Quiz 2 T2 AP Biology Chapters 11 and 12
Name __________________________Date___________
Multiple Choice Questions (2 points each)
1. In garden peas, a single gene controls stem length. The recessive allele (t) produces short stems
when homozygous. The dominant allele (T) produces long stems. A short-stemmed plant is crossed
with a heterozygous long-stemmed plant. Which of the following represents the expected
phenotypes of the offspring and the ratio in which they will occur?
(A) 3 long-stemmed plants: 1 short-stemmed plant
(B) 1 long-stemmed plant: 1 short-stemmed plant
(C) 1 long-stemmed plant: 3 short-stemmed plants
(D) Long-stemmed plants only
B
Questions 2-3
A male fruit fly (Drosophila melanogaster) with red eyes and long wings was mated with a female
with purple eyes and vestigial wings. All of the offspring in the F1 generation had red eyes and long
wings.
These F1 flies were test crossed with purple-eyed, vestigial-winged flies. Their offspring, the F2
generation, appeared as indicated below.
F2 Generation
125 red eyes, long wings
purple eyes, vestigial
124
wings
18 purple eyes, long wings
16 red eyes, vestigial wings
283 Total
2.
If in the F1 and F2 generations the same characteristics appeared in both males and females, it would
be safe to assume that these traits for eye color and wing length
(A) are sex-linked
(B) vary in dominance according to sex
(C) are sex-influenced characteristics
(D) are autosomal characteristics
D
3.
In the F2 generation, the results are best explained by the fact that
(A) the test cross with the F1 flies resulted in sterile offspring
(B) these genes for eye color and wing shape do not pass through the F1 generation
(C) these genes for eye color and wing shape are found on the same chromosome
(D) crossing-over decreases variability
C
4. An allele on the X chromosome is responsible for the recessive trait of red-green color blindness
in humans. The daughter of a woman who is color-blind has normal vision and marries a man
who is color-blind. This couple has a son, what is the probability that the son will be color-blind?
(A) 0
(B) 1/4
(C) 1/2
(D) 3/4
C
5. In the pedigree above, circles represent females, squares represent males, and shaded figures represent
individuals expressing a specific trait. The expression of this trait is most likely due to which of the
following?
(A) Sex-linked dominant inheritance
(B) Sex-linked recessive inheritance
(C) Autosomal dominant inheritance
(D) Autosomal recessive inheritance
D
Directions: Each group of questions below consists of five lettered headings followed by a list of
numbered phrases or sentences. For each numbered phrase or sentence, select the one heading to which it
is most closely related and fill in the corresponding circle on the answer sheet. Each heading may be used
once, more than once, or not at all in each group.
Questions 6-9 refer to the following genetic crosses, in which R and T represent dominant
alleles and r and t represent recessive alleles for two genes that are located on different
chromosomes.
(A) TtRr x ttRr
(B) TtRr x TtRr
(C) TTRr x TtRR
(D) TtRr x ttrr
6. Offspring from this cross are expected to exhibit a 1 : 1 : 1 : 1 phenotypic ratio.
D
7. Offspring from this cross are expected to exhibit a 3 : 3 : 1 : 1 phenotypic ratio.
A
8. All offspring from this cross are expected to exhibit both dominant traits.
C
9. This cross is an example of a dihybrid cross in which both parents are heterozygous.
B
10. If a single locus controls wing shape, then the alleles for this gene act as
(A)
(B)
(C)
(D)
dominant-recessive alleles
incomplete-dominance alleles
codominant alleles
multiple alleles
A
Questions 11-13 refer to the following information.
Achondroplastic dwarfism is a dominant genetic trait that causes severe malformation of the skeleton.
Homozygotes for this condition are spontaneously aborted (hence, the homozygous condition is lethal)
but heterozygotes will develop to be dwarfed.
Matthew has a family history of the condition, although he does not express the trait. Jane is an
achondroplastic dwarf. Matthew and Jane are planning a family of several children and want to know the
chances of producing a child with achondroplastic dwarfism.
11. The genotypes of Matthew and Jane are best represented as
Matthew
Jane
(A)
AA
Aa
(B)
Aa
aa
(C)
aa
aa
(D)
aa
Aa
D
12. The probability that Matthew and Jane’s first child will be an achondroplastic dwarf is
(A) 0%
(B) 25%
(C) 50%
(D) 75%
C
13. If three children are born to Matthew and Jane, what are the chances that the first two children will not
express the trait but that the third child will be an achondroplastic dwarf?
(A) 5/8
(B) 4/8
(C) 3/8
(D) 1/8
D
Questions 14-15 refer to the following genetic terms
(A) Codominance
(B) Incomplete Dominance
(C) Multiple alleles
(D) Pleitotropy
14. A red flower and a white flower produce a pink flower
B Incomplete dominance shows as a phenotypic blending of traits.
15. A red flower and a white flower produce a red and white streaked flower
A In codominant inheritance both alleles are expressed.
16. Hemophilia is a sex-linked recessive trait. A male hemophiliac and phenotypically normal female
have a girl that is a hemophiliac. All of the following statements are correct EXCEPT:
(A) The daughter inherited a recessive gene from each parent.
(B) The daughter inherited a dominant allele from her mother.
(C) The mother is a carrier of hemophilia
(D) The genotype of the mother is XH Xh
B
17. What is the likelihood that a couple will have 4 offspring all of which are female?
(A) 1/16
(B) ¼
(C) 1/3
(D) ¾
A
18. Assume that a cross is made between AaBb and aabb plants and all of the offspring are
eitherAaBb or aabb. These results are consistent with the following circumstance:
A. Complete linkage
B. Alternation of generations
C. Codominance
D. Incomplete dominance
A
Reason: Genes that are completely linked are so close together that recombination rarely
if ever occurs.
19. Assume that a cross is made between AaBb plants and aabb plants and the offspring
occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are
consistent with the following circumstance:
A. Sex-linked inheritance with 30% crossing over
B. Linkage with 50% crossing over
C. Linkage with approximately 33 map units between the two gene loci
D. Independent assortment
C
Reason: 48 + 52/ 300 = 33% = 33 map units
20. The recessive mutations s (scute bristles) and rb (ruby eyes) identify two linked
autosomal genes of Drosophila melanogaster. When females heterozygous for these
genes were crossed with scute bristled, ruby eyed males, the following classes and
numbers of progeny (out of 1000) were obtained:
wildtype bristles, wildtype eyes 188
scute bristles, wildtype eyes 307
wildtype bristles, ruby eyes 313
scute bristles, ruby eyes 192
Based upon these results, the map distance between the s and rb genes is estimated to be:
A. 31.3 map units
B. 38 map units
C. 30.7 map units
D. greater than 50 units because all four classes of offspring were observed
B
Free Response Questions
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.
A. What is the simplest explanation for the inheritance of these colors in chickens? (5 points)
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
B. What offspring would you predict from the mating of a gray rooster and a black hen? Punnett square to
answer (5 points)
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 )
2. In some plants, a true-breeding, red-flowered strain gives all pink flowers when crossed with a whiteflowered strain: RR (red) x rr (white) ---> Rr (pink).
A. 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 terminalwhite? (5 points)
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 axialpink.
B. What will be the ratios in the F2 generation? Use the Punnett square to answer(5 points)
The ratio of genotypes can be determined by examining the Punnett square below:
The ratio of phenotypes will be:
6 axial-pink
3 axial-red
3 axial-white
8 pink
4 red
4 white
2 terminal-pink
1 terminal-white
1 terminal-red
12 axial
4 terminal
3. The recessive mutations m (mahogany eyes) and eb (ebony body) identify two autosomal
linked genes of Drosophila melanogaster. When females heterozygous for these genes
were crossed with mahogany eyed, ebony bodied males, the following classes and
numbers of progeny (out of 1000) were obtained:
wildtype eyes, wildtype body 370
mahogany eyes, wildtype body 130
wildtype eyes, ebony body 120
mahogany eyes, ebony body 380
Based upon these results, the map distance between the m and eb genes is estimated to
be:
Show all your work Answer: 25 map units.
Reason: 130 + 120/1000 = 250/1000 = 25%
4. A newly identified fruit fly mutant, cyclops eye (large and single in the middle of the head), is
hypothesized to be autosomal dominant. The experimenter started with homozygous wild type
females (yes, virgins) and homozygous cyclops males. The data from the F2 generation was 44
wild type males, 60 wild type females, 110 cyclops males and 150 cyclops females. Does this
data support or reject the hypothesis? Use chi square to prove your position. Once you have your
answer check my answer below.
Answer
The chi square for the cyclops eye problem approximates 2.4 (depending on how your round
your numbers) with one degree of freedom. The difference is insignificant and the hypothesis is
accepted.
Explanation:
Since it is hypothesized that cyclops eye is autosomal we can lump the wild type males and wild
type females together (likewise for the cyclops flies). Therefore, we observed:
104 wild type flies and 260 cyclops flies
The expecteds for this problem are determined from the total number of flies observed
44 + 60 + 110 + 150 = 364
If Cyclops eye is in fact autosomal dominant 1/4 of the F2 generation should be wild type and
3/4 of the F2 generation should be cyclops eye. Therefore:
1/4 x 364 = 91 wild type flies expected, and
3/4 x 364 = 273 cyclops eye flies.
Filling out the X2 equation then:
X2 = [(expected wild - observed wild)^2]/expected wild + [(expected cyclops - observed
cyclops)^2]/expected cyclops
X2 = [(91-104)^2]/91 + [(273-260)^2]/273
X2 = [(-13)^2]/91 + [(13)^2]/273
X2 =169/91 + 169/273
X2 =1.8 + 0.6
X2 = 2.4
Since there are two groups of data (and two terms) there are 2-1 = 1 degree of freedom.
From the Chi-square table the standard must frequently used to divide significant from
insignificant is 0.05 (5%) we can say that this deviation (difference) is insignificant.