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AP Biology – Unit 3 Review
Terms: use the index cards provided to review any of the terms below that you do not fully understand
Somatic cells
Gametic cells
Sister chromatids
Centromere
Mitosis
Cytokinesis
Interphase
G1 phase
S phase
G2 phase
Prophase
Prometaphase
Metaphase
Anaphase
Telophase
Kinetochores
Cleavage
Cell plate
Binary fission
G1 checkpoint
G0 phase
Growth factors
Density-dependent
inhibition
Gene
Locus
Asexual reproduction
Sexual reproduction
Karyotype
Homologous chromosomes
Autosomes
Sex chromosomes
Diploid
Haploid
Fertilization
Zygote
Meiosis I
Meiosis II
Prophase I
Synapsis
Crossing over
Chiasmata
Metaphase I
Anaphase I
Telophase I
Prophase II
Metaphase II
Anaphase II
Telophase II
Recombinant chromosomes
True-breeding
P generation
F1 generation
F2 generation
Dominant allele
Recessive allele
Law of segregation
Punnett square
Homozygous
Heterozygous
Phenotype
Genotype
Testcross
Monohybrid
Dihybrid
Law of independent
assortment
Complete dominance
Incomplete dominance
Codominance
Pleiotropy
Epistasis
Polygenic
Multifactorial
Pedigree
Carriers
Wild-type
Sex-linked gene
Barr body
Linked genes
Genetic recombination
Nondisjunction
Aneuploidy
Monosomy
Trisomy
Polyploidy
Deletion
Duplication
Inversion
Translocation
Transformation
Bacteriophage
Double helix
Antiparallel
Semiconservative model
Origin of replication
Replication fork
Helicase
Topoisomerase
RNA primer
Primase
DNA polymerase
Leading strand
Lagging strand
Okazaki fragments
DNA ligase
Nuclease
Telomerase
Heterochromatin
Euchromatin
Plasmid
Recombinant DNA
Restriction enzymes
Gel electrophoresis
Sticky ends
PCR
DNA sequencing
Transcription
mRNA
Translation
Ribosome
Codon
Reading frame
RNA polymerase
Promoter
TATA box
Intron
Exon
RNA splicing
Spliceosome
Ribozyme
tRNA
anticodon
rRNA
P site
A site
E site
Mutation
Point mutation
Silent mutation
Missense mutation
Nonsense mutation
Frameshift mutation
Mutagen
Carcinogen
Operator
Operon
Repressor
Regulatory gene
trp operon
lac operon
Inducer
Activator
Epigenetics
miRNAs
siRNAs
RNAi
Biotechnology
Reverse transcriptase
Genetically modified
Transgenic organisms
Gene therapy
STRs
SNP
Proteomics
Reading: Read through ALL of the information below and look up any information you do not fully understand.
Mitosis produces two genetically identical daughter cells, while meiosis occurs in sexually reproducing
organisms and results in haploid cells. The cell cycle consists of five major phases: G1, S, and G2, which
comprise interphase, and mitosis and cytokinesis, which make up the cell division phase.
Meiosis results in genetic variation:
 Independent assortment of chromosomes: homologous pairs of chromosomes separate depending
on the random way they line up on the metaphase plate during metaphase I. There is an equal
chance that a particular gamete will receive a maternal chromosome or a paternal chromosome.
 Crossover: crossover produces recombinant chromosomes, combining genes inherited from both
parents.
 Random fertilization: any sperm can fertilize any egg
Cyclins and cyclin-dependent kinases are responsible for controlling the cell cycle. Density-dependent
growth factors prevent cells from continuing to divide if there are no longer any sites upon which to
anchor. Cancer cells do not exhibit such inhibition and have escaped form cell cycle controls.
Heredity (8%)
See page 3 for information about genetic variation through meiosis. In spermatogenesis, four mature
sperm cells will result. In oogenesis, only one daughter cell results from meiosis, with the rest becoming
polar bodies that will degenerate.
DNA molecules are packaged into chromosomes. DNA is first wrapped around proteins called histones.
Each DNA-wrapped histone and the DNA around it will form a nucleosome. The string of nucleosomes
coils to form a chromatin fiber, which then become looped domains attached to a scaffold of nonhistone
proteins. The chromatin folds further to result in the chromosome.
Important heredity concepts:
 Incomplete dominance is the blending of two characteristics to form an intermediate
characteristic, such as in red and white carnations.
 Codominance: both traits will show, such as in roan cattle, with patches of red and white hair.
 Multiple alleles: occurs when there are more than two allelic forms of a gene, such as in ABO
blood types
 Pleiotropy: the ability of one single gene to affect an organism in several or many ways, causing a
“cascade” of symptoms
 Epistasis: two separate genes control one trait, with one gene masking the expression of the other
gene. For example, a gene for production of melanin can be epistatic to one for the deposition of
melanin
 Polygenic Inheritance: characters varying along a continuum, such as height or skin tone. There
is no either-or option.
Genes that are on the same chromosome are linked genes. One map unit distance on a
chromosome is the distance within which recombination occurs 1 percent of the time.
Molecular Genetics (9%)
DNA is a double helix, consisting of two strands running in opposite directions (antiparallel). One runs
5’ to 3’, while the other is 3’ to 5’. Each nucleotide consists of a five-carbon sugar (deoxyribose), a
phosphate, and a nitrogen base. The nucleotides are connected by phosphodiester linkages. Adenine
and guanine are purines, with two rings, while thymine and cytosine are pyrimidines, with one ring.
Adenine always pairs with thymine (uracil in RNA), and guanine always pairs with cytosine.
Nitrogenous bases are connected by hydrogen bonds.
DNA replication is semiconservative, as proved by Meselsohn and Stahl. Each strand of a DNA
double helix will be incorporated into a new strand of DNA.
1. Replication begins at the origins of replication, forming replication bubbles.
2. Replication proceeds in both directions, forming a replication fork.
3. DNA polymerase catalyzes the elongation of the DNA strands, going in the 5’ to 3’ direction.
This is the leading strand.
4. Elongation of the lagging strand, in the 3’ to 5’ direction, must be accomplished with the assistance
of Okazaki fragments. These are small segments of DNA synthesized, with ligase going back to join
together the Okazaki fragments.
DNA synthesis is primed using RNA primer (RNA nucleotides joined together by primase).
The machinery that uses DNA to synthesize proteins read nucleotide sequences in triplet code, with each
three nucleotides being one codon. Each codon will code for an amino acid. In transcription, DNA is
transcribed into messenger RNA (mRNA). The RNA will be processed, given a 5’ cap and poly (A) tail
after having its introns excised by snRNPs (small nuclear ribonucleoproteins) and spliceosomes.
Translation is the process by which the codons are changed into a sequence of amino acids. Transfer RNA
(tRNA) holds an amino acid corresponding to the anticodon that will match to the codon on a strip of
mRNA. As a ribosomal RNA (rRNA) unit proceeds along a piece of mRNA, tRNAs will bring their amino
acids to the translation machinery as necessary, forming a polypeptide chain.
Gene mutations include point mutations, in which only one base pair is changed. An insertion or
deletion is more detrimental, as it will cause a frameshift mutation, altering the series of codons
downstream of the mutation. This can either cause a missense (mutated polypeptide formed) or
nonsense mutation (no polypeptide formed).
A virus consists of DNA or RNA enclosed in a protein coat called a capsid. Viruses can only reproduce
within a host cell. The bacteriophage can reproduce using the lytic cycle or the lysogenic cycle. In the lytic
cycle, the phage will enter a host cell, replicate itself, and cause the cell to lyse, releasing more infectious
phages. In the lysogenic cycle, the phage DNA will integrate with the host genome, becoming a prophage
that is replicated each time the host cell replicates. At a certain point, the prophage will switch to the lytic
phase.
Biotechnology uses recombinant DNA techniques for practical purposes. Scientists have been able to clone
genes by isolating a gene of interest, inserting it into a plasmid, and then inserting the plasmid into a vector,
such as a bacterium. As the bacteria reproduce themselves, copies of the plasmid and gene of interest will
also be reproduced. Restriction enzymes can be used to cut out desired genes. Gel electrophoresis
separates large molecules of DNA based on their rate of movement through agarose gel in an electric field.
The polymerase chain reaction (PCR) is an automated technique to rapidly copy a small piece of DNA.
Restriction fragment length polymorphisms (RFLPs) are noncoding regions in human DNA that are as a
DNA fingerprint.
Practice FRQ’s
1. A difference between prokaryotes and eukaryotes is seen din the organization of their genetic
material.
a. Discuss the organization of the genetic material in prokaryotes and eukaryotes
b. Contrast the following activities in prokaryotes and eukaryotes
o Replication of DNA
o Transcription or Translation
o Gene regulation
o Cell division
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1. Information flow in cells can be regulated by various mechanisms.
a. Describe the role of THREE of the following in the regulation of protein synthesis:
o RNA splicing
o Repressor proteins
o Methylation
o siRNA
b. Information flow can be altered by mutation. Describe THREE different types of mutations
and their effect on protein synthesis.
c. Identify TWO environmental factors that increase the mutation rate in an organism, and
discuss their effect on the genome of the organism.
d. Epigenetics is the study of heritable changes in the phenotype caused by mechanisms other
than changes in the DNA sequence. Describe ONE example of epigenetic inheritance.
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Practice Multiple Choice
1. In a species that has five different alleles for a
3. The occurrence of a particular genetic
gene at a particular locus, how many
condition in a family is shown in the
different alleles may be present in the
pedigree above. Which of the following is the
somatic cells of one diploid individual?
most likely inheritance pattern for the
a. One
individuals with the condition? Squares
b. Two
represent males, circles represent females,
c. Three
and shaded symbols represent individuals
d. Four
who exhibit the condition.
e. Five
2. How many different genotypes are possible
from the cross show below?
AaBb x AaBb
a. 2
b. 4
a. Autosomal dominant
c. 7
b. Sex-linked dominant
d. 9
c. Y linked
e. 16
d. Autosomal recessive
e. Sex-linked recessive
4. In the experiments by Meselson and Stahl that demonstrated the semiconservative replication of
DNA, the researchers cultured bacteria in a medium containing a heavy isotope of nitrogen, 15N.
They then moved the bacteria to a medium containing 14N, the lighter, more common isotope of
nitrogen. After each round of replication, the researhers extracted the DNA and centrifuged the
solution to separate the DNA bands by density. The test tubes below illustrate the possible banding
pattern found after two bacterial generations (two rounds of DNA replication). Which test tube best
illustrates the bands predicted by the semiconservative model of DNA replication?
Questions 5-8 refer to the following
a. Nonsense codon
5. Protein synthesis termination triplet
b. Anticodon
6. Site of protein synthesis
c. Ribosome
7. Base sequence on messenger RNA that aids
d. Exon
its transport across the nuclear envelope
e. Poly-A tail
8. Triplet on tRNA
Questions 9-12 refer to the following types of hereditary material
a.
Unpaired unreplicated linear chromosomes
9. Typical of prokaryotic cells after fission
b.
Unpaired replicated linear chromosomes
10. Eukaryotic cells at prophase of mitosis
c.
Paired replicated linear chromosomes
11. Eukaryotic cells of metaphase I
d.
Circular chromosomes
12. Plasmid exchanged by conjugating bacteria
e.
Extra-chromosomal circular DNA
Questions 13-15
The graph below shows the results of a study of a culture of cells in active mitotic proliferation. The cells
were stained with a dye specific to nuclear DNA. The cells were then scanned to determine how much
DNA was present per cell. The x-axis of the graph presents the relative amount of DNA per cell; the y-axis
presents the number of cells for each value of DNA content.
13. The cells found in the region of the graph labeled Q
are involved in what major cell cycle activity?
a.
Cell division
b.
Active cell motility
c.
Cell differentiation
d.
Tetrad formation
e.
DNA synthesis
14. The region on the graph labeled P represents cells in
what stage of the cell cycle?
15.
a.
G1
b.
S
c.
G2
d.
Mitosis
e.
Cytokinesis
Cells in which region of the graph are ready to
enter mitosis as the next step in the cell cycle?
a.
P only
b.
Q only
c.
R only
d.
P and Q only
e.
Q and R only
Questions 16-20: Allele T codes for the ability to taste phenylthiocarbamide (PTC), and the gene locus is
identified as TAS2R38. One combination of 3 single nucleotide polymorphisms in noncoding sequences (SNPs)
correlates in humans with the ability to taste PTC. The restriction site for enzyme X includes one of the SNPs
associated with the ability to taste PTC. In the lab, students collect samples of cheek cells, extract the DNA, and
amplify a portion of the TAS2R38 locus using polymerase chain reaction (PCR). This, in addition to restriction
enzyme analysis and electrophoresis, will enable a prediction about the ability of the cell donors to taste PTC.
16. Which of the following is the most appropriate method to prepare a sample of cheek cells for PCR?
a. Obtain cells with saline mouthwash, boil, and chelate out any contaminating metals
b. Scrape cheek cells from inside the mouth; treat with nucleases
c. Using sterile technique, obtain a small punch biopsy specimen and keep at room
temperature
d. Use cheek cells obtained by rinsing the mouth with an antibacterial wash, and subject the
cells in culture to a strong antibiotic
e. Use a swab to obtain cells from the outer surface of the cheek after carefully washing the area
17. In order to amplify the sequence with part of the TAS2R38 gene, which of the following would be
the appropriate primers to use?
a. Primers that complement the telomeric sequences of the chromosomes
b. Primers that correspond to the 5’ and 3’ ends of the sequence near the SNP sites
c. RNA primers that complement the cDNA of the sequence of the whole gene
d. Degenerate DNA primers that will amplify both wild-type and SNP sequences
e. DNA primers for the chromosome known to include the TAS2R38 gene
18. Why does digesting the DNA with enzyme X enable a prediction of the cell donor’s ability to taste?
a. The nontaster allele (t) will not be cut and will therefore generate a larger fragment
b. Tasters will have three cutting sites instead of one
c. The taster allele )T_ will produce a larger protein than the nontaster allele (t)
d. Nontasters lack recognition sites for any restriction enzymes
e. Enzyme X will cut within the coding sequence of gene TAS2R38
19. Following amplification and restriction enzyme digestion, the presence of the T allele is recognized
on the electrophoresis gel by two bands: one of 177 base pairs (bp) and another of 44 bp. The
heterozygote (Tt) will show three bands on the gel. Which of the following describes their sizes?
a. 44, 133, and 177
c. 44, 133, and 221
e. 89, 133, and 221
b. 44, 89, 133
d. 44, 177, and 221
20. The students further studied DNA sequences data from various primates in order to explore the
evolution of this gene sequence. The study revealed that each of the other primates has the same SNP
pattern as human tasters. Which of the following is the most reasonable inference that can be made about
the evolution of this gene sequence?
a. The ability to taste bitter foods gave humans a selective advantage over other primates
b. Since the ability to taste is dominant, it has to have evolved first
c. Ancestral humans must have acquired the nontaster SNPs from nonprimates
d. Lack of the ability to taste must have a selective advantage for humans
e. The ability to detect bitter taste must have some selection advantages for nonhuman primates