March 1, 2002
IB 31 Midterm I Name:_KEY_______________________
I.
Multiple Choice (20 points).
1. Who
is most commonly thought of as a proponent
of group selection.
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A. V. Wynne-Edwards
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B. P. Grant
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C. J. Maynard Smith
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D. G. C. Williams
2.
Wallace was to Darwin as:
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A. Mendel was to Galton
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B. Maynard Smith was to Price
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C. Tolman was to Tryon
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D. Bouchard was to Galton
3. Which
is not one of Niko Tinbergen’s “Four Questions”?
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A. How does it develop?
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B. What is the survival value?
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C. How did it evolve?
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D. What will it become?
4. An
ethogram is basically
_xx__
A. A complete list and description of behaviors exhibited by a species
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B. A weighing of the importance of
various behaviors in a species.
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C. A description of the habitat in which a
species is found.
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D. A diagram of the neural circuitry
underlying a behavior.
5. A
direct measure of your evolutionary personal fitness might be:
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A. How long you live.
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B. How many children you have.
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C. How fast you can run a mile.
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D. How old you were when you had your
first child.
6. Darwin
was to the Galapagos as Wallace was to
___
A. The Canary Islands
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B. The Isthmus of Panama
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C. The Malay Archipelago
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D. The Society Islands
7. Stabilizing
selection changes
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A. The mean and the variance
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B. The mean, but not the variance
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C. The variance, but not the mean
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D. Neither the mean or the variance
Name:___________________
8. The
hypothesis that infanticide can increase and individual’s direct fitness by
providing greater opportunity to produce young is not supported by work on
which animal.
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A. Jacana
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B. Giant Water Bug
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C. Cat
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D. Lionfish
9. Two
traits do not assort independently.
This could be due to:
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A. The genes producing these traits are on
the same chromosome.
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B. The traits are pleiotropic.
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C. Neither A or B n is likely.
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D. Either A or B could be
correct.
10. A
color blind woman marries a man who can see color. Which statement is true for their son and daughter?
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A. The son will see color and
so will his sister.
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B. The son and the daughter will both be
color blind.
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C. The son will be color blind,
the daughter will see color.
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D. The son will see color, but the daughter will be color blind.
II. Fill
in the Blank (20 points):
1. DARWIN_____________________
wrote Origin of the Species.
2. EPISTASIS
(OR INTERACTION)_____________________ occurs when one gene influences the
expression of another gene.
3. DIZYGOTIC_____________________
or fraternal twins are formed from two fertilized eggs.
4. Darwin’s
reading of a book by _MALTHUS_________________ gave him the idea of the
tremendous reproductive
potential of organisms.
5. Down’s
Syndrome is due to an extra ___CHROMOSOME (21)___________________________.
6. How
many phenotypes are there for ABO human blood types? FOUR (A, B, O AND AB)_________________
7. The
allele that produces sickle cell helps protect
a heterozygous individual from MALARIA_______________.
8. The
ratio of total genetic variance to total phenotypic variance for a trait is
called HERITABILITY BROAD__________________.
9. Male
honeybees develop from a/an UNFERTILIZED (OR HAPLOID)_______________________
egg.
III.
GALTON_______________ was the founder of behavioral genetics.
Name:
__________________
III.
Obvious
Question (10 points)
Briefly outline the theory of Natural Selection
as deduced by Darwin and Wallace.
·
TREMENDOUS
REPRODUCTIVE POTENTIAL
·
POPULATIONS
REMAIN RELATIVELY STABLE
·
LOST OF
MORTALITY
·
VARIATION
IN POPULATIONS
·
SOME
VARIATION IS HERITABLE
·
ENVIRONMENTAL
SUITABILITY INCREASES THE REPRODUCTION OF SOME ANIMALS THAT ARE BETTER SUITED
TO THE ENVIRONMENT
·
THE
CONSEQUENCE IS NATURAL SELECTION (DIFFERNTIAL REPRODUTIVE SUCCESS) WHICH
RESULTS IN CHANGES IN GENE FREQUENCY LEADING TO EVOLUTION.
IV.
Define in the space provided five (5) of the following (4 points each;
total of 20 points):
1. Altruism
SAY SOMETHING ABOUT BEHAVIOR THAT OCCURS AT A
COST TO SELF BUT BENEFIT TO OTHERS. A
NICE DEFINITION WOULD BE “ACTING TO INCREASE ANOTHER INDIVIDUAL’S LIFETIME
NUMBER OF OFFSPRING AT A COST TO ONE’S OWN SURVIVAL AND REPORDUCITON … ACTING TO INCREASE ANOTHER’S DIRECT
FITNESS AT THE EXPENSE OF YOUR OWN.”
2. Simultaneous hermaphrodite
ALL INDIVIDUALS HAVE BOTH FEMALE AND MALE GONADS
(SEX ORGANS). THEY ARE FERTILE.
3. Haplo-diploid
A SEX DETERMINATION SYSTEM WHERE SEX IS
DETERMINED BY THE NUMBER OF CHROMOSOMES (INSTEAD OF X AND Y CHROMOSOMES
LIKE IN HUMANS). A MALE IS HAPLOID
(FROM UNFERTILIZED EGG) AND A FEMALE IS DIPLOID (FROM FERTILIZED EGG).
4.
ESS
A STRATEGY THAT CAN NOT BE BEATEN BY (OR INVADED
BY) ANY OTHER STRATEGY … THE BEST STRATEGY
5. Evolution
BIG E EVOLUTION – DESCENT FROM A COMMON
PRIMORIDAL ANCESTOR
LITTLE E EVOLUTION – CHANGES IN ALLELE
FREQUENCIES IN A POPULATION OVER TIME.
5.
Hardy-Weinberg
Equilibrium
A
COMPLETE ANSWER INCLUDES THE FOLLOWING:
·
REQUIREMENTS
FOR H-W EQUILIBIRUM (LIST THESE – I.E RANDOM MATING, NO GENE FLOW, NO MUTAIONS
ETC …)
·
EQUATION
P2 + 2PQ + Q2 = 1
·
MAIN
POINT: IF A POPULATION IS IN H-W EQUILIBIUM, THEN THE ALLELE FREQUENCIES ARE
CONSTANT OVER TIME (I.E. NO LITTLE E EVOLUTION)
V. In
the space provided, answer four (4)
of the following (5 points each; 20 points total):
1. List
four primary reasons that gene frequencies might change in a population.
·
GENETIC
DRIFT
·
GENE FLOW
·
MUTATION
·
NATURAL
SELECTION
2. Using
a single behavior, give examples of proximate and ultimate explanations.
MAKE THE PONT THAT PROXIMATE ANSWERS HOW AND
ULTIMATE ANSWERS WHY. YOU COULD HAVE
EXPLAINED INFANTICIDE IN LANGUR MONKEYS.
ULTIMATE: INFANTICIDE BROUGHT THE FEMALES INTO ESTRUS MORE QUICKLY;
PROXIMATE: UNFAMILLIAR ODOR OF THE CUBS INDUCES THE MALES TO DESTORY THEM
3. Given
an example of a gene-environment interaction for a behavioral trait.
EXAMPLES INCLUDED INTELLIGENCE, OR EXPLORATORY
BEHAVIOR IN MICE (SEE LECTURE NOTES ON THE ENRICHED ENVIRONMENT VS. NORMAL VS.
DEPRIVED)
4. What
is wrong with “good of the species” explanations?
INDIVIDUALS NORMALLY ACT TO INCREASE THEIR OWN
REPRODUCTIVE SUCCESS. GOS ARGUMENTS DO
NOT ACCOUNT FOR CHEATERS, AND SO A MUTANT CHEATER COULD GET AWAY WITH “SELFISH
BEHAVIOR’ AND THUS THE GOS SYSTEM IS NOT AN ESS. REMEMBER THE LEMING IN THE INNERTUBE CARTOON? THE MAIN POINT: CHEATERS WILL ‘INVADE’ THE
SYSTEM, HAVE LOST OF OFFPSRING, AND PASS THE “CHEATER GENES.”
5. Given
the following pay-off matrix for a Hawk - Dove game, is there a single
ESS? Explain.
Hawk Dove
Hawk 10 20
Dove 5 15
IF YOU CARE ABOUT THE MATH: V = 20; Cost of
Fighting = 0; Cost of Display = +10
ANALZE INVASION VERTICALLY:
WITHIN AN ALL HAWK POPULATION, DOVES DO NOT
INVADE BECAUSE 10 > 5.
WITHIN AN ALL DOVE POP, HAWKS DO INVADE BECAUSE
20> 15.
THUS, THE ESS IS HAWK LIKE BEHAVIOR.
NOTE: THESE ARE NOT ACTUAL HAWKS AND DOVES. THE HAWK IS A METAPHOR FOR “HAWK-LIKE” OR
AGGRESSIVE INDIVIDUALS, AND “DOVE” FOR INDIVIDUALS THAT ARE LESS AGGRESSIVE.
Main point:
Game Theory here allows us to ask:
If a mutant was born in a population, would it be able to do well and
reproduce, thus spreading its genes? If
yes, then the “mutant” (hawk or dove) can “invade” the population.
VI.
Quantitative Questions. Answer
one (1) of the following questions on this page. (10 points)
1. You
have available two inbred lines of unicorn to determine the heritability of
horn length in this species. Using the
data from the crosses below, calculate the heritability for this trait for
these lines. Be sure and show your
calculations.
STRAIN N MEAN VARIANCE
Bright-eyed 10 1 cm 2 cm
Bushy-tailed 50 20 cm 3 cm
F1 60 10 cm 10
cm
F2 20 11. cm 5 cm
Vtp
= Ve + Vg + V I
Ve
= 2 + 3 + 10 / 3 = 5
If we breed only the shortest horned male and
female F2’s to one another, approximately how many generations will
it take until mean horn length is equal to the Bushy-tailed parental stock?
NEVER.
YOU CAN NOT DO THIS BECAUSE HERITABILITY = 0.
2. You
decide to select for horn length in a group of unicorns taken straight from the
wild. They are not inbred. The population mean horn length is 10
cm. You breed several of the longest
horned unicorns to one another. The
mean horn length of this group of parents is 20 cm. When the young from these crosses grow up, you measure their
horns and find that they have a mean length of 15 cm. What is the heritability of horn length based on this
selection experiment? After 10
generations of selection, the mean population horn length is 30 cm. You again breed the longest horned males and
females to one another (mean length 35 cm) and their offspring have a mean horn
length of 30 cm. What is the
estimate of heritability for this generation of selection? Why might the value be changing?
FIRST
QUESTION:
S = 20 –
10 = 10
R = 10 –
15 = 5
SECOND QUESTION (for second selection expt):
H2N = 0 / 5 = 0%
THIRD QUESTION:
THE VALUE IS CHANGING BECAUSE GENETIC VARIATION IS EXHAUSTED! (YOUR GENES CAN ONLY MAKE THE HORN SO LONG …)