Biology - Unit 9: Inheritance Patterns PRACTICE TEST
____
1. Two similar chromosomes that you inherit from your parents (one from your mother, one from your father)
are called
a. homologous chromosomes.
c. sex chromosomes.
b. sister chromatids.
d. homozygous alleles.
____
2. A distinguishing characteristic that can be inherited is a(n)
a. cross.
b. allele.
c. gene.
____
3. Which of the following phrases describes the Punnett square in Figure 6.1?
a. 1/4 probability of heterozygous offspring
b. 3/4 probability that dominant trait is
expressed
____
d. trait.
4. Homozygous alleles are_______
a. always inherited together.
b. different forms of the same trait.
c. 3/4 probability of homozygous offspring
d. heterozygous-heterozygous cross
c.
d.
identical forms of the same gene.
identical forms of two different genes.
____
5. Mendel knew that the variations in the offspring generations resulted from his experiments because he
a. allowed plants to cross-pollinate.
c. controlled the fertilization process.
b. ensured that plants self-pollinated.
d. changed the growing conditions.
____
6. Which pair of genes in Figure 6.2 would be most likely to be inherited together?
a. A and B
b. A and C
c. A and D
d. A and E
____
7. Recessive alleles may not be expressed because they are
a. masked by a dominant allele.
c. the most common allele in a population.
b. the least common allele in a population.
d. less likely to have crossing over.
____
8. Mendel's second law of genetics, the law of independent assortment, is one explanation of the
a. random fertilization of gametes.
c. greater strength of dominant alleles.
b. genetic variation within species.
d. final stages of gametogenesis.
____
9. Which of the following events is an important factor in increasing variety among sexually reproducing
organisms?
a. testcross
b. gene linkage
c. crossing over
d. mitosis
____ 10. Imagine two heterozygous parents. Each has a dominant allele B for brown eyes and a recessive allele b for
blue eyes. The phenotypic ratio for brown:blue eyes in their children is
a. 1:2:1.
b. 3:1.
c. 9:3:3:1.
d. 1:3:1.
____ 11. A person who has a disorder caused by a recessive allele is
a. considered a carrier of the disorder.
c. unable to pass the allele to offspring.
b. homozygous for the recessive allele.
d. certain to have offspring with the disorder.
____ 12. The Punnett square in Figure 7.1 shows a cross between two parents who have the genotype Ss for a genetic
disorder caused by a recessive allele. Which of the following will have the genetic disorder?
a. Ss parent
b. Ss offspring
c. SS offspring
d. ss offspring
____ 13. Two parents have the genotype Gg for a genetic disorder caused by a dominant allele. What is the chance that
any of their children will inherit the disorder?
a. 25%
b. 50%
c. 75%
d. 100%
____ 14. Human height occurs in a continuous range because it is affected by the interaction of several genes, making
it a
a. autosomal trait.
b. sex-linked trait.
c. polygenic trait.
d. codominant trait.
____ 15. A female is born with attached earlobes, which is a recessive phenotype. Which of the following genotypes
could her parents have?
a. RR and RR
b. Rr and RR
c. Rr and rr
d. RR and rr
____ 16. What is the main reason that sex-linked disorders are most often observed in males?
a. The X chromosome only has genes for genetic disorders.
b. The Y chromosome cannot have genes that cause genetic disorders.
c. The Y chromosome cannot mask alleles on the X chromosome.
d. The X chromosome has genes only for sex determination.
____ 17. The numbers in Figure 10-1 represent the chromosome number found in each of the dog cells shown. The
processes that are occurring at A and B are ____.
Figure 10-1
a. mitosis and fertilization
b. meiosis and fertilization
c. mitosis and pollination
d. meiosis and pollination
____ 18. A white mouse whose parents are both white produces only brown offspring when mated with a brown
mouse. The white mouse is most probably ____.
a. homozygous recessive
c. homozygous dominant
b. heterozygous
d. haploid
____ 19. In mink, brown fur color is dominant to silver-blue fur color. If a homozygous brown mink is mated with a
silver-blue mink and 8 offspring are produced, how many would be expected to be silver-blue?
a. 0
c. 6
b. 3
d. 8
____ 20. The diagram in Figure 10-2 shows a diploid cell with two homologous pairs of chromosomes. Due to
independent assortment, the possible allelic combinations that could be found in gametes produced by the
meiotic division of this cell are ____.
Figure 10-2
a. Bb, Dd, BB, and DD
b. BD, bD, Bd, and bd
c. BbDd and BDbd
d. Bd and bD only
Figure 10-5
____ 21. What is the genotype of generation 1 in Figure 10-5?
a. II
c. ii
b. Ii
d. I
Figure 10-7
____ 22. What fraction of this cross will be recessive for both traits?
a. 1/2
c. 1/8
b. 1/4
d. 1/16
Figure 10-8
____ 23. In Figure 10-8, which set of chromatids illustrates the result of a single crossover of the homologous
chromosomes?
a. A
c. C
b. B
d. D
____ 24. Suppose an animal is heterozygous AaBb, and the traits are not linked. When meiosis occurs, what is the total
number of possible combinations of gametes that can be made for these traits?
a. 2
c. 6
b. 4
d. 8
____ 25. Mendel took the pollen from a tall pea plant and pollinated the flower of a short pea plant. When he did this,
he removed the male parts of the flower on the short plant. Why was it important that he remove the male
parts from the flower of the short plant?
a. because he wanted to prevent self pollination
b. because the short plant was heterozygous
c. because he wanted to prevent the development of seeds
d. because the short plant was not a true-breeding plant
____ 26. Mendel crossed a true-breeding plant that produced green seeds with a true-breeding plant that produced
yellow seeds to produce an F1 generation. The entire F1 generation produced yellow seeds. Then he crossed
the F1 offspring with each other to produce the F2 generation. From the F2 generation, he counted 6022
yellow seeds.Which of these is the most likely estimate of the number of green seeds he collected from the F2
generation?
a. 0
c. 6000
b. 2000
d. 18000
____ 27. A heterozygous organism is best described as which of these?
a. dominant
c. hybrid
b. genotype
d. true-breeding
Figure 10-9
____ 28. Consider the cell labeled X in Figure 10-9 containing 4 chromosomes. Which of the four cells below it
represents a healthy gamete that could be produced from this cell?
a. A
c. C
b. B
d. D
Use the exhibit to answer the questions that follow.
29. In this plant species, the allele for purple flowers is dominant and the allele for white flowers is recessive.
Write the genotype for the offspring that have white flowers.
30. Write the genotype for the offspring that have purple flowers.
31. What is the genotype of the unknown parent?
Short Answer
32. Which type of interaction is shown by the alleles for flower color?
33. Suppose the first generation pea plants had flowers that were partly red and partly white instead of all pink.
What type of relationship between alleles would this be? Write one statement to support your answer. Use the
word phenotype in your answer.
Biology - Unit 9: Inheritance Patterns PRACTICE TEST
Answer Section
1. ANS: A
PTS: 1
DIF: Level A
REF: act0976aaf18007e102_14
TOP: Ch 6 Test - A
NOT: 978-0-618-78317-5
2. ANS: D
PTS: 1
DIF: Level A
REF: act0976aaf18007e102_38
TOP: Ch 6 Test - A
NOT: 978-0-618-78317-5
3. ANS: B
PTS: 1
DIF: Level A
REF: act0976aaf18007e102_46
STA: FL.NGSSS.SCI.08.9-12.SC.912.N.1.1.6 | FL.NGSSS.SCI.08.9-12.SC.912.L.16.1 |
FL.NGSSS.SCI.08.9-12.SC.912.L.16.2
TOP: Ch 6 Test - A
NOT: 978-0-618-78317-5
4. ANS: C
PTS: 1
DIF: Level A
REF: act0976aaf18007e102_54
STA: FL.NGSSS.SCI.08.9-12.SC.912.L.16.2
TOP: Ch 6 Test - A
NOT: 978-0-618-78317-5
5. ANS: C
PTS: 1
DIF: Level A
REF: act0976aaf18007e102_79
STA: FL.NGSSS.SCI.08.9-12.SC.912.L.16.17
TOP: Ch 6 Test - A
NOT: 978-0-618-78317-5
6. ANS: A
PTS: 1
DIF: Level A
REF: act0976aaf18007e102_70
STA: FL.NGSSS.SCI.08.9-12.SC.912.L.15.15 | FL.NGSSS.SCI.08.9-12.SC.912.L.16.17
TOP: Ch 6 Test - A
NOT: 978-0-618-78317-5
7. ANS: A
PTS: 1
DIF: Level A
REF: act0976aaf18007e102_95
STA: FL.NGSSS.SCI.08.9-12.SC.912.L.16.2
TOP: Ch 6 Test - A
NOT: 978-0-618-78317-5
8. ANS: B
PTS: 1
DIF: Level A
REF: act0976aaf18007e102_103
TOP: Ch 6 Test - A
NOT: 978-0-618-78317-5
9. ANS: C
PTS: 1
DIF: Level A
REF: act0976aaf18007e102_111
TOP: Ch 6 Test - A
NOT: 978-0-618-78317-5
10. ANS: B
PTS: 1
DIF: Level A
REF: act0976aaf18007e102_127
STA: FL.NGSSS.SCI.08.9-12.SC.912.N.1.1.6
TOP: Ch 6 Test - A
NOT: 978-0-618-78317-5
11. ANS: B
PTS: 1
DIF: Level A
REF: act0976aaf18007e110_194
STA: FL.NGSSS.SCI.08.9-12.SC.912.L.16.2
TOP: Ch 7 Test - A
NOT: 978-0-618-78317-5
12. ANS: D
PTS: 1
DIF: Level A
REF: act0976aaf18007e110_210
STA: FL.NGSSS.SCI.08.9-12.SC.912.N.1.1.6 | FL.NGSSS.SCI.08.9-12.SC.912.L.16.1 |
FL.NGSSS.SCI.08.9-12.SC.912.L.16.2
TOP: Ch 7 Test - A
NOT: 978-0-618-78317-5
13. ANS: C
PTS: 1
DIF: Level A
REF: act0976aaf18007e110_223
STA: FL.NGSSS.SCI.08.9-12.SC.912.N.1.1.6 | FL.NGSSS.SCI.08.9-12.SC.912.L.16.2
TOP: Ch 7 Test - A
NOT: 978-0-618-78317-5
14. ANS: C
PTS: 1
DIF: Level A
REF: act0976aaf18007e110_240
STA: FL.NGSSS.SCI.08.9-12.SC.912.L.16.2
TOP: Ch 7 Test - A
NOT: 978-0-618-78317-5
15. ANS: C
PTS: 1
DIF: Level A
REF: act0976aaf18007e110_280
STA: FL.NGSSS.SCI.08.9-12.SC.912.L.16.2 | FL.NGSSS.SCI.08.9-12.SC.912.L.16.4
TOP: Ch 7 Test - A
NOT: 978-0-618-78317-5
16. ANS: C
PTS: 1
DIF: Level A
REF: act0976aaf18007e110_312
TOP: Ch 7 Test - A
NOT: 978-0-618-78317-5
17. ANS: B
Meiosis involves the reduction in chromosome number prior to fertilization.
Feedback
A
B
C
D
If this were mitosis, then step A would not lead to a reduction in chromosome number.
Correct.
Pollination occurs in plants.
Pollination occurs in plants.
PTS: 1
DIF: Bloom's Level C | DOK 3-MOD
REF: 275
NAT: LS_2b
STA: SC.912.L.16.16
TOP: 10-2
18. ANS: A
The most likely scenario is that the white mouse displays the recessive trait. If this is the case, then the white
mouse must be homozygous. If the white mouse were either homozygous dominant or heterozygous, then it
would likely produce white offspring when mated with a brown mouse.
Feedback
A
B
C
D
Correct.
If the white mouse is heterozygous, then this would mean white is dominant and its
brown mate is homozygous recessive. Such a cross would yield about half white and
brown offspring.
If the white mouse were homozygous dominant, then its offspring would have to be
white.
Sperm and eggs are haploid. Individual mice are not.
PTS: 1
DIF: Bloom's Level D | DOK 2-MOD
REF: 277–281
NAT: LS_2b
STA: SC.912.L.16.2
TOP: 10-6
19. ANS: A
The brown mink will always donate a dominant allele to its offspring, which means all the offspring will have
the dominant phenotype. None of the offspring would be silver-blue, which is the recessive color.
Feedback
A
B
C
D
Well done.
Remember, the brown mink is homozygous.
Remember, the brown mink is homozygous.
Brown is dominant to silver-blue.
PTS: 1
DIF: Bloom's Level D | DOK 3-MOD
REF: 277–281
NAT: LS_2b
STA: SC.912.L.16.2
TOP: 10-6
20. ANS: B
Chromosomes separate so that one of each gamete contains one member of each pair. Thus, B and b would
not be found together, nor would D and d. However B can be combined with either D or d, and similarly b can
be combined with either D or d.
Feedback
A
B
C
D
Homologous pairs would separate from one another.
Correct.
Each gamete would have one copy of each homologous pair.
Four combinations are possible.
PTS: 1
DIF: Bloom's Level E | DOK 3-MOD
NAT: LS_2b
STA: SC.912.L.16.2
21. ANS: B
The first generation are all heterozygotes, or Ii.
REF: 275–276
TOP: 10-7
Feedback
A
B
C
D
The individuals in P1 are homozygous.
Well done.
Review pages 277–281.
The individuals have two alleles, not one.
PTS: 1
DIF: Bloom's Level D | DOK 3-MOD
REF: 277–281
NAT: LS_2b
STA: SC.912.L.16.2
TOP: 10-6
22. ANS: D
This is a classic F1 cross for two independent traits. The ratio is 9:3:3:1. Of the sixteen possible combinations,
only one or 1/16 has a genotype with completely recessive alleles.
Feedback
A
B
C
D
Check page 282.
See page 282.
Only one in 16 are expected to have both recessive traits.
Well done.
PTS: 1
DIF: Bloom's Level C | DOK 3-MOD
REF: 277–281
NAT: LS_2b
STA: SC.912.L.16.2
TOP: 10-6
23. ANS: A
A single crossover will cause B and b to change positions relative to A and a on just two chromatids. The
other two chromatids will be unaffected, one would remain AB and the other aa. The chromatids that
experience the crossover will contain Ab and aB.
Feedback
A
B
C
D
This is correct.
This shows the effect of zero crossovers.
This shows the effect of two crossovers.
There should be 2 copies of B.
PTS: 1
DIF: Bloom's Level E | DOK 3-MOD
NAT: LS_2b
STA: SC.912.L.16.16
24. ANS: B
There are four combinations: AB, Ab, aB, and ab.
REF: 283–284
TOP: 10-8
Feedback
A
B
C
D
Remember the traits are not linked.
Correct.
Check page 275.
Find all possible combinations of A with B and b and a with B and b.
PTS: 1
DIF: Bloom's Level D | DOK 3-MOD
REF: 275
NAT: LS_2b
STA: SC.912.L.16.2
TOP: 10-5
25. ANS: A
He removed the male parts on the short plant because he wanted to avoid letting the plant self-pollinate. He
wanted the pollen to come only from a tall plant.
Feedback
A
B
C
D
Correct
Both plants were true-breeding homozygous.
Mendel used the seeds.
Mendel used only true-breeding plants.
PTS: 1
DIF: Bloom's Level D | DOK 3-MOD
REF: 277
NAT: IS_1a
STA: SC.912.L.16.1
TOP: 10-4
26. ANS: B
In the F2 generation, we expect a 3 to 1 ratio of yellow to green seeds. Thus, if there are approximately 6000
yellow seeds, then we expect about 2000 green seeds.
Feedback
A
B
C
D
Check page 278.
Correct.
Try again.
There should be 3 times more yellow seeds than green seeds.
PTS: 1
DIF: Bloom's Level D | DOK 3-MOD
REF: 278
NAT: LS_2b
STA: SC.912.L.16.1
TOP: 10-4
27. ANS: C
A heterozygous (Aa) individual is produced by a cross of two different true-breeding (AA and aa)
individuals.Thus the heterozgous individual is a hybrid.
Feedback
A
B
C
D
This description refers to a trait or allele.
This description refers to the genetic makeup of an individual.
Correct.
This refers to a homozygous individual.
PTS: 1
DIF: Bloom's Level B | DOK 2-MOD
REF: 279
NAT: LS_2b
STA: SC.912.L.16.2
TOP: 10-5
28. ANS: A
The healthy gamete would have half the original number of chromatids. In this case, the number is A.
Feedback
A
B
C
D
Correct.
The chromatids in the gamete would not be joined at the centromere.
There should be two chromatids in the gamete.
Try again.
PTS: 1
NAT: LS_2b
29. ANS:
DIF: Bloom's Level C | DOK 3-MOD
STA: SC.912.L.16.16
REF: 275
TOP: 10-1
ff
PTS: 1
DIF: Level A
TOP: Ch 6 Test - A
30. ANS:
Ff
REF: act0976aaf18007e102_176
NOT: 978-0-618-78317-5
PTS: 1
DIF: Level A
TOP: Ch 6 Test - A
31. ANS:
Ff
REF: act0976aaf18007e102_180
NOT: 978-0-618-78317-5
PTS: 1
32. ANS:
1 = C1C1, 2 = C1C2, 3 = C1C2, 4 = C2C2; incomplete dominance
PTS: 1
DIF: Level A
REF: act0976aaf18007e110_354
STA: FL.NGSSS.SCI.08.9-12.SC.912.L.16.2
TOP: Ch 7 Test - A
NOT: 978-0-618-78317-5
33. ANS:
Codominance. The phenotypes of both parents are expressed fully and separately in the offspring.
PTS: 1
DIF: Level A
REF: act0976aaf18007e110_362
STA: FL.NGSSS.SCI.08.9-12.SC.912.L.16.2
TOP: Ch 7 Test - A
NOT: 978-0-618-78317-5