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.

 

                  xx___ A. V. Wynne-Edwards

                  ___ B.  P. Grant

                  ___ C. J. Maynard Smith

                  ___ D. G. C. Williams

 

2.  Wallace was to Darwin as:

 

                  ___ A.             Mendel was to Galton

                  xx___ B.          Maynard Smith was to Price

                  ___ C. Tolman was to Tryon

                  ___ D. Bouchard was to Galton

 

3.  Which is not one of Niko Tinbergen’s “Four Questions”?

 

                  ___ A.             How does it develop?

                  ___ B.  What is the survival value?

                  ___ C. How did it evolve?

                  xX___ D.         What will it become?

 

4.               An ethogram is basically

                  _xx__ A.          A  complete list and description of behaviors exhibited by a species

                  ___ B.  A weighing of the importance of various behaviors in a species.

                  ___ C. A description of the habitat in which a species is found.

                  ___ D. A diagram of the neural circuitry underlying a behavior.

 

5.  A direct measure of your evolutionary personal fitness might be:

 

                  ___ A. How long you live.

                  xx___ B.          How many children you have.

                  ___ C. How fast you can run a mile.

                  ___ 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

                  ___ B.  The Isthmus of Panama

                  xx___ C.          The Malay Archipelago

                  ___ D. The Society Islands

 

7.               Stabilizing selection changes

 

                  ___ A.             The mean and the variance

                  ___ B.  The mean, but not the variance

                  xx___ C.          The variance, but not the mean

                  ___ 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.

 

                  ___ A. Jacana

                  ___ B.  Giant Water Bug

                  ___ C. Cat

                  xx___ D.          Lionfish

 

9.  Two traits do not assort independently.  This could be due to:

                  ___ A. The genes producing these traits are on the same chromosome.

                  ___ B.  The traits are pleiotropic.

                  ___ C. Neither  A or B n is likely.

                  xx___ 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?

 

                  ___ A.             The son will see color and so will his sister.

                  ___ B.  The son and the daughter will both be color blind.

                  xx___ C.          The son will be color blind, the daughter will see color.

                  ___ 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

 

H2B = Vg / Vtp = 0/5 = 0 %

                  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

H2 n = r / s  = 5 / 10 = 50%

 

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 …)