Download Biology Final Exam Review Questions Answer Section SHORT

Biology Final Exam Review Questions
Answer Section
SHORT ANSWER
1. Vocab
2. ANS:
As a cell grows larger, more demands are placed on its DNA, and the cell has more trouble moving enough
nutrients and wastes across the cell membrane.
PTS: 1
DIF: L2
REF: p. 274 | p. 276
OBJ: 10.1.1 Explain the problems that growth causes for cells. STA: CA.BIO.1.a
TOP: Foundation Edition
BLM: comprehension
3. ANS:
Because the offspring of asexual reproduction are genetically identical to parents, they have the characteristics
that help them survive in the conditions in which the parent cells survived. They might not have
characteristics to survive should the conditions change.
PTS: 1
DIF: L3
REF: p. 278
OBJ: 10.1.2 Compare asexual and sexual reproduction.
STA: CA.BIO.2.a
TOP: Foundation Edition
BLM: evaluation
4. ANS:
Packaging genetic material into chromosomes helps the cell separate the DNA precisely during cell
division. If the genetic material was spread out into smaller pieces, some of the material might get lost
more easily when the cell divided into two cells.
PTS: 1
DIF: L2
REF: p. 280
OBJ: 10.2.1 Describe the role of chromosomes in cell division.
STA: CA.BIO.1.c TOP: Foundation Edition
BLM: analysis
5. ANS:
Chromatids are two identical DNA strands joined by a centromere, and chromatin is the material (DNA and
proteins) that makes up chromosomes.
PTS: 1
DIF: L3
REF: p. 280 | p. 282
OBJ: 10.2.1 Describe the role of chromosomes in cell division.
STA: CA.BIO.1.c TOP: Foundation Edition
BLM: synthesis
6. ANS:
A: G1 phase, cell growth; B: S phase, DNA replication; C: G2 phase, preparation for mitosis; D: M phase, cell
division (mitosis and cytokinesis).
PTS: 1
DIF: L2
REF: p. 281 | p. 282
OBJ: 10.2.2 Name the main events of the cell cycle.
STA: CA.BIO.1.c
TOP: Foundation Edition
BLM: analysis
7. ANS:
1 is anaphase. 2 is prophase. 3 is interphase (or G2 phase). 4 is telophase. 5 is metaphase.
They occur in the following order: 3, 2, 5, 1, and 4 (or: 2, 5, 1, 4, 3).
PTS: 1
DIF: L3
REF: p. 280 | p. 282
OBJ: 10.2.3 Describe what happens during the four phases of mitosis.
STA: CA.BIO.1.a TOP: Foundation Edition
BLM: analysis
8. ANS:
In plant cells, a cell plate forms in the cytoplasm midway between each new nucleus. The cell plate gradually
develops into a separating membrane, and a cell wall begins to appear in the cell plate. In animal cells, there is
no cell plate. The cell membrane is drawn inward until the cytoplasm is pinched into two nearly equal parts.
PTS: 1
DIF: L3
REF: p. 284
OBJ: 10.2.4 Describe the process of cytokinesis.
STA: CA.BIO.1.a
TOP: Foundation Edition
BLM: synthesis
9. ANS:
A cell that lacked cyclins would probably not undergo mitotic division, and then it would continue to grow,
have DNA overload, and exchange materials inefficiently until it dies.
PTS: 1
DIF: L3
REF: p. 286
OBJ: 10.3.1 Describe how the cell cycle is regulated.
STA: CA.BIO.1.h
BLM: evaluation
10. ANS:
Cancer cells do not respond to the signals that control the growth of normal cells. As a result, cancer cells
form tumors and can spread throughout the body.
PTS: 1
DIF: L2
REF: p. 289
OBJ: 10.3.2 Explain how cancer cells are different from other cells.
TOP: Foundation Edition
BLM: comprehension
11. ANS:
Cancer cells are not constrained by crowding and would probably continue to grow after forming a thin layer
covering the bottom of the petri dish.
PTS: 1
DIF: L3
REF: p. 289
OBJ: 10.3.2 Explain how cancer cells are different from other cells.
BLM: synthesis
12. ANS:
Embryonic stem cells come from embryos and are pluripotent, whereas adult stem cells come from adults and
are only multipotent.
PTS: 1
DIF: L2
REF: p. 295
OBJ: 10.4.2 Define stem cells and explain their importance.
STA: CA.BIO.4.d
TOP: Foundation Edition
BLM: analysis
13. ANS:
Harvesting adult stem cells do not generally harm the donor, whereas harvesting embryonic stem cells usually
destroys the embryo.
PTS: 1
DIF: L2
REF: p. 297
OBJ: 10.4.3 Identify the possible benefits and issues relating to stem cell research.
STA: CA.IE.1.m
BLM: evaluation
14. ANS:
.
PTS: 1
DIF: L3
REF: p. 308 | p. 309
OBJ: 11.1.1 Describe Mendel's studies and conclusions about inheritance.
STA: CA.BIO.3.a | CA.BIO.3.b
TOP: Foundation Edition
BLM: synthesis
15. ANS:
Allowing the F1 pea plants to self-pollinate caused the recessive phenotype to reappear in the F2 generation.
Self-pollination of the F1 plants also allowed the 3:1 phenotype ratios to occur, supporting Mendel’s theory.
Self-pollination showed that traits controlled by recessive alleles could reappear in the F2 generation.
PTS: 1
DIF: L3
REF: p. 311 | p. 312
OBJ: 11.1.2 Describe what happens during segregation.
STA: CA.BIO.3.b
BLM: synthesis
16. ANS:
Segregation happens when the alleles for each gene separate. Each gamete gets only one allele for each gene.
The principle of independent assortment states that the way alleles for one pair of genes segregate does not
affect the segregation of other alleles.
PTS: 1
DIF: L2
REF: p. 312 | p. 317
OBJ: 11.1.2 Describe what happens during segregation.
STA: CA.BIO.3.b
TOP: Foundation Edition
BLM: analysis
17. ANS:
Thirty of the offspring are expected to be tall and have yellow seeds.
PTS: 1
DIF: L3
REF: p. 316 | p. 317
OBJ: 11.2.2 Explain the principle of independent assortment.
STA: CA.BIO.3.a | CA.BIO.3.b
BLM: analysis
18. ANS:
Mendel’s principles of heredity apply to all organisms.
PTS: 1
DIF: L1
REF: p. 318
OBJ: 11.2.3 Explain how Mendel's principles apply to all organisms.
STA: CA.BIO.3.b TOP: Foundation Edition
BLM: knowledge
19. ANS:
100%
PTS: 1
DIF: L2
REF: p. 314 | p. 315 | p. 316
OBJ: 11.2.3 Explain how Mendel's principles apply to all organisms.
STA: CA.BIO.3.b TOP: Foundation Edition
BLM: analysis
20. ANS:
Keep an arctic fox warm when its native environment would be cool.
PTS: 1
DIF: L3
REF: p. 321
OBJ: 11.3.2 Explain the relationship between genes and the environment.
STA: CA.BIO.6.g*
BLM: evaluation
21. ANS:
The number of chromosomes is cut in half.
PTS: 1
DIF: L1
REF: p. 324
OBJ: 11.4.2 Summarize the events of meiosis.
STA: CA.BIO.2.a
TOP: Foundation Edition
BLM: knowledge
22. ANS:
Mitosis produces diploid body cells, whereas meiosis produces haploid gametes.
PTS: 1
DIF: L2
REF: p. 327
STA: CA.BIO.2.a | CA.BIO.2.d | CA.BIO.2.e
BLM: analysis
23. ANS:
sex cells, gametes
OBJ: 11.4.3 Contrast meiosis and mitosis.
TOP: Foundation Edition
PTS: 1
DIF: L1
REF: p. 323 | p. 325
OBJ: 11.4.3 Contrast meiosis and mitosis.
STA: CA.BIO.2.a | CA.BIO.2.d | CA.BIO.2.e
TOP: Foundation Edition
BLM: knowledge
24. ANS:
The genes that Mendel studied were located on different chromosomes or were located far apart on the same
chromosome.
PTS: 1
DIF: L2
REF: p. 328 | p. 329
OBJ: 11.4.4 Describe how alleles from different genes can be inherited together.
STA: CA.BIO.2.c BLM: evaluation
25. ANS:
The harmless living bacteria took in pneumonia-causing DNA (genes) from the heat-killed, pneumoniacausing bacteria, as a result of which the harmless bacteria changed into bacteria that cause pneumonia.
He concluded that a chemical factor, a gene, had transformed the bacteria.
PTS: 1
DIF: L3
REF: p. 350 | p. 351
OBJ: 12.1.1 Summarize the process of bacterial transformation.
STA: CA.BIO.5.a BLM: synthesis
26. ANS:
P-32 & S-35
PTS: 1
DIF: L1
REF: p. 339
OBJ: 12.1.1 Summarize the process of bacterial transformation.
STA: CA.BIO.5.a TOP: Foundation Edition
BLM: comprehension
27. ANS:
DNA stores, copies, and transmits information.
PTS: 1
DIF: L1
REF: p. 342 | p. 343
OBJ: 12.1.3 Identify the role of DNA in heredity.
STA: CA.BIO.5.a
TOP: Foundation Edition
BLM: knowledge
28. ANS:
It is most important during the formation of reproductive cells, because the loss of any genetic material then
means the loss of valuable information for offspring.
PTS: 1
DIF: L2
REF: p. 343
OBJ: 12.1.3 Identify the role of DNA in heredity.
STA: CA.BIO.5.a
BLM: synthesis
29. ANS:
The circles are the phosphate group, the pentagons are deoxyribose, and the A and T (adenosine and thymine)
are the bases.
PTS: 1
DIF: L2
REF: p. 345
OBJ: 12.2.1 Identify the chemical components of DNA.
STA: CA.BIO.5.a
TOP: Foundation Edition
BLM: application
30. ANS:
The nucleotides in a strand of DNA are joined by covalent bonds between their sugar and phosphate groups,
and by hydrogen bonds between the complimentary bases.
PTS: 1
DIF: L3
REF: p. 344
OBJ: 12.2.1 Identify the chemical components of DNA.
STA: CA.BIO.5.a
TOP: Foundation Edition
BLM: synthesis
31. ANS:
He systematically destroyed all the other kinds of molecules besides DNA in the dead-cell mixture before
using the mixture to successfully transform harmless bacteria into helpful bacteria.
PTS: 1
DIF: L2
REF: p. 340
OBJ: 12.2.2 Discuss the experiments leading to the identification of DNA as the molecule that carries the
genetic code.
STA: CA.BIO.5.a TOP: Foundation Edition
BLM: synthesis
32. ANS:
Rosalind Franklin used powerful X-ray beams to make diffraction photographs that gave Watson and Crick
the clues they needed to determine DNA’s structure.
PTS: 1
DIF: L2
REF: p. 346 | p. 347
OBJ: 12.2.3 Describe the steps leading to the development of the double-helix model of DNA.
STA: CA.BIO.5.a | CA.IE.1.g
TOP: Foundation Edition
BLM: synthesis
33. ANS:
The percentage of adenine would have increased by about 5 percent.
PTS: 1
DIF: L3
REF: p. 345
OBJ: 12.2.3 Describe the steps leading to the development of the double-helix model of DNA.
STA: CA.BIO.5.a | CA.IE.1.g
BLM: analysis
34. ANS:
The hydrogen bonds between the base pairs must be broken, and the molecule must unwind.
PTS: 1
DIF: L2
REF: p. 350| p. 351
OBJ: 12.3.1 Summarize the events of DNA replication.
STA: CA.BIO.1.b | CA.BIO.5.a | CA.BIO.5.b
TOP: Foundation Edition
BLM: analysis
35. ANS:
In prokaryotes, DNA replication starts in one place, and in eukaryotes DNA replication starts in many places.
PTS: 1
DIF: L1
REF: p. 353
OBJ: 12.3.2 Compare DNA replication in prokaryotes with that of eukaryotes
STA: CA.BIO.1.c | CA.BIO.5.a
TOP: Foundation Edition
BLM: knowledge
36. ANS:
A ribose molecule, a phosphate group, and a nitrogenous base are the three main parts of an RNA nucleotide.
PTS: 1
DIF: L1
REF: p. 362
OBJ: 13.1.1 Contrast RNA and DNA.
STA: CA.BIO.5.a TOP: Foundation Edition
BLM: knowledge
37. ANS:
RNA polymerase might be unable to bind to the promoter, and, as a result, the gene would not be transcribed.
PTS: 1
DIF: L3
REF: p. 364 | p. 365
OBJ: 13.1.2 Explain the process of transcription.
STA: CA.BIO.5.a | CA.BIO.5.b
BLM: synthesis
38. ANS:
The DNA molecule must be separated into two strands.
PTS: 1
DIF: L2
REF: p. 364
OBJ: 13.1.2 Explain the process of transcription.
STA: CA.BIO.5.a | CA.BIO.5.b
TOP: Foundation Edition
BLM: comprehension
39. ANS:
There could be 16 combinations of nucleotides, which is too few combinations for all 20 amino acid to have a
unique code.
PTS: 1
DIF: L3
REF: p. 366 | p. 367
OBJ: 13.2.1 Identify the genetic code and explain how it is read.
STA: CA.BIO.4.b BLM: evaluation
40. ANS:
A stop codon on the mRNA causes translation to stop.
PTS: 1
DIF: L1
REF: p. 367
OBJ: 13.2.2 Summarize the process of translation.
STA: CA.BIO.4.a
TOP: Foundation Edition
BLM: knowledge
41. ANS:
Messenger RNA provides the code for the translation, ribosomal RNA reads the code, and a tRNA molecule
brings the next amino acid specified by the code.
PTS: 1
DIF: L2
REF: p. 368 | p. 369
OBJ: 13.2.2 Summarize the process of translation.
STA: CA.BIO.4.a
BLM: comprehension
42. ANS:
Information is transferred from DNA to RNA to protein.
PTS: 1
DIF: L1
REF: p. 363
OBJ: 13.2.3 Describe the "central dogma" of molecular biology.
STA: CA.BIO.1.d TOP: Foundation Edition
BLM: comprehension
43. ANS:
Sample answer: Mutations in humans have resulted in stronger bones and better disease resistance.
PTS: 1
DIF: L2
REF: p. 376
OBJ: 13.3.2 Describe the effects mutations can have on genes. STA: CA.BIO.4.c
TOP: Foundation Edition
BLM: knowledge
44. ANS:
The lactose binds to the lac repressors, causing the repressors to release the operator.
PTS: 1
DIF: L2
REF: p. 377 | p. 378
OBJ: 13.4.1 Describe gene regulation in prokaryotes.
STA: CA.BIO.1.c
BLM: application
45. ANS:
External forces (such as a drying pond) trigger internal factors (such as hormonal changes) that change the
rate of metamorphosis.
PTS: 1
DIF: L2
REF: p. 383
OBJ: 13.4.3 Relate gene regulation to development in multicellular organisms.
STA: CA.BIO.4.d TOP: Foundation Edition
BLM: comprehension
46. ANS:
A sperm that has 23 chromosomes fertilizes an egg that has 23 chromosomes resulting in a 23 pairs of
chromosomes (46 total) in the autosomal cells of the individual.
PTS: 1
DIF: L2
REF: p. 392 | p. 393
OBJ: 14.1.1 Identify the types of human chromosomes in a karyotype.
STA: CA.BIO.2.f | CA.BIO.2.g
TOP: Foundation Edition
BLM: comprehension
47. ANS:
The sex chromosomes are homologous because one sex chromosome is inherited from one parent, and the
other is inherited from the other parent.
PTS: 1
DIF: L3
REF: p. 393
OBJ: 14.1.1 Identify the types of human chromosomes in a karyotype.
STA: CA.BIO.2.f | CA.BIO.2.g
BLM: synthesis
48. ANS:
Males have just one X chromosome.
PTS: 1
DIF: L2
REF: p. 395
OBJ: 14.1.2 Describe the patterns of the inheritance of human traits.
STA: CA.BIO.3.a TOP: Foundation Edition
BLM: comprehension
49. ANS:
The probability that their son will be colorblind is 50%.
PTS: 1
DIF: L2
REF: p. 395
OBJ: 14.1.2 Describe the patterns of the inheritance of human traits.
STA: CA.BIO.3.a TOP: Foundation Edition
BLM: analysis
50. ANS:
A person who has type AB blood can safely receive transfusions of all ABO blood types.
PTS: 1
DIF: L2
REF: p. 394
OBJ: 14.1.2 Describe the patterns of the inheritance of human traits.
STA: CA.BIO.3.a TOP: Foundation Edition
BLM: application
51. ANS:
The symbol should be shaded if the individual has the trait.
PTS: 1
DIF: L2
REF: p. 396 | p. 397
OBJ: 14.1.3 Explain how pedigrees are used to study human traits.
STA: CA.BIO.3.c*
TOP: Foundation Edition
BLM: application
52. ANS:
The DNA sequence of the allele that causes cystic fibrosis has a deletion of three bases.
PTS: 1
DIF: L2
REF: p. 399
OBJ: 14.2.1 Explain how small changes in DNA cause genetic disorders.
STA: CA.BIO.4.c TOP: Foundation Edition
BLM: comprehension
53. ANS:
The frequency of the sickle cell allele would probably decrease because the allele would no longer be
beneficial in heterozygous individuals.
PTS: 1
DIF: L3
REF: p. 400
OBJ: 14.2.1 Explain how small changes in DNA cause genetic disorders.
STA: CA.BIO.4.c BLM: evaluation
54. ANS:
If human cells have a Y chromosome, the person is a male regardless of how many X chromosomes are in the
cells.
PTS: 1
DIF: L2
REF: p. 401
OBJ: 14.2.2 Summarize the problems caused by nondisjunction.
STA: CA.BIO.4.c TOP: Foundation Edition
BLM: analysis
55. ANS:
You can tell that nondisjunction occurred because the gametes have an abnormal number of chromosomes.
PTS: 1
DIF: L1
REF: p. 401
OBJ: 14.2.2 Summarize the problems caused by nondisjunction.
STA: CA.BIO.4.c BLM: analysis
56. ANS:
The goal of the Human Genome Project is to attempt to sequence all human DNA.
PTS: 1
DIF: L1
REF: p. 406
OBJ: 14.3.2 State the goals of the Human Genome Project and explain what we have learned so far.
STA: CA.BIO.5.c BLM: knowledge
57. ANS:
In hybridization, organisms with dissimilar traits are crossed. In inbreeding, organisms with similar traits are
crossed.
PTS: 1
DIF: L3
REF: p. 419 | p. 420
OBJ: 15.1.1 Explain the purpose of selective breeding.
STA: CA.BIO.5.c
BLM: analysis
58. ANS:
Dolly and the sheep from which she was cloned have identical genes.
PTS: 1
DIF: L2
REF: p. 427
OBJ: 15.2.3 Define transgenic and describe the usefulness of some transgenic organisms to humans.
STA: CA.BIO.5.c TOP: Foundation Edition
BLM: comprehension
59. ANS:
Viruses are used in gene therapy because they can transfer genes into human cells.
PTS:
OBJ:
STA:
BLM:
1
DIF: L2
REF: p. 431
15.3.2 Explain how recombinant DNA technology can improve human health.
CA.BIO.5.d* | CA.IE.1.m
TOP: Foundation Edition
analysis
60. ANS:
No. The Genetic Information Nondiscrimination Act protects against discrimination based on genetic
information.
PTS:
OBJ:
STA:
BLM:
1
DIF: L1
REF: p. 437
15.4.3 Describe some of the ethical issues relating to biotechnology.
CA.BIO.5.c | CA.IE.1.m
TOP: Foundation Edition
comprehension