sphere
w
c c
ms md
cy cw
md md
300
100
cw cw m s m d
2
yellow sphere X
y
w
s
c c mm
d
green sphere
300
w
100
w
w
c c
ms m-
w
c c
md md
y
w
c c
ms m-
cy cw
md md
cw cw m s m d
a. Which color form is dominant? __yellow_____
For part b, fill in the blank from the choices below. No explanation needed.
b. Dominance for this trait could be determined from cross 1 only
cross 2 only either cross
c. Which fruit shape is dominant ? ____sphere__________
d. Dominance for this trait could be determined from cross 2 only
cross 1 only cross 2 only either cross
d. For each cross indicate the genotypes of the parents and progeny in the space underneath
the phenotype. Use these allele designations OR define your own alleles symbols below.
cy= yellow
cw = green
md = disk
ms = sphere
OPTIONAL: Use your own alleles symbols. Define them here:
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4. (4 pts.) Examine the diagrams shown below which were taken from one your assigned
problems in Chapter 3. Each line represents a single molecule of double-stranded DNA, which
will segregate as indicated.
A. 3 pts: +1pt for each correct answer; -1 pt for each incorrect answer
Diagram #1 (below). Choose all correct interpretations.
a) This drawing shows a 1n=2 cell undergoing mitosis
b) This drawing shows a 2n=2 cell undergoing mitosis
c) In meiosis II in an organism of genotype AABB, all MII divisions will look like this
d) In meiosis II in an organism of genotype AaBB, ½ of MII divisions will look like this
e) In meiosis II in an organism of genotype AaBb, ½ of MII divisions will look like this
Ó
Ô
B. 1 pt Examine Diagram #2 (below). This drawing is consistent with which of the scenarios
listed below? Choose all correct interpretations.
a) Mitosis in a 2n=4 cell of genotype AaBb
b) Mitosis in a 4n=8 organism of genotype AaBb
c) Meiosis I in a 2n=4 organism of genotype AaBb
d) Meiosis I in a 4n=8 organism of genotype AaBb
e) None of these interpretations are correct
Ó
Ô
5. (4 pts.) a. A friend of yours needs to determine the genotype (AA or Aa) of a plant that
shows the dominant phenotype for a single-gene trait. He allows the plant to self and looks at
10 offspring. If the plant is heterozygous (Aa), what is the probability that all 10 progeny show
the dominant phenotype? Show your work and circle your answer. Just set the answer up.
Aa X Aa à ¾ chance dominant phenotype vs ¼ chance recessive
Probability that all 10 offspring show the dominant phenotype is [3/4]10
b. You argue that a test cross would be better at revealing the presence of a recessive allele
among a small number of progeny. Briefly but explicitly explain your argument. Be sure to
state what a test cross is.
Test cross = Aa X aa (tester) This cross is more likely to reveal the presence of a recessive
allele in any given offspring since the probability of seeing a recessive offspring is ½ versus the
¼ probability of seeing a recessive offspring in the self progeny of an Aa plant. The probability
that an Aa X aa cross will reveal only dominant phenotypes among 10 offspring is [½]10 -- much
lower than [3/4]10
NOTE: Many of you indicated that the probability of seeing a recessive
offspring in an Aa X aa cross was ½, but didn’t state that this probability is higher that what
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you would expect from a self-cross. On future quizzes and exams, but sure to make these
connections explicitly.
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Page (pts)
1 (8 )
Score
2 (9)
3 (8 )
Total (25)
5