AP® BIOLOGY

BIOLOGY

2010 SCORING GUIDELINES

Visit the College Board on the Web: www.collegeboard.com.

Question 3

A new species of fly was discovered on an island in the South Pacific. Several different crosses

were performed, each using 100 females and 100 males. The phenotypes of the parents and the

resulting offspring were recorded.

Cross I: True-breeding bronze-eyed males were crossed with true-breeding red-eyed females. All

the F

offspring had bronze eyes. F

flies were crossed, and the data for the resulting F

flies are

given in the table below.

Phenotype

Male F

emale

Bronze eyes 3,720 3,800

Red eyes 1,260 1,320

Cross II: True-breeding normal-winged males were crossed with true-breeding stunted-winged

females. All the F

offspring had stunted wings. F

flies were crossed, and the data for the

resulting F

flies are given in the table below.

Phenotype Male Female

Normal wings 1,160 1,320

Stunted wings 3,600 3,820

Cross III: True-breeding bronze-eyed, stunted-winged males were crossed with true-breeding red-

eyed, normal-winged females. All the F

offspring had bronze eyes and stunted wings. The F

flies

were crossed with true-breeding red-eyed, normal-winged flies, and the results are shown in the

table below.

Phenotype M

ale Female

Bronze eyes, stunted wings 2,360 2,220

Bronze eyes, normal wings 220 300

Red eyes, stunted wings 260 220

Red eyes, normal wings 2,240 2,180

BIOLOGY

2010 SCORING GUIDELINES

Visit the College Board on the Web: www.collegeboard.com.

Question 3 (continued)

(a) What conclusions can be drawn from cross I and cross II? Explain how the data support your

conclusions for each cross. (4 points maximum)

Conclusion for cross I

(1 point maximum)

Possible explanations for cross I

(1 point maximum)

• Bronze dominant/red recessive

• Autosomal (non-sex-linked)

• All F

/heterozygotes express dominant trait

(bronze).

• F

shows 3:1 ratio

(bronze:red/dominant:recessive).

• Equal distribution of F

phenotypes for both

genders.

Conclusion for cross II

(1 point maximum)

Possible explanations for cross II

(1 point maximum)

• Stunted dominant/normal recessive

• Autosomal (non-sex-linked)

• All F

/heterozygotes express dominant trait

(stunted).

• F

shows 3:1 ratio

(stunted:normal/dominant:recessive).

• Equal distribution of F

phenotypes for both

genders.

(b) What conclusions can be drawn from the data from cross III? Explain how the data support your

conclusions. (4 points maximum)

Conclusion for cross III

(1 point per bullet; 2 points maximum)

Explanation for cross III

(1 point per bullet; 2 points maximum)

• Genes linked

• Crossing over

• Genes 10 map units apart

• Not a 1:1:1:1 ratio (as predicted by

independent assortment).

• Not a 1:1 ratio/two recombinant

phenotypes (unexpected).

• Frequency of recombinant phenotypes

was 10 percent (setup equation

OK)/parental phenotypes

(bronze/stunted and red/normal) are

represented in 90 percent of offspring.

BIOLOGY

2010 SCORING GUIDELINES

Visit the College Board on the Web: www.collegeboard.com.

Question 3 (continued)

(c)

Identify and discuss TWO different factors that would affect whether the island’s fly population is in

Hardy-Weinberg equilibrium for the traits above. (4 points maximum)

dentification

(1 point per bullet; 2 points maximum)

Discussion of effect

(1 point per bullet; 2 points maximum)

• Large population • Minimized genetic drift.

• Random mating • No gene pool change due to mate

preferences.

• No mutation • No new alleles in population.

• No immigration/emigration/

migration (no gene flow)

• No gene pool change by addition/loss

of alleles.

• No natural selection • No alleles favored or disfavored by

environment.

Visit the College Board on the Web: www.collegeboard.com.

BIOLOGY

2010 SCORING COMMENTARY

Visit the College Board on the Web: www.collegeboard.com.

Question 3

Overview

This question offered an opportunity to demonstrate fundamental knowledge about

the Mendelian

inheritance of single gene traits with complete dominance as well as the opportunity to recognize and

explain the effects of gene linkage on phenotype. The question further provided an opportunity to project

an understanding of genetics from the level of individual flies to the level of population genetics by

discussing the effects of genetic change on Hardy-Weinberg equilibrium. Data tables containing the

phenotypic results of three different fly crosses were provided. Cross I showed the F

data of a cross

between two heterozygotes for eye color (bronze versus red). Cross II showed the F

data of a cross

between two heterozygotes for wing type (stunted versus normal wings). Both sets of data indicated a

typical autosomal dominant form of inheritance. In part (a) students were asked to draw conclusions from

the cross I and cross II data and then explain how the data supported their conclusions. Data from a third

cross showed the results of crossing a heterozygote for both traits with a fly that was recessive for both.

The data clearly indicated linkage between the genes for eye color and wing type. In part (b) students were

again asked to draw conclusions from the data and to explain how the data supported their conclusions. In

part (c) students were asked to identify and discuss two factors that would affect the Hardy-Weinberg

equilibrium of the fly population.

Sample: 3A

Score: 10

In part (a) 1 point was earned for the conclusion that in cross I “bronze eyes are the dominant trait.” One

point was earned fo

r the explanation of the cross I conclusion by stating that the cross between homozygous

parents “produces an F

generation w/ all bronze eyes.” The response earned the 2-point maximum from

cross I but also could have been awarded a point for noting that the F

generation had “an approximate 3:1

ratio of bronze eyes to red eyes.” For cross II, 1 point was earned for the conclusion that “stunted wings are

dominant.” One point was earned for the explanation that the parental cross “leads to an F

generation of

solely stunted-winged offspring.” Again, the 2-point maximum was reached for this section; however,

another point could have been earned for explaining how the F

data support the conclusion of the

dominance of stunted wings. Furthermore, if the maximum points for part (a) had not already been earned, all

4 points in part (a) could have been earned by the response that “[b]oth crosses also show that … eye color

and wing shape are not sex linked as equal ratios of each … trait appear in male and female flies.”

In part (b) 1 point was earned for the statement that “crossing over occurs between these two loci,” and

1 point was earned

with the response that these loci “are probably on the same chromosome.” After stating

and demonstrating with a Punnett square that the expected phenotype ratio for cross III should be 1:1:1:1,

the student notes that, “[h]owever, the F

offspring have way more bronze/stunted and red/normal flies when

compared with bronze/normal and red/stunted flies.” This response earned 2 points: 1 point for noting that

there was not a 1:1:1:1 ratio as would be predicted by independent assortment, and 1 point for explaining

that the frequency of parental phenotypes is much greater than that of the recombinant phenotypes.

In part (c) 2 points were earned for identifying “population size and the amount of gene flow” as two factors

that affect Hardy-Weinberg equilibrium

. An additional point could have been awarded for discussion of the

effect of gene flow on the population, had the response not already earned the maximum 10 points.

BIOLOGY

2010 SCORING COMMENTARY

Question 3 (continued)

Sample: 3B

Score: 8

In part (a) 1 point was earned for the cross I conclusion that “the bronze eyed allele is do

minant.” One point

was earned for the explanation that when the heterozygous offspring of the original parents mated, the F

“ratio of phenotypes would be 3:1, 3 being the dominant phenotype, and the ratio of the results was

roughly 3 bronze:1 red.” One point was earned for the cross II conclusion that “stunted wings were

dominant,” and 1 point was earned for the explanation that “[a]ll of the F

generation … had stunted

wings.”

No points were earned in part (b).

In part (c) 1 point was earned for identifying that mutations could affect Hardy-Weinberg equilibrium, and

1 point was earned for explaining that a mutation such as “white eyes” could introduce new genes into the

gene pool. One point was earned for identifying random mating as a second factor that could affect Hardy-

Weinberg equilibrium, and 1 point was earned for the discussion that “[i]f random mating does not occur,

… [t]he allele that doesn’t attract mates will begin to become less frequent in the population … and

equilibrium will be thrown off.”

Sample: 3C

Score: 6

In part (a) 1 point was earned for explaining the conclusion to

cross I by stating that “in F

all of them had

bronze eyes,” and 1 point was earned for the cross I conclusion with the statement, “so that must be the

dominant allele.” One point was earned for explaining the cross II conclusion — “in F

all the offspring had

stunted wings” — and 1 point was earned for the cross II conclusion, “so that must be the dominant allele.”

No points were earned in part (b).

In part (c)

2 points were earned for identifying “no mutations” and “random mating” as two factors that

affect Hardy-Weinberg equilibrium.

Visit the College Board on the Web: www.collegeboard.com.