Biology Practice Test
Biology Test- Chapter 11: Introduction to Genetics TPC Snoqualmie Ridge
Multiple Choice
Identify the choice that best completes the statement or answers the question.
1. Gregor Mendel used pea plants to study
|
a. |
flowering. |
|
b. |
gamete formation. |
|
c. |
the inheritance of traits. |
|
d. |
cross-pollination. |
2. Offspring that result from crosses between true-breeding parents with different traits
|
a. |
are true-breeding. |
|
b. |
make up the F2 generation. |
|
c. |
make up the parental generation. |
|
d. |
are called hybrids. |
3. The chemical factors that determine traits are called
|
a. |
alleles. |
|
b. |
traits. |
|
c. |
genes. |
|
d. |
characters. |
4. Gregor Mendel concluded that traits are
|
a. |
not inherited by offspring. |
|
b. |
inherited through the passing of factors from parents to offspring. |
|
c. |
determined by dominant factors only. |
|
d. |
determined by recessive factors only. |
5. When Gregor Mendel crossed a tall plant with a short plant, the F1 plants inherited
|
a. |
an allele for tallness from each parent. |
|
b. |
an allele for tallness from the tall parent and an allele for shortness from the short parent. |
|
c. |
an allele for shortness from each parent. |
|
d. |
an allele from only the tall parent. |
6. The principle of dominance states that
|
a. |
all alleles are dominant. |
|
b. |
all alleles are recessive. |
|
c. |
some alleles are dominant and others are recessive. |
|
d. |
alleles are neither dominant nor recessive. |
7. When Gregor Mendel crossed true-breeding tall plants with true-breeding short plants, all the offspring were tall because
|
a. |
the allele for tall plants is recessive. |
|
b. |
the allele for short plants is dominant. |
|
c. |
the allele for tall plants is dominant. |
|
d. |
they were true-breeding like their parents. |
8. A tall plant is crossed with a short plant. If the tall F1 pea plants are allowed to self-pollinate,
|
a. |
the offspring will be of medium height. |
|
b. |
all of the offspring will be tall. |
|
c. |
all of the offspring will be short. |
|
d. |
some of the offspring will be tall, and some will be short. |
9. The principles of probability can be used to
|
a. |
predict the traits of the offspring produced by genetic crosses. |
|
b. |
determine the actual outcomes of genetic crosses. |
|
c. |
predict the traits of the parents used in genetic crosses. |
|
d. |
decide which organisms are best to use in genetic crosses. |
10. In the P generation, a tall plant is crossed with a short plant. The probability that an F2 plant will be tall is
|
a. |
50%. |
|
b. |
75%. |
|
c. |
25%. |
|
d. |
100%. |
11. Organisms that have two identical alleles for a particular trait are said to be
|
a. |
hybrid. |
|
b. |
homozygous. |
|
c. |
heterozygous. |
|
d. |
dominant. |
|
Tt |
||||
|
T |
t |
|||
|
TT |
T |
TT |
Tt |
|
|
T |
TT |
Tt |
||
|
T |
= |
tall |
|
t |
= |
short |
Figure 11-1
12. In the Punnett square shown in Figure 11-1, which of the following is true about the offspring resulting from the cross? (Tt x TT)
|
a. |
About half are expected to be short. |
|
b. |
All are expected to be short. |
|
c. |
About half are expected to be tall. |
|
d. |
All are expected to be tall. |
13. The genotypic ratio of the offspring in Figure 11-1 is:
|
a. |
2TT:2Tt |
c. |
1TT:2Tt:1tt |
|
b. |
2tall:2short |
d. |
3tall:1short |
14. The phenotypic ratio of the offspring in Figure 11-1 is:
|
a. |
2TT:2Tt |
c. |
1TT:2Tt:1tt |
|
b. |
2tall:2short |
d. |
4 tall |
15. A Punnett square shows all of the following EXCEPT
|
a. |
all possible results of a genetic cross. |
|
b. |
the genotypes of the offspring. |
|
c. |
the alleles in the gametes of each parent. |
|
d. |
the actual results of a genetic cross. |
16. If you made a Punnett square showing Gregor Mendel’s cross between true-breeding tall plants and true-breeding short plants, the square would show that the offspring had
|
a. |
the genotype of one of the parents. |
|
b. |
a phenotype that was different from that of both parents. |
|
c. |
a genotype that was different from that of both parents. |
|
d. |
the genotype of both parents. |
17. What principle states that during gamete formation genes for different traits separate without influencing each other’s inheritance?
|
a. |
principle of dominance |
|
b. |
principle of independent assortment |
|
c. |
principle of probabilities |
|
d. |
principle of segregation |
18. How many different allele combinations would be found in the gametes produced by a pea plant whose genotype was RrYY?
|
a. |
2 |
|
b. |
4 |
|
c. |
8 |
|
d. |
16 |
19. If a pea plant that is heterozygous for round, yellow peas (RrYy) is crossed with a pea plant that is homozygous for round peas but heterozygous for yellow peas (RRYy), how many different phenotypes are their offspring expected to show?
|
a. |
2 |
|
b. |
4 |
|
c. |
8 |
|
d. |
16 |
20. Situations in which one allele for a gene is not completely dominant over another allele for that gene are called
|
a. |
multiple alleles. |
|
b. |
incomplete dominance. |
|
c. |
polygenic inheritance. |
|
d. |
multiple genes. |
21. A cross of a red cow (RR) with a white bull (WW) produces all roan offspring (RRWW). This type of inheritance is known as
|
a. |
incomplete dominance. |
|
b. |
polygenic inheritance. |
|
c. |
codominance. |
|
d. |
multiple alleles. |
22. The number of chromosomes in a gamete is represented by the symbol
|
a. |
Z. |
|
b. |
X. |
|
c. |
N. |
|
d. |
Y. |
23. If an organism’s diploid number is 12, its haploid number is
|
a. |
12. |
|
b. |
6. |
|
c. |
24. |
|
d. |
3. |
24. Gametes have
|
a. |
homologous chromosomes. |
|
b. |
twice the number of chromosomes found in body cells. |
|
c. |
two sets of chromosomes. |
|
d. |
one allele for each gene. |
25. Gametes are produced by the process of
|
a. |
mitosis. |
|
b. |
meiosis. |
|
c. |
crossing-over. |
|
d. |
replication. |
Figure 11-3
26. What is shown in Figure 11-3? (Figure 11-16 in your book)
|
a. |
independent assortment |
|
b. |
anaphase I of meiosis |
|
c. |
crossing-over |
|
d. |
replication |
27. Chromosomes form tetrads during
|
a. |
prophase of meiosis I. |
|
b. |
metaphase of meiosis I. |
|
c. |
interphase. |
|
d. |
anaphase of meiosis II. |
28. What happens between meiosis I and meiosis II that reduces the number of chromosomes?
|
a. |
Crossing-over occurs. |
|
b. |
Metaphase occurs. |
|
c. |
Replication occurs twice. |
|
d. |
Replication does not occur. |
29. Unlike mitosis, meiosis results in the formation of
|
a. |
diploid cells. |
|
b. |
haploid cells. |
|
c. |
2N daughter cells. |
|
d. |
body cells. |
30. Unlike mitosis, meiosis results in the formation of
|
a. |
two genetically identical cells. |
|
b. |
four genetically different cells. |
|
c. |
four genetically identical cells. |
|
d. |
two genetically different cells. |
31. In a 2 factor cross where both parents are heterozygous for both traits (TtYy x TtYy), the expected phenotypic ratio would be:
|
a. |
1:1:1:1 |
c. |
3:1 |
|
b. |
12:4 |
d. |
9:3:3:1 |
32. When you flip a coin, what is the probability that it will come up tails?
|
a. |
1/2 |
|
b. |
1/4 |
|
c. |
1/8 |
|
d. |
1 |
33. The wide range of skin colors in humans comes about because more than four different genes control this trait. This is an example of:
|
a. |
multiple alleles |
c. |
codominance |
|
b. |
polygenic traits |
d. |
incomplete dominance |
34. Human blood type alleles of A and B are equally dominant to each other and are both expressed. This is an example of:
|
a. |
codominance |
c. |
polygenic traits |
|
b. |
incomplete dominance |
d. |
multiple alleles |
35. Human blood types are produced by alleles A, B, and O. Having more than 2 alleles control a trait is called:
|
a. |
incomplete dominance |
c. |
polygenic traits |
|
b. |
codominance |
d. |
multiple alleles |
36. When the heterozygous phenotype is a combination or an intermediate of the two homozygous phenotypes, it is called
|
a. |
incomplete dominance |
c. |
polygenic traits |
|
b. |
codominance |
d. |
multiple alleles |
37. If the sex cell of an organism has 20 chromosomes, then the body cells will have:
|
a. |
20 chromosomes |
c. |
15 chromosomes |
|
b. |
10 chromosomes |
d. |
40 chromosomes |
True/False
Indicate whether the statement is true-A or false-B.
38. A trait is a specific characteristic that varies from one individual to another. _________________________
39. An organism with a dominant allele for a particular form of a trait will sometimes show that trait. _________________________
40. When alleles segregate from each other, they join. _________________________
41. The probability that a gamete produced by a pea plant heterozygous for stem height (Tt) will contain the recessive allele is 100%. _________________________
42. Mitosis results in two cells, whereas meiosis results in one cell. _________________________
43. The different forms of a gene are called:
|
a. |
factors |
c. |
traits |
|
b. |
alleles |
d. |
gametes |
44. When two heterozygous tall pea plants are crossed, the expected genotype ratio of the offspring is:
|
a. |
3:1 |
c. |
2:2 |
|
b. |
1:2:1 |
d. |
4:0 |
45. The principle of independent assortment states that ____________________ for different traits can segregate independently during the formation of gametes.
|
a. |
sex cells |
c. |
characteristics |
|
b. |
chromosomes |
d. |
genres |
46. If pea plants that are homozygous for round, yellow seeds (RRYY) were crossed with pea plants that are heterozygous for round, yellow seeds (RrYy), the expected phenotype(s) of the offspring would be:
|
a. |
round, yellow |
c. |
round, green |
|
b. |
wrinkled, yellow |
d. |
wrinkled, green |
47. Red, white, and pink phenotypes in flowers called four o’clocks are an example of:
|
a. |
codominance |
c. |
incomplete dominance |
|
b. |
multiple alleles |
d. |
incomplete dominance |
48. An organism’s _______ have half the number of chromosomes found in the organism’s body cells.
|
a. |
gametes |
c. |
phenotype |
|
b. |
genotype |
d. |
diploid cells |
49. Genetics is:
|
a. |
the study of genes |
c. |
study of traits |
|
b. |
the study of heredity |
d. |
study of Gregor Mendel |
50. How many recessive alleles for a trait must an organism inherit in order to show that trait?
|
a. |
none |
c. |
2 |
|
b. |
1 |
d. |
4 |
Essay
51. You wish to determine whether a tall pea plant is homozygous or heterozygous for tallness. What cross should you perform to arrive at your answer? Explain your choice of cross.
52. Explain the difference between incomplete dominance and codominance.
53. Contrast the cells produced by mitosis with those produced by meiosis.
54. Complete the following dihybrid cross: GgRr x GgRr. Make the Punnett square and list the phenotype ratios.
Biology Test- Chapter 11: Introduction to Genetics
Answer Section
MULTIPLE CHOICE
1. ANS: CPTS: 1DIF: BOBJ: 11.1.1
2. ANS: DPTS: 1DIF: AOBJ: 11.1.1
3. ANS: CPTS: 1DIF: BOBJ: 11.1.2
4. ANS: BPTS: 1DIF: AOBJ: 11.1.2
5. ANS: BPTS: 1DIF: EOBJ: 11.1.2
6. ANS: CPTS: 1DIF: BOBJ: 11.1.3
7. ANS: CPTS: 1DIF: AOBJ: 11.1.3
8. ANS: DPTS: 1DIF: BOBJ: 11.1.4
9. ANS: APTS: 1DIF: AOBJ: 11.2.1
10. ANS: BPTS: 1DIF: EOBJ: 11.2.1
11. ANS: BPTS: 1DIF: BOBJ: 11.2.2
12. ANS: DPTS: 1DIF: EOBJ: 11.2.2
13. ANS: APTS: 1
14. ANS: DPTS: 1
15. ANS: DPTS: 1DIF: AOBJ: 11.2.2
16. ANS: CPTS: 1DIF: EOBJ: 11.2.2
17. ANS: BPTS: 1DIF: BOBJ: 11.3.1
18. ANS: APTS: 1DIF: AOBJ: 11.3.1
19. ANS: APTS: 1DIF: EOBJ: 11.3.1
20. ANS: BPTS: 1DIF: BOBJ: 11.3.2
21. ANS: CPTS: 1DIF: AOBJ: 11.3.2
22. ANS: CPTS: 1DIF: BOBJ: 11.4.1
23. ANS: BPTS: 1DIF: AOBJ: 11.4.1
24. ANS: DPTS: 1DIF: EOBJ: 11.4.1
25. ANS: BPTS: 1DIF: BOBJ: 11.4.2
26. ANS: CPTS: 1DIF: AOBJ: 11.4.2
27. ANS: APTS: 1DIF: AOBJ: 11.4.2
28. ANS: DPTS: 1DIF: EOBJ: 11.4.2
29. ANS: BPTS: 1DIF: BOBJ: 11.4.3
30. ANS: BPTS: 1DIF: AOBJ: 11.4.3
31. ANS: DPTS: 1
32. ANS: APTS: 1DIF: BOBJ: 11.2.1
33. ANS: BPTS: 1
34. ANS: APTS: 1
35. ANS: APTS: 1
36. ANS: APTS: 1
37. ANS: DPTS: 1
MODIFIED TRUE/FALSE
38. ANS: T
PTS: 1DIF: BOBJ: 11.1.1
39. ANS: F, always
PTS: 1DIF: BOBJ: 11.1.3
40. ANS: F, separate
PTS: 1 DIF: B OBJ: 11.1.4
41. ANS: F
50%
PTS: 1DIF: EOBJ: 11.2.1
42. ANS: F, four cells
PTS: 1DIF: BOBJ: 11.4.3
COMPLETION
43. ANS: b. alleles
PTS:1 DIF: B11.1.2
44. ANS: b. 1 TT : 2 Tt : 1 tt
PTS:1 DIF: E11.2.2
45. ANS: d. genes
PTS:1 DIF: B11.3.1
46. ANS: a. round, yellow seeds
PTS:1 DIF: E11.3.1
47. ANS: c. incomplete dominance
PTS:1 DIF: E11.3.2
48. ANS: a. gametes
PTS:1 DIF: A11.4.1
SHORT ANSWER
49. ANS:
b. study of heredity.
PTS:1 DIF: B11.1.1
50. ANS:
d. two recessive alleles
PTS:1 DIF: A11.1.3
ESSAY
51. ANS:
The tall pea plant should be crossed with a short pea plant. If the tall pea plant is homozygous, all of the offspring will be tall. If the tall pea plant is heterozygous, it is likely that about half of the offspring will be tall and half will be short.
PTS:1 DIF: A11.2.2
52. ANS:
In incomplete dominance, one allele is not completely dominant over another. As a result, the heterozygous phenotype is intermediate between the two homozygous phenotypes. In codominance, both alleles are dominant. As a result, the heterozygous phenotype is a combination of each homozygous phenotype.
PTS:1 DIF: A11.3.2
53. ANS:
agf
PTS: 1
54. ANS:
9:3:3:1
PTS: 1