## Ap Genetics Solutions

Problem 1
A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15 chicks are gray, 6 are black, and 8 are white.

What is the simplest explanation for the inheritance of these colors in chickens?

Incomplete or codominance. Feather color is controlled by 2 genes B = black and b = white. The third phenotype is the result of a 50-50 mix of black and white to produce gray.

The 15 gray, 6 black, and 8 white birds represent a 2:1:1 ratio&emdash;the result of mating two heterozygous individuals: (Bb x Bb)

1 BB : 2 Bb : 1 bb

What offspring would you predict from the mating of a gray rooster and a black hen?

A gray rooster (Bb ) mated to a black hen (BB ) can be represented by the following Punnett square:

50% of the offspring should be gray (Bb ) and 50% black (BB )

Problem 2
In some plants, a true-breeding, red-flowered strain gives all pink flowers when crossed with a white-flowered strain: RR (red) x rr (white) —> Rr (pink).

If flower position (axial or terminal) is inherited as it is in peas what will be the ratios of genotypes and phenotypes of the generation resulting from the following cross: axial-red (true-breeding) x terminal-white?

Note: Axial (A ) is dominant over terminal (a ).

The genotypes of the parents are AARR and aarr. Therefore the gametes of the parents must be AR and ar so the genotype for all the offspring in the F1 generation will be AaRr, and their phenotype will be axial-pink.

What will be the ratios in the F2 generation?

The ratio of genotypes can be determined by examining the Punnett square below:

The ratio of phenotypes will be:

 6 axial-pink 8 pink 3 axial-red 4 red 3 axial-white 4 white 2 terminal-pink 12 axial 1 terminal-white 4 terminal 1 terminal-red

## Problem 3

Flower position, stem length, and seed shape were three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows:

 Character Dominant Recessive Flower position Axial (A ) Terminal (a ) Stem length Tall (T ) Dwarf (t ) Seed shape Round (R ) Wrinkled (r)

If a plant that is heterozygous for all three characters were allowed to self-fertilize, what proportion of the offspring would be expected to be as follows: (Note – use the rules of probability (and show your work) instead of huge Punnett squares)

a) homozygous for the three dominant traits

AATTRR = 1/4 x 1/4 x 1/4 = 1/64

b) homozygous for the three recessive traits

aattrr = 1/4 x 1/4 x 1/4 = 1/64

c) heterozygous (assumed for each trait)

AaTtRr = 1/2 x 1/2 x 1/2 = 1/8

d) homozygous for axial and tall, heterozygous for seed shape

AATTRr = 1/4 x 1/4 x 1/2 = 1/32

Problem 4
A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino was crossed with a second one, 7 blacks and 5 albinos were obtained.

What is the best explanation for this genetic situation?

Black is dominant over white

Write genotypes for the parents, gametes, and offspring.

First cross:

 Parent’s genotypes = BB (black) x bb (white) gametes = B b F1 offspring = all Bb

Second cross:

 Parent’s genotypes = Bb (black) x bb (white) gametes = B or b b F1 offspring = Bb or bb

There should be 50% black to 50% white offspring in this cross.

Problem 5
In sesame plants, the one-pod condition (P ) is dominant to the three-pod condition (p ), and normal leaf (L ) is dominant to wrinkled leaf (l) . Pod type and leaf type are inherited independently. Determinine the genotypes for the two parents for all possible matings producing the following offspring:

a. 318 one-pod normal, 98 one-pod wrinkled

Parental genotypes: PPLl x PPLl or PpLl x PPLl

b. 323 three-pod normal, 106 three-pod wrinkled

Parental genotypes: ppLl x ppLl

c. 401 one-pod normal

Parental genotypes: PPLL x PpLL or PPLl x PPLL or PPLL x PpLl etc (nine possible genotypes).

d. 150 one-pod normal, 147 one-pod wrinkled, 51 three-pod normal, 48 three-pod wrinkled. (a 3: 3: 1: 1 ratio)

Parental genotypes: PpLl x Ppll (see below for details)

3 One-pod normal (PPLl , PpLl , PpLl)

3 One-pod wrinkled (PPll , Ppll , Ppll)

1 Three-pod normal (ppLl)

1 Three-pod wrinked (ppll)

e. 223 one-pod normal, 72 one-pod wrinkled, 76 three-pod normal, 27 three-pod wrinkled (a 9: 3: 3: 1 ratio)

Parental genotypes: PpLl x PpLl

Problem 6
A man with group A blood marries a woman with group B blood. Their child has group O blood. What are the genotypes of these individuals?

Father = AO (or IAi)

Mother = BO (or IBi)

First Child = OO (or ii)

What other genotypes and in what frequencies, would you expect in offspring from this marriage?

Examine the Punnett square to determine the other genotypes possible.

The other genotypes for children are (according to Campbell’s system): 1/4 IAIB, 1/4 IAi, 1/4 IBi

Problem 7
Color pattern in a species of duck is determined by one gene with three alleles. Alleles H and I are codominant, and allele i is recessive to both. Note: this situation is similar to the ABO blood system.

How many phenotypes are possible in a flock of ducks that contains all the possible combinations of these three alleles?

As in the ABO blood system 4 phenotypes are possible in this case:

 Genotype Phenotype HH, Hi (H) II, Ii (I) HI (HI) ii (i)

Problem 8
Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband are both carriers, what is the probability of each of the following?

Under these circumstances assume the following Punnett square to be true.

Where NN or Nn = normal conditions and nn = PKU

a. all three of their children will be of normal phenotype

3/4 x 3/4 x 3/4 = 27/64

b. one or more of the three children will have the disease (x)

1 – 27/64 = 37/64

 All three have x 2 out of 3 has x 1 out of 3 has x + + = x x o o o x 3 Combinations x o x o x o o x x x o o + 3(3/4 x 1/4 x 1/4) + 3(3/4 x 3/4 x 1/4) =

Note: the probability of the disease (x) = 1/4 & the probability of being normal (o) = 3/4

c. all three children will have the disease

1/4 x 1/4 x 1/4 = 1/64

d. at least one child out of three will be phenotypically normal

(Note: Remember that the probabilities of all possible outcomes always add up to 1)

1 – 1/64 = 63/64

Problem 9
The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring would have the following genotypes?

a. aabbccdd = x x x = 1/256

b. AaBbCcDd = x x x = 1/16

c. AABBCCDD = x x x = 1/256

d. AaBBccDd = x x x = 1/64

e. AaBBCCdd = x x x = 1/128

Just remember that the probability of a heterozygote (Xx) = 2/4 or 1/2 and the probability of a homozygote XX or xx = 1/4

Problem 10
In 1981, a stray black cat with unusual rounded curled-back ears was adopted by a family in California. Hundreds of descendants of the cat have since been born, and cat fanciers hope to develop the “curl” cat into a show breed. Suppose you owned the first curl cat and wanted to develop a true breeding variety.

How would you determine whether the curl allele is dominant or recessive?

Mate the stray to a non-curl cat. If any offspring have the “curl” trait it is likely to be dominant. If the mutation is recessive, then on ly non-curl offspring will result.

How would you select for true-breeding cats?

You know that cats are true-breeding when curl crossed with curl matings produce only curl offspring.

How would you know they are true-breeding?

A pure-bred “curl cat” is homozygous.

1. If the trait is recessive any inividual with the “curl” condition is homozygous recessive.
2. If the trait is dominant you can determine if the individual in question is true breeding (CC) or heterozygous (Cc) with a test cross (to a homozygous recessive individual).

Problem 11
What is the probability that each of the following pairs of parents will produce the indicated offspring (assume independent assortment of all gene pairs?

a. AABBCC x aabbcc —-> AaBbCc

(1)(1)(1) = 1

b. AABbCc x AaBbCc —–> AAbbCC

()()() = 1/32

c. AaBbCc x AaBbCc —–> AaBbCc

()()() = 1/8

d. aaBbCC x AABbcc —-> AaBbCc

(1)()(1) = 1/2

Problem 12
Karen and Steve each have a sibling with sickle-cell disease. Neither Karen, Steve, nor any of their parents has the disease, and none of them has been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple should have another child, the child will have sickle-cell anemia.

In order for Karen and Steve to have siblings with sickle cell anemia their parents must be carriers (Nn). We also know that John and Carol are not homozygous recessive (nn) because they do not have the disease. Therefore the chance that Karen is a carrier is 2/3 (NN, Nn, nN) and the chance that Steve is a carrier is also 2/3. If they have a child and both Karen and Steve are carriers then the child has one chance in 4 of having sickle cell anemia. Since each event is independent of one another the overall probability of the child having sickle cell anemia is:

2/3 x 2/3 x 1/4 = 1/9.

Problem 13
Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the disease and blood group are independently inherited. A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time. What is the probability that the second child will also have the disease? Assume both parents are heterozygous for the “disease” gene.

 Genotype Father AO Mother BO

OO expresses the disease

The second child’s chance of having the disease is = x = 1/16

Problem 14
In tigers, a recessive allele causes an absence of fur pigmentation (a “white tiger”) and a cross-eyed condition. If two phenotypically normal tigers that are heterozygous at this locus are mated, what percentage of their offspring will be cross-eyed? What percentage will be white?

Using the Punnett square below where P = normal pigmenation and p = white

then 25% will be white (pp) and all of the white offspring will also be cross-eyed

Problem 15
In corn plants, a dominant allele I inhibits kernel color, while the recessive allele i permits color when homozygous. At a different locus, the dominant gene P causes purple kernel color, while the homozygous recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the F1 generation?

Phenotypic ratios: White (I _ _ _) = 12

Purple (i i P_ ) = 3

Red ( i i p p) = 1

The dominant allele I is epistatic to the p locus, and thus the F1 generation will be:

9 I_P_ : colorless

3 I_pp : colorless

3 i iP_ : purple

1 i i pp: red

Problem 16
The pedigree below traces the inheritance of alkaptonuria, a biochemical disorder. Affected individuals, indicated here by the filled-in circles and squares, are unable to break down a substance called alkapton, which colors the urine and stains body tissues. Does alkaptonuria appear to be caused by a dominant or recessive allele?

Recessive

Fill in the genotypes of the individuals whose genotypes you know. What genotypes are possible for each of the other individuals?

If alkaptonuria is recessive George must be a carrier. See below.

If alkaptonuria is dominant Carla could not have the disease, as indicated in the pedigree chart, since the parents do not express the trait. See Below.

Problem 17
A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits (5). Extra digits is a dominant trait. What fraction of this couple’s children would be expected to have extra digits?

Because the daughter is normal the man’s genotype must be heterozygous for the trait so:

if X = extra digits and x = normal (5) digits then:

50% of the offspring will be polydactylic

Problem 18
Imagine you are a genetic counselor, and a couple planning to start a family came to you for information. Charles was married once before, and he and his first wife had a child who has cystic fibrosis. The brother of his current wife Elaine died of cystic fibrosis. Cystic Fiborsis is a lethal recessive condition (a person with CF cannot have children).

What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine has the disease)

The Probability that Elaine is a carrier is 2/3 (She does not have Cystic Fibrosis which eliminates one of the 4 possibilities. She does have 2 chances out of three of being a carrier. [Nn, nN (carriers) or NN]

The probability that the baby (?) has the disease (if Elaine is a carrier) is 1/4

The total probability is 2/3 x 1/4 or 1/6.

Problem 19
In mice, black color (B ) is dominant to white (b ). At a different locus, a dominant allele (A ) produces a band of yellow just below the tip of each hair in mice with black fur. This gives a frosted appearance known as agouti. Expression of the recessive allele (a ) results in a solid coat color. If mice that are heterozygous at both loci are crossed, what will be the expected phenotypic ratio of their offspring?

B = Black – dominant

A = Agouti – dominant

BbAa x BbAa =

 Genotype Phenotype 1 BBAA agouti 2 BbAA agouti 2 BBAa agouti 4 BbAa agouti 1 BBaa black 2 Bbaa black 1 bbAA white 2 bbAa white 1 bbaa white

The phenotypic ratio is:

9 agouti: 4 white: 3 black

Problem 20
The pedigree below traces the inheritance of a vary rare biochemical disorder in humans. Affected individuals are indicated by filled-in circles and squares. Is the allele for this disorder dominant or recessive?

The allele is most likely dominant because the #2 individual (see below) with the trait marries a woman with the trait and 50% of their offspring are normal. If the trait were recessive one would expect the following:

 100% of offspring would have the disease, which is not the case.

What genotypes are possible for the individuals marked 1, 2, and 3?

1. Bb (heterozygous)
2. Bb ( if 2’s genotype were bb he would not have the disease and if BB all his children would have the condition.)
3. bb (all normal individuals are homozygous recessive)

## AP Sample 5 Lab 2

 ENZYME CATALASE

Introduction:

Enzymes are proteins one of the four macromolecules: carbohydrates, proteins, lipids, and nucleic acids.  They are produced by cells to act as a catalyst in chemical reactions.  Enzymes are made by the ribosomes within a cell.  The enzymes that need to stay within the cell are made by the free ribosomes in the cytoplasm.  The enzymes that are made to be exported to other cells are created by the ribosomes on the Rough E.R. The purpose of enzymes is to lower the activation energy (the energy that is put in at the beginning of the reaction) of a reaction; therefore speeding the reaction rate up.

Catalase is one of the enzymes found in the cells of the human body. It is made up of five hundred amino acids in four polypeptide chains. It is found in the perioxisome microbody in the cells of eukaryotic organisms.  The perioxisomes make hydrogen peroxide (H2O2). It is a byproduct of cellular respiration. This is very toxic to cells.  That is why catalase is there.  Catalase breaks down the hydrogen peroxide before it damages the cell. The formula for the reaction of decomposition of hydrogen peroxide by catalase is:

H2O2 H2O2 H2O + O2 H2O + O2

Substrates are the substances that attach to the enzyme.  They attach at the enzyme’s active site with induced fit.  The combining of and enzyme and substrate requires ATP.  The reaction when an enzyme and a substrate join is reversible.  The reaction can make more products or it can make more substrate.

Enzymes are affected by many things; such as salt concentration, pH, temperature, substrate concentration, product concentration, activators, and inhibitors.  These things can cause the enzyme to denature (change shape).  If the enzyme denatures, then the substrate isn’t able to attach to the active site of the enzyme.

The typical salt concentration for an enzyme is intermediate.  This allows the substrate to attach to the enzyme correctly.  If the concentration is lowered the R groups of the enzyme attract to each other and pull the enzyme into a different shape (changing the active site).  If the concentration of salt is raised, then the active site of the enzyme is blocked.  Therefore the substrate can’t attach to the enzyme.

Enzymes function best in an environment with a pH of seven.  When the pH of the enzyme’s environment become more acidic (H+ ions increase) the enzyme denatures.  This causes the active site to change its shape and the substrate can’t attach.  The same thing happens when the enzyme’s environment gains H+ ions (becomes more basic).

When heat is added to anything the molecules increase is movement.  With an enzyme, when ten degrees Celsius is added the reaction rate is doubled.  This will continue until optimum temperature is reached.  Then the enzyme is destroyed.  Enzymes denature at approximately forty-fifty degrees Celsius.

Enzymes follow the Law of Mass Action.  This determines the rate and direction of an enzyme and substrate reaction.  If there is a high amount of substrate and a low amount of products the reaction will continue to make more products.  If there is a low amount of substrate and a high amount of products, then the reaction will stop to get more substrate to continue the reaction.  This is true except when products are immediately metabolized or exported from the cell.

Activators enable the substrate to fit the active site better.  They increase the reaction.  Inhibitors block the active site or cause the enzyme to denature. They slow down the reaction.

In this lab assaying is performed.  An assay measures the amount of a substance that is left after a reaction.  In this case, hydrogen peroxide.

Hypothesis:

The enzyme catalase will decompose H2O2 (hydrogen peroxide) best when its environment is ideal.

Materials:

The materials used in the first part of exercise 2A are 10mL of 1.5% H2O2, one 50mL beaker, and 1mL of fresh catalase. The materials used in the second part of exercise 2A are one test tube, 5mL of fresh catalase, a hot plate, 10mL of 1.5% H2O2, and one 50mL beaker.  The materials used in the third part of exercise 2A are a potato, a knife, a ruler, one 50mL beaker, and 10mL of 1.5% H2O2.  The materials used in exercise 2B are 10mL of 1.5% H2O2, 1mL of water, 10mL of H2SO4 (1.0 M), a stirring rod, two 50mL beakers, 5mL syringe, and 5mL of 2% KMnO4.   The materials used in exercise 2C are a 50mL beaker, 10-15mL of 1.5% H2O2, 10mL syringe, and 5-10mL of 2% KMnO4.   The materials used in exercise 2D are 60mL of 1.5% H2O2, 60mL of H2SO4 (1.0M), 6mL of fresh catalase, six 50mL beakers, three 10mL syringes, six 10mL beakers, and 30mL of 2% KMnO4.

Methods:

Exercise 2A:  To observe the reaction of catalase and hydrogen peroxide, transfer 10mL of 1.5% H2O2 into a 50mL beaker and add 1mL of fresh catalase.  Observe the reaction and record results on Table 1.  Next, the effect of boiling catalase needs to be demonstrated.  Place 5mL of fresh catalase into a test tube and boil it over a hot plate for 5 minutes.  Allow the boiled catalase to cool very well.  Then place 10mL of 1.5% H2O2 into a 50mL beaker and add 1mL of the cooled, boiled catalase.  Observe the reaction and record results on Table 1.  Finally, to see the presence of catalase in living tissue, cut 1 cm3 of potato and macerate it.  Then, place it in a 50mL beaker with 10mL of 1.5% H2O2.  Observe the reaction and record results on Table 1.

Exercise 2B:  This is the procedure for establishing a baseline.  First, put 10mL of 1.5% H2O2 in a beaker.  Add 1mL of water (instead of catalase).  Add 10mL of H2SO4 (1.0M). Mix the solution well.  Remove a 5mL sample from the solution and place it into a 50mL beaker.  Using the 5mL syringe, add 2% KMnO4 one drop at a time (swirling after each drop) to the sample until a persistent pink or brown color is obtained.  Record the results on Table 2.

Exercise 2C:  This will determine how much H2O2 will decompose spontaneously in an uncatalyzed reaction. Place 10-15mL of H2O2 in a 50mL beaker and store it uncovered for approximately 24 hours  Determine the amount of H2O2 remaining after 24 hours by running a baseline test.  Use a 10mL syringe to get 10mL of 2% KMnO4.  Add the KMnO4 to the solution one drop at a time (swirling after each drop) until a pink or brown color is permanently obtained.  Record the results in Table 3.

Exercise 2D:  If a day or more has passed since exercise 2B has been performed, reestablish the baseline by performing exercise 2B again.  Record the results in Table 4.  This exercise will determine the amount of H2O2 is disappearing over 10, 30, 60, 120, 180, and 360 seconds. First, place 10mL of 1.5% H2O2 in six 50mL beakers. Label the beakers 10, 30, 60, 120, 180, and 360 seconds.  Obtain 6mL of fresh catalase and 60mL of H2SO4 (1.0M). Add 1mL of catalase to the 10 second beaker with one syringe. At 10 seconds, add 10mL of H2SO4 (1.0M) with another syringe.  For each of the times, repeat the adding of the catalase and H2SO4.  Allow the reactions to proceed for 30, 60, 120, 180, and 360 seconds respectively.  After the times have passed remove a 5mL sample from each of the six beakers and place each of the samples into a separate 10mL beaker.  Use 5ml of 2% KMnO4 for each of the six beakers to determine the amount of H2O2 that is left after the reaction.  Add KMnO4 to the samples one drop at a time (swirling after each drop) until a pink or brown color is permanently obtained. Record results in Table 5. Graph the results.

Results:

Table 1

Enzyme Activity

ActivityObservations

Bubbles

Nothing

## Presence of Catalase

Bubbles

Table 2

Establishing a Baseline

Volume

5mL

1.6mL
Baseline (final volume – initial volume)

3.4mL

Table 3

Rate of Hydrogen Peroxide Spontaneous Decomposition

Volume
Initial KMnO4

6mL

### Final KMnO4

1mL
Amount of KMnO4 used after 24 hours

5mL
Amount of H2O2 spontaneously decomposed

(ml baseline – ml after 24 hours)

1.6mL

### Percent of H2O2 spontaneously decomposed

(ml baseline – ml after 24 hours/ baseline)

53%

Table 4

Establishing a Second Baseline

Volume

5mL

1mL
Baseline (final volume – initial volume)

4mL

Table 5

Rate of Hydrogen Peroxide Decomposition by Catalase

Time ( Seconds)

10 30 60 120 180 360
Baseline  KMnO4

4mL4mL4mL4mL4mL4mL
Initial volume KMnO4

5mL5mL5mL5mL5mL5mL
Final volume KMnO4

2mL3mL4mL4.2mL4mL4.2mL
Amount KMnO4 used

(baseline – final)

3mL2mL1mL.8mL1mL.8mL

### Amount H2O2 used

(KMnO4 – initial)

1mL2mL3mL3.2mL3mL3.2mL

1.      Determine the initial rate of the reaction and the rates between each of the time points.

0-10 –  .1

10-30  –  .05

30-60  –  .03

60-120  –  .003

120-180  –  .003

180-360  –  .001

2. When is the rate the highest? Explain.

They are the highest at the beginning of the reaction because; less hydrogen          peroxide had been spontaneously decomposed.  Therefore, there was more to be          broken down.

3.      When is the rate the lowest? For what reasons?

They are the lowest at the end of the reaction because more hydrogen peroxide     has been decomposed.

4.      Explain the inhibiting effect of sulfuric acid on the function of catalase. Relate this to enzyme structure and chemistry.

The acid lowered the pH and the change made catalase gain H+ ions which            caused the active site to change shape.

5.      Predict the effect lowering the temperature would have on the rate of enzyme activity. Explain.

It would decrease the reaction because changing the temperature causes the          enzyme to denature.  This causes the active site to change place, which slows      down the reaction.

6.      Design a control experiment to test the effect of varying pH, temperature, or enzyme concentration.

Label 3 test tubes 0oC, rt, and 100oC.  Place an equal amount of enzyme in each    tube.  Put the 0oC on ice, leave the rt out and boil the 100oC tube.  Observe the changes in all 3 tubes.

Error Analysis:

Some possible errors that could have occurred in this lab include taking inaccurate measurements of the substances used, making wrong calculations, and the times of testing the experiments could have been delayed (ex: instead of 60 seconds it took place at 65 seconds).

Conclusion:

The enzyme catalase is created to break down H2O2.  If the environment of catalase is changed in any way; such as pH and temperature; then it can not do its job of breaking down H2O2.  This is proven when catalase broke down H2O2 in 2A and produced water and oxygen when it wasn’t boiled.  When the catalase was boiled, it had been denatured; therefore it wouldn’t break the H2O2 down.  In 2D the sulfuric acid kept catalase from breaking down H2O2 also. It lowered the pH of the solution so the catalase denatured.  You can tell by looking at Table 5.  The lower times have less H2O2 used than the higher times.  The higher times had more time with out the H2SO4, so the catalase was able to break down more H2O2.

The rates of the reaction in this lab were very high at the beginning.  Then, they started to slow down.  This is because at the beginning of the reaction there is more H2O2 to decompose.  Towards the end of the reaction there is less to decompose, so the rate slows down quite suddenly.

## AP Lab Results Objectives

The “Dirty Dozen” Objectives and Results

 1 2 3 4 5 6 7 8 9 10 11 12

Laboratory 1: Diffusion and Osmosis

Overview

In this lab you will investigate the movement of water across semi-permeable membranes. You will also examine the effect of solute
concentrations on water potential as it relates to living plant tissues.

Objectives

• Describe the mechanisms of diffusion and osmosis
• Describe how solute size and molar concentration affect the process of diffusion through a selectively permeable membrane
• Design an experiment to demonstrate and measure water potential
• Relate osmotic potential to solute concentration and water potential (Mr. Knight’s Notes on Water Potential)
• Describe how pressure affects the water potential of a solution
• Describe the effects of water gain or loss in animal or plant cells
• Calculate the water potential of living plant cells from experimental data

Results

When a solution such as that inside a potato cell is separated from pure water by a selectively permeable membrane water will move
by osmosis from the surrounding area where the water potential is higher into the cell where water potential is lower due to the
presence of solute. The movement of water into the cell, causes the cell to swell and the cell membrane pushes against the cell wall
to produce an increase in pressure (turgor). This process will continue until the water potential of the cell equals the water
potential of the pure water outside the cell. At this point, a dynamic equilibrium is reached and net water movement will cease.

Sample Multiple – Choice Questions

1. A dialysis bag is filled with a 3% starch solution. The bag is immersed in a beaker of water containing a 1% IKI solution.
All of the following observations are correct EXCEPT:

A. When the bag is first placed in the beaker, the water potential inside the bag is negative.

B. When the bag is first placed in the beaker, the solution in the beaker is yellow brown.

C. The starch solution inside the bag is hypertonic relative to the solution in the beaker.

D. After fifteen minutes, the solution in the bag turns blue.

E. After fifteen minutes, the mass of the dialysis bag has decreased.

*****************************************************************************************

Laboratory 2: Enzyme Catalysis

Overview

In this lab you will measure the rate of a reaction in the presence and absence of a catalyst. The catalyst, (catalase), is an enzyme in cells that
catalyzes the breakdown of toxic H202.

 Lab 2. Enzyme Catalysis Data 10s 30s 60s 90s 120s 180s 360s Baseline .364 .340 .449 .574 .694 .699 1.06 3.471

 Lab 2. Enzyme Catalysis Data [Sample] 10s 30s 60s 90s 120s 180s 360s .9 1.8 2.4 2.8 3.1 3.3 3.4

Objectives

• Graph data from an enzyme experiment
• Determine the rates of enzymatically catalyzed reactions
• Discuss the method for determining enzyme activity
• Discuss the relationship between dependent and independent variables
• Discuss the effect of initial reaction rates produced by changes in temperature, pH, enzyme concentrations, and substrate concentrations
• Design an experiment to measure the effect of enzyme activity produced by changes in temperature, pH, enzyme concentrations,
and substrate concentrations

Results

In the first few minutes of an enzymatic reaction, the number of substrate molecules is usually so large compared to the number of
enzyme molecules that changing the substrate concentration does not (for a short period at least) affect the number of successful collisions
between substrate and enzyme. During this early period, the enzyme is acting on substrate molecules at a constant rate. The slope of the
graph line during this early period is called the initial velocity of the reaction. The initial velocity, or rate, of any enzyme catalyzed reaction
is determined by the characteristics of the enzyme molecule. It is always the same for an enzyme and its substrate as long as temperature
and pH are constant and substrate is present in excess. Also, in this experiment the disappearance of the substrate, H202 is essential in this
reaction. Once all the H202 has reacted, any more KMnO4 added will be in excess an will not be decomposed.

Sample Multiple – Choice Questions

1. Which of the following is LEAST likely to increase the forward rate of an enzyme-mediated reaction?
1. An increase in the substrate concentration
2. An increase in the enzyme concentration
3. An increase in the product concentration
4. An increase in pH
5. An increase in the temperature

*****************************************************************************************

Laboratory 3: Mitosis and Meiosis

Overview

In this lab you will study plant mitosis using prepared slides of onion root tips and will calculate the relative period of the phases of mitosis
in the meristem of root tissue. You will also study the crossing over and recombination that occurs during meiosis.

Objectives

• Compare the events of mitosis in plant cells with those of animal cells
• Demonstrate the procedure to stain tissue for the identification of cells in the various stages of mitosis
• Calculate the relative duration of mitosis

Results

The relative length of mitotic stages are, 53.4% of prophase, 17.4% of metaphase, 16.8% of anaphase and 12.4% of telophase. Meiosis
is important for sexual reproduction because it reduces the chromosome number by half and it also results in new combinations of genes
through independent assortment and crossing over, followed by the random fertilization of eggs by sperm.

*****************************************************************************************

Laboratory 4: Plant Pigments and Photosynthesis

Overview

In this lab you will separate plant pigments using paper chromatography. You will also measure the rate of photosynthesis in isolated
chloroplasts using a measurement technique that involves the reduction of the dye, DPIP. The transfer of electrons during the
light-dependent reactions of photosynthesis reduces DPIP and changes its color from blue to colorless.

Objectives

• Understand the principles of chromatography
• Calculate Rf values
• Design an experiment in which chromatography is used as a separation technique
• Describe a technique for determining photosynthetic rate
• Understand the relationship between dependent and independent variables
• Describe how light intensity, light wavelength, and temperature can affect photosynthesis
• Design an experiment to measure how light intensity, light wavelength, and temperature can affect photosynthetic rates

Results

The solvent moves up the paper by capillary action, which occurs as a result of the attraction of solvent molecules to the paper and the
attraction of solvent molecules to one another. As the solvent moves up the paper, it carries along any substances dissolved in it, in
this case pigments. The pigments are carried along at different rates because they are not equally soluble in the solvent and because
they are attracted, to different degrees, to the cellulose in the paper through the formation of hydrogen bonds. Also, as the DPIP is
reduced and becomes colorless, the resultant increase in light transmittance is measured over a time course using a spectrophotometer.

Sample Multiple – Choice Questions

1. In a paper chromatography procedure, molecules with which of the following characteristics migrate the fastest up the
chromatography paper?

A. High solubility in solvent and weak hydrogen bonding to cellulose.

B. High solubility in solvent and strong hydrogen bonding to cellulose.

C. Low solubility in solvent and strong hydrogen bonding to cellulose.

D. Low solubility in solvent and weak hydrogen bonding to cellulose.

E. Insoluble in solvent.

*****************************************************************************************

Laboratory 5: Cell Respiration

Overview

In this lab you will measure oxygen consumption during respiration as a change in gas volume in germinating and nongerminating peas
at two different temperatures.

Objectives

• Discuss the gas laws as they apply to the function of a respirometer
• Interpret data related to the effects of temperature on cell respiration
• Interpret data related to the effects of germination or nongermination on cell respiration
• Explain or determine the significance of a control
• Explain the relationship between dependent and independent variables
• Calculate a rate of cell respiration by utilizing graphed data
• Design an experiment to use a respirometer to measure cellular respiration

Results

Germinating peas respire and need to consume oxygen in order to continue the growing process. Pea seeds are nongerminating and do not
respire actively. These seeds are no longer the site of growth and thus do not need oxygen for growth. In consideration to temperature,
at higher temperatures more oxygen is consumed which means more respiration is occurring. 686 kilocalories are released during respiration.
When temperature decreases molecular motion slows down and respiration decreases because less energy is made available.

*****************************************************************************************

Laboratory 6: Molecular Biology

Overview

In this lab you will investigate some basic principles of genetic engineering. Plasmids containing specific fragments of foreign DNA will be
used to transform E. coli cells, conferring antibiotic (ampicillin) resistance and the Lac + phenotype (ability to metabolize lactose).
Restriction enzyme digests of phage lambda DNA will also be used to demonstrate techniques for separating and identifying DNA fragments
using gel electrophoresis.

Objectives

• Discuss the principles of bacterial transformation
• Describe how to prepare competent E. coli cells
• Discuss the mechanisms of gene transfer using plasmid vectors
• Discuss the transfer of antibiotic resistance genes and tell how to select positively for transformed cells that are antibiotic resistant
• Discuss the mechanisms of action for restriction endonucleases
• Discuss how a plasmid can be engineered to include a piece of foreign DNA that alters the phenotype of the transformed cells
• Understand and be able to explain the principles of electrophoresis as they pertain to separating and identifying DNA fragments

Results

Bacterial Transformation-Ampicillin resistance: In this exercise, we will introduce competent E. Coli cells to take up the plasmid pAMP
which contains a gene for ampicllin resistance. Normally, E. Coli cells are destroyed by the antibiotic ampicillin, but E. Coli cells that
have been transformed will be able to grow on agar plates containing ampicillin. Thus, we can select for transformants; those cells
that are not transformed will be killed by ampicillin; those that have been transformed will survive.

Restriction Enzyme Cleavage of DNA: Restriction endonuclease recognizes specific DNA sequences in double-stranded DNA and digests
the DNA at these sites. The result is the production of fragments of DNA of various lengths corresponding to the distance between
identical DNA sequences within the chromosome. By taking DNA fragments and systematically reinserting the fragments into an
organism with minimal genetic material, it is possible to determine the function of particular gene sequences

Electrophoresis: fragments of DNA can be separated by gel electrophoresis when any molecule enters the electrical field, the
mobility or speed at which it will move is influenced by the charge (negative charges travel to positive/top pole of gel), the density
of the molecule, (the smaller the molecule, the faster it travels), the strength of the electrical field, and the density of the medium (gel)
which it is migrating.

Sample Multiple – Choice Questions

1. After growth on ampicillin to select bacteria transformed with a mixture of recombinant DNA containing plasmids, you must identify
a clone containing a specific gene sequence. You would:
1. Blot transfer clones to membranes, and screen using a radioactive probe complementary to the gene.
2. Re-grow bacteria in ampicillin where only transformants containing the gene of interest can grow.
3. Culture bacteria in both ampicillin and tetracycline to select for bacteria containing the gene of interest.
4. Digest DNA from the plasmid to isolate the gene fragment.
5. Do a restriction map of plasmid DNA to identify the correct clone

*****************************************************************************************

Laboratory 7: Genetics Of Drosophila

Overview

In this lab you will use the fruit fly Drosophila melanogaster to do genetic crosses. You will learn how to collect and manipulate fruit flies,
collect data from F1 and F2 generations, and analyze the results from a monohybrid, dihybrid or sex link cross.

Objectives

• Conduct a genetics experiment for a number of generations
• Compare predicted results with actual results
• Explain the importance of Chi-Square analysis
• Design genetic crosses in an experiment to illustrate independent assortment and sex linkage
• Discuss the life cycle of the fruit fly, recognize the sex of fruit flies, and recognize several types of classic mutations

Results

From this lab, you will be able to find genotypes and phenotypic expression within a fruit fly. Also, recessive genes and mutations will be
revealed as the student crosses a variety of Drosophila alleles. For example, if a female carrier for an x-linked, recessive trait, was crossed
with a male without the recessive trait the results would be:

½ males with x-linked trait ½ males without

½ females carriers ½ females without

0 females express sex linked traits

Sample Multiple – Choice Questions

1. A population consists of 20 individuals of which 64% are homozygous dominant for a particular trait and the remaining individuals
are all heterozygous. All of the following can explain the situation except
1. Genetic drift is occurring
2. The recessive allele is deleterious
3. All homozygous recessive individuals emigrate
4. The population is very small
5. Only heterozygous individuals mate

*****************************************************************************************

Laboratory 8: Population Genetics and Evolution

Overview

In this lab you will learn about Hardy-Weinberg law of genetic equilibrium and study the relationship between evolution and changes
in allele frequency by using your class as a sample population.

Objectives

• Calculate allele and genotype frequencies using the Hardy-Weinberg theorem
• Discuss the effect of natural selection on allelic frequencies
• Explain and predict the effect of allelic frequencies of selection against the homozygous recessive
• Discuss the relationship between evolution and changes in allele frequencies, as measured by deviations from the Hardy-Weinberg
law of genetic equilibrium

Results

Assuming that Hardy-Weinberg equilibrium is maintained allele and genotype frequencies should remain constant from generation to
generation. For this to happen the five following situations must all occur:

1. Population is very large. The effects of chance on changes in allele frequencies is thereby greatly reduced.
2. Individuals show no mating preference, i.e. random mating.
3. There is no mutation of alleles.
4. No differential migration occurs, (no immigration or emigration).
5. All genotypes have an equal chance of surviving and reproducing, i.e. there is no natural selection.

In humans, several genetic diseases have been well characterized. Some of these diseases are controlled by a single allele where the
homozygous recessive genotype has a high probability of not reaching reproductive maturity. If this were to occur both the homozygous
dominant and heterozygous individuals will survive while the homozygous recessive will become extinct.

Sample Multiple – Choice Questions

1. Which of the following generates the formation of adaptations?
1. Genetic drift
2. Mutations
3. Gene flow
4. Sexual reproduction
5. Natural selection

2. All of the following are examples of evolution, except:

1. Mutations in an individual
2. Changes in an allele frequency in a population
3. Changes in an allele frequency in a species
4. Divergence of a species into two species

*****************************************************************************************

Laboratory 9: Transpiration

Overview

In this lab you will apply what you learned about water potential from Lab 1 (Diffusion and Osmosis) to the movement of water within
the plant. You will study the organization of the plant stem as it relates to these processes by observing sections of fresh tissue.

Objectives

• Describe how differences in water potential affect the transport of water from roots to stems to leaves
• Relate transpiration to the overall process of water transport in plants
• Discuss the importance of properties of water – including hydrogen bonding, adhesion, and cohesion – to the transport of water in plants
• Quantitatively demonstrate the effects of different environmental conditions on the rate of transpiration in plants
• Identify the vascular tissues of the plant stem and describe their functions

Results

Conditions that cause a decreased rate of water loss from leaves result in a decreased water potential gradient from stem to leaf and
therefore in a decreased rate of water movement up the stem to the leaves. Conditions that cause an increased rate of water loss from
leaves result in an increase in the water potential gradient from stem to leaf and therefore in an increase in the rate of water movement
up the stem to the leaves.

1. Room Conditions
1. When you expose a plant to room conditions nothing is supposed to happen. The reasoning for this is room conditions
don’t cause drastic changes in the plants environment for major transpiration or even water gain to occur. The plant
under room conditions is considered to be your control
1. Floodlight
1. When light is absorbed by the leaf, some of the light energy is converted to heat and remember that transpiration
rate increases with temperature. We learned in Unit One of the Campbell’s edition that when the temperature of
liquid water rises, kinetic energy of the water molecules increases. As a result, the rate at which liquid water is
converted to water vapor increases. When the water is turned into water vapor, it easily passes out through stomata
out into the outer atmosphere.
2. The floodlight is an example of a plant near the sun (which is why the plant is one meter away from the light) Due
to the aforementioned properties of plants you should see a loss of water
1. Fan
1. An increase in wind speed results in an increase in the rate of leaf water loss because increased wind decreases
the boundary layer of still air at the leaf surface. This boundary layer acts to slow leaf water loss. Increased
wind also causes the rapid removal of evaporating water molecules from the leaf surface. This results in
a low water potential in the air immediately and the water level should drop.
1. Mist
1. Increased humidity in the air surrounding the leaf decreases the water potential gradient between the saturated
air in the leaf air spaces and the air surrounding the leaf, resulting in a decreased rate of leaf water loss.
However, when the humidity of the air surrounding the leaf if very low, the water potential of the air is low.
Therefore, the water potential gradient between the air spaces of the leaf and the surrounding air is high, and
the rate of leaf water loss increases.

Or maybe this explanation is better

1. When there is a great amount of humidity, transpiration decreases because of water potential. When the humidity is
at a low or normal, the mesophyll cells in the plant are much higher in water potential than the relatively drier
surrounding air. Due to the properties of water potential, which states that water tends to evaporate from the leaf
surface moving from an area of higher water potential to an area of lower water potential, transpiration occurs.
But, because of the high humidity, the surrounding air has a higher water potential than the mesophyll cells and water
loss is at a minimum.

Adaptations to reduce leaf water loss include a reduced number of stomates, an increase in the thickness of the leaf cuticle, a decrease
in leaf surface area, and adaptations that decrease air movements around stomates, such as dense hairs and sunken stomates. Because leaves
are all different in size, reporting the water loss without considering a unit area would provide non-comparable data.

Sample Multiple – Choice Questions

1. Which of the following series of terms correctly indicates the gradient of water potential from lowest water potential to highest
water potential?
1. Air, leaf, stem, root, soil
2. Soil, root, stem, leaf, air
3. Root, leaf, stem, air, soil
4. Air, soil, root, leaf, stem
5. Stem, leaf, root, soil, air

*****************************************************************************************

Laboratory 10: Physiology of the Circulatory System

Overview

You will learn how to measure blood pressure and measure pulse rate under different physiological conditions: standing, reclining,
after the baroreceptor reflex, and during and immediately after exercise. The blood pressure and pulse rate will be analyzed and
related to a relative fitness index. You will also measure the effect of temperature on the heart rate of the Daphnia magna, and
calculate a Q10 for the relationship between temperature and heart rate.

Objectives

• Measure pulse rate
• Measure blood pressure
• Describe the relationship between the changes in heart rate and blood pressure relative to changes in body position
• Describe the relationship between changes in heart rate and exercise
• Determine the “fitness index” for an adult human
• Perform statistical analysis on class data
• Define Q10
• Determine the Q10 of heart rate in a living organism such as Daphnia

Results

The sphygmomanometer measures the blood pressure. The blood pressure cuff is inflated so that blood flow stops to through the brachial
artery in the upper arm. A stethoscope is used to listen to blood flow entering the brachial artery. When blood first enters the artery,
snapping sounds called the sounds of Korotkoff are generated.

1. Blood pressure and heart rate increase when you move from a reclining to a standing position counteracting gravitational pull
on the blood
2. Elevated arterial blood pressure indicates increased arterial resistance to blood flow
3. Fit individuals can pump a larger volume of blood with each contraction and deliver more oxygen to muscle tissue than the
hearts of unfit individuals. As a result, blood pressure and heart rate increases are smaller for fit individuals, and the time
required to return to normal conditions is shorter for fit individuals than unfit individuals.
4. For the Daphnia, remember that ectothermic animals use behavior to regulate their body temperatures and that Q10
cannot be determined for endothermic animals because body temperatures remain constant regardless of environmental temperatures.

Sample Multiple – Choice Questions

1. A Q10 value of 3 in an ectothermic animal means that the metabolic rate
1. Triples when body temperature triples
2. Triples when body temperature increases by 10° C
3. Doubles when the body temperature increases by 3° C
4. Doubles when the body temperature increases by 10° C
5. Triples when the body temperature decreases by 10° C

If you’re having trouble understanding what blood pressure really is, come here and learn. Great for an overview of how blood pressure works,
how bad high blood pressure can be, and much more. Look here for a better understanding of your blood pressure.

*****************************************************************************************

Laboratory 11: Behavior: Habitat Selection

Overview

In this lab, you will examine the habitat preferences of the brine shrimp, Artemia. You will use controlled experimentation to determine
the thermal, pH, and light environments selected by Artemia. Based on your experience with this lab, you will design an experiment that
could be used to survey other variables and other organisms.

Objectives

• Describe the relationship between dependent and independent variables
• Discuss the value of comparing experimental results with control results
• Graph an interpret histogram data
• Measure the volumes, distances, and temperature using metric scales
• Design and conduct an experiment to measure the effect of environmental variables on habitat selection

Results

When conducting this experiment, a couple of things should be understood.

1. Artemia are able to survive in a wide range of salty environments by they tend to live in really salty environments.
2. Three variables are tested: light, temperature and pH. The control is exposed to room light, room temperature, and neutral pH is
also prepared. For each of the variables, a gradient is established providing a continuous variation from weak to strong intensities. Each habitat variable is applied to a 100 cm clear plastic tube filled with water and brine shrimp. The ends of the plastic tube
possess extremes of a continuous variation in the habitat property. After the brine shrimp have been exposed to the habitat
condition, they are divide into four groups by tightening three clamps around the tube at equal intervals. Each group is individually removed from the tube at equal intervals. Each group is individually removed from the tube and the number of live shrimp is counted.
3. The histograms are prepared showing the number of brine shrimp in each of the four intensities. A histogram is prepared for each
of the three variables and the control. From the data in the histograms, conclusions cab be made describing the habitat preferences
of the brine shrimp

Sample Multiple – Choice Questions

1. Artemia brine shrimp are rarely found in bodies of water with salt concentrations below 5%.
1. This is probably because the brine shrimp prefer low levels of salt concentration
2. The brine shrimp prefer high levels of salt concentration
3. The brine shrimp cannot survive in fresh water
4. The brine shrimp cannot survive in temperatures found in bodies of water with fresh water or water with low salinity
5. Predators of the brine shrimp are common in fresh water and water with low salinity

*****************************************************************************************

Laboratory 12: Dissolved Oxygen and Aquatic Primary Productivity

Overview

You will measure and analyze the dissolved oxygen concentration in water samples using the Winkler technique. You will also measure and
analyze the primary productivity of natural waters or lab cultures

Objectives

• Describe the physiological importance of carbon and oxygen in an ecosystem
• Understand the physical and biological factors that affect the solubility of dissolved gases in aquatic ecosystems
• Describe a technique for measuring dissolved oxygen
• Define primary productivity
• Describe the relationship between dissolved oxygen and the processes of photosynthesis and respiration as they affect primary
productivity in an ecosystem
• Design an experiment to measure primary productivity in an aquatic ecosystem
• Understand the effect of light and nutrients on photosynthesis

Results

The amount of oxygen dissolved in natural water samples is measured and analyzed to determine the primary productivity of the sample.
The amount of dissolved oxygen is dependent upon many factors.

A. Temperature

B. Salinity

C. Photosynthesis

D. Respiration

Primary productivity is a measure of the amount of biomass produced by autotrophs through photosynthesis per unit time. It can be
examined by the following factors:

1. Gross Primary Productivity
2. Net Primary Productivity
3. Respiratory Rate

These determine primary productivity:

1. The Winkler Method—this is use to measure dissolved oxygen using a titration technique. Titration is the process of adding a substance of known concentration to a solution containing a substance of unknown concentration until a specific reaction is completed and a color change occurs
2. The light and dark bottle method
3. Initial bottle
4. Light bottle
5. Dark bottle

Sample Multiple – Choice Questions

1. The net primary productivity for a temperate forest was measured at 2000 mg C/m2/day. The respiratory rate of the community
was determined to be 1000 mg C/m2/day. The gross primary productivity for this community is
1. 1000 mg C/m2/day
2. 2000 mg C/m2/day
3. 3000 mg C/m2/day
4. 4000 mg C/m2/day
5. 5000 mg C/m2/day

2. (C) Since enzyme-mediated reactions are reversible (they convert product back to substrate), increasing the concentration of the
product will slow the forward direction of the reaction and accelerate the reverse reaction. Conversely, and increase in the
substrate concentration will increase the forward rate of the reaction. Increasing the enzyme concentration will not slow the
reaction rate but may increase it if the substrate concentration is high enough to utilize additional enzyme. An increase
in pH or temperature may change the rate of reaction, but the nature of the enzyme must be known in order to determine whether
the rate is increased or decreased

1. (A) Water potential is highest in soil, decreases from root to leaf, and is lowest in the air. Water moves from the soil into the roots
and through the plant an transpires from the leaf because water moves from the area of greatest water potential to the area of lowest
water potential
2. (B) The Q10 is the ratio of the metabolic rate at one temperature to the metabolic rate at a temperature 10° colder. A Q10 equal
to 3 indicates that he metabolic rate triples when the body temperature increases by 10° C
3. (E) The predators of brine shrimp cannot survive in bodies of water with high concentrations of sale. Thus, brine shrimp survive in
bodies of water with a high salt concentration because predators are absent. In waters with low concentrations of salt, predators
eliminate the brine shrimp
4. (C) The gross primary productivity is the sum of the net primary productivity and the respiratory rate.

BACK

## AP Lecture Guide 24 – The Origin of Species

 AP Biology: CHAPTER 24: THE ORIGIN OF SPECIES 1. Define the term species.____________________________________________________________________________________________________________________________________________________2. How do the patterns of speciation differ?a. anagenesis _____________________________________________________________b. cladogenesis ___________________________________________________________3. What is thought to be essential for the formation of distinct species rather than a continuumfrom one form of life to another?____________________________________________________________________________________________________________________________________________________4. Define and give an example for each of the following barriers that cause isolation.a. prezygotic Barriers _________________________________________________________________________________________________________________________________b. habitat Isolation ___________________________________________________________________________________________________________________________________c. behavioral Isolation ________________________________________________________________________________________________________________________________d. temporal Isolation __________________________________________________________________________________________________________________________________e. mechanical isolation ________________________________________________________________________________________________________________________________f. gamete isolation ___________________________________________________________________________________________________________________________________g. postzygotic barriers ________________________________________________________________________________________________________________________________h. hybrid inviability ___________________________________________________________________________________________________________________________________i. hybrid sterility _____________________________________________________________________________________________________________________________________5. Define the Modes of Speciationa. allopatric speciation ________________________________________________________________________________________________________________________________b. sympatric speciation ________________________________________________________________________________________________________________________________6. How does the antelope squirrel demonstrate allopatric speciation?____________________________________________________________________________________________________________________________________________________7. What does the concept of “ring species” demonstrate? Give an example.____________________________________________________________________________________________________________________________________________________8. How do island chains encourage adaptive radiation?____________________________________________________________________________________________________________________________________________________9. What are the two intrinsic factors that result in sympatric speciation?____________________________________________________________________________________________________________________________________________________10. How can polyploidy lead to speciation?____________________________________________________________________________________________________________________________________________________11. Why are allopolyploid hybrids are usually sterile?____________________________________________________________________________________________________________________________________________________12. What did Hugo de Vries discover in the evening primrose?____________________________________________________________________________________________________________________________________________________13. What is thought to be the two factors demonstrating sympatric speciation in the cichlids ofLake Victoria, in East Africa?____________________________________________________________________________________________________________________________________________________14. Compare gradualism and punctuated equilibrium.____________________________________________________________________________________________________________________________________________________15. How does microevolution differ from macroevolution?____________________________________________________________________________________________________________________________________________________16. Identify a couple of factors that could lead to the pattern of evolution we see as divergence?____________________________________________________________________________________________________________________________________________________17. What does the Mollusk eye demonstrate?____________________________________________________________________________________________________________________________________________________18. What does the evolution of the horse demonstrate?____________________________________________________________________________________________________________________________________________________19. Define each of the following evolutionary trends:a. convergent evolution _______________________________________________________________________________________________________________________________b. analogous traits ___________________________________________________________________________________________________________________________________c. parallel evolution __________________________________________________________________________________________________________________________________d. co-evolution _______________________________________________________________________________________________________________________________________

## AP Powerpoints 7th

AP Biology PowerPoints
Biology 7th edition
By
Neil A. Campbell, Jane B. Reece, Lawrence G. Mitchell