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AP BIOLOGY > AP BIO UNIT 6 > Flashcards

AP BIO UNIT 6 Flashcards

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1
Q

Rosalind Franklin

A

Performed x-ray crystallography of DNA in the 1950s. Her work revealed a pattern that was regular and repetitive.

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2
Q

Edwin Chargaff

A

During the same era as Rosalind Franklin, Edward Chargaff analyzed DNA samples from different species. He found that the amount of adenine always equaled the amount of thymine and that the amount of cytosine always equaled the amount of guanine.

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3
Q

Purines

A

Double ring structure (A, G)

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4
Q

Pyrimidines

A

Single ring structure (C, U, T)

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5
Q

Nucleotide Pairing

A

Base pairs are held together by hydrogen bonds. Adenine and thymine have two hydrogen bonds while cytosine and guanine have three.

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6
Q

Watson and Crick

A

Combines the findings of Franklin (helix shape) and Chargaff (base pairing) to create the first 3D, double helix model of DNA.

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7
Q

Key Features of DNA Structure

A

DNA is a double-stranded helix. The backbone is a sugar-phosphate and the center is nucleotide pairs. DNA strands are antiparallel. One strand runs in the 5’ to 3’ direction while the other runs in the opposite, 3’ to 5’.

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8
Q

DNA

A

The primary source of heritable information. Genetic information is stored and passed from one generation to the next through DNA. EXCEPTION: RNA is the primary source of heritable information in some viruses.

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9
Q

DNA in Eukaryotic Cells

A

DNA is found in the nucleus. Linear chromosomes.

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10
Q

DNA in Prokaryotic Cells

A

DNA is in the nucleoid region. Chromosomes are circular. Prokaryotes (and some eukaryotes) also contain plasmids.

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11
Q

Plasmids

A

Small, circular DNA molecules that are separate from the chromosomes. Plasmids replicate independently from the chromosomal DNA. They are primarily found in prokaryotes. They contain genes that may be useful to the prokaryote when it is in a particular environment, but may not be required for survival. Plasmids can be manipulated in laboratories. Plasmids can be removed from bacteria, then a gene of interest can be inserted into the plasmid to form recombinant plasmid DNA. When the recombinant plasmid is inserted back into the bacteria the gene will be expressed. Bacteria can exchange genes found on plasmids with neighboring bacteria. Once DNA is exchanged, the bacteria can express the genes acquired. Helps with the survival of prokaryotes.

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12
Q

RNA vs. DNA

A

RNA: ribonucleic acid, single-stranded, A=U & C=G
DNA: deoxyribonucleic acid, double-stranded, A=T & C=G

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13
Q

DNA Replication

A

DNA replicates during the S phase of the cell cycle.

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14
Q

Models of DNA Replication

A

There were 3 alternative models for DNA replication. Conservative, Semi-Conservative, & Dispersive.

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15
Q

Conservative Model

A

The parental strands direct the synthesis of an entirely new double-stranded molecule. The parental strands are fully “conserved.”

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16
Q

Semi-Conservative Model

A

The two parental strands each make a copy of itself. After one round of replication, the two daughter molecules each have one parental and one new strand.

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17
Q

Dispersive Model

A

The material in the two parental strands is dispersed randomly between the two daughter molecules. After one round of replication, the daughter molecules contain a random mix of parental and new DNA.

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18
Q

Which Model is Correct?

A

In 1954, Meselson and Stahl performed an experiment using bacteria. Process:
1. Bacteria was cultured with a heavy isotope, 15N.
2. Bacteria was transferred to a medium with 14N, a lighter isotope.
3. DNA was centrifuged and analyzed after each replication.
By analyzing samples of DNA after each generation, it was found that the parental strands were following the semi-conservative model.

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19
Q

Steps in DNA Replication

A
  1. DNA replication begins at sites called origins of replication. Various proteins attach to the origin of replication and open the DNA to form a replication fork.
  2. Helicase will unwind the DNA strands at each replication fork. To keep the DNA from re-bonding with itself, proteins called single-stand binding proteins (SSBPs) bind to the DNA to keep it open. Topoisomerase will help prevent strain ahead of the replication fork by relaxing supercoiling.
  3. The enzyme primase initiates replication by adding short segments of RNA, called primers, to the parental DNA strand. The enzymes that synthesize DNA can only attach new DNA nucleotides to an existing strand of nucleotides. Primers serve as the foundation for DNA synthesis.
  4. Antiparallel Elongation: DNA Polymerase III (DNAP III) attaches to each primer on the parental strand and moves in the 3’ to 5’ direction. As it moves, it adds nucleotides to the new strand in the 5’ to 3’ direction The DNAP III that follows helicase is known as the leading strand and it only requires one primer. The DNAP III on the other parental strand that moves away from helicase is known as the lagging strand and requires many primers.
  5. The leading strand is synthesized in one continuous segment, but since the lagging strand moves away from the replication fork it is synthesized in chunks.
  6. After DNAP III forms an Okazaki fragment, DNAP I replaces RNA nucleotides with DNA nucleotides. DNA ligase joins the Okazaki fragments together forming a continuous DNA strand.
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20
Q

Okazaki Fragments

A

segments of the lagging strand

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21
Q

Problems at the 5’ End

A

Since DNAP III can only add nucleotides to a 3’ end, there is no way to finish replication on the 5’ end of a lagging strand. Over many replications, this would mean that DNA would become shorter and shorter. How are genes protected from this? Telomeres form a cap at the end of DNA to help postpone erosion. The enzyme telomerase added telomeres to DNA.

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22
Q

Telomeres

A

Repeating units of short nucleotide sequences that do not code for genes.

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23
Q

Proofreading and Repair

A

As DNA polymerase adds nucleotides to the new DNA strand, it proofreads the bases added. If errors still occur, mismatch repair will take place. Enzymes remove and replace the incorrectly paired nucleotides. If segments of DNA are damaged, nuclease can remove segments of nucleotides, and DNA polymerase and ligase can repair the segments.

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24
Q

Proteins

A

Proteins are polypeptides made up of amino acids. Amino acids are linked by peptide bonds.

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25
Q

Gene Expression

A

The process by which DNA directs the synthesis of a protein. Includes two stages: transcription and translation. Occurs in all organisms.
DNA –> Transcription –> RNA –> Translation –> Protein

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26
Q

Transcription

A

The synthesis of RNA using information from DNA. Allows for the “message” of DNA to be transcribed. Occurs in the nucleus.

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27
Q

Translation

A

The synthesis of polypeptides using information from RNA. Occurs at the ribosome. A nucleotide sequence becomes an amino acid base sequence.

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28
Q

Types of RNA

A

Messenger RNA (mRNA), Ribosomal RNA (rRNA), Transfer RNA (tRNA)

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29
Q

Messenger RNA

A

Messenger RNA is synthesized during transcription using a DNA template. mRNA carries information from the DNA (at the nucleus) to the ribosomes in the cytoplasm.

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30
Q

Transfer RNA

A

Transfer RNA molecules are important in the process of translation. Each tRNA can carry a specific amino acid. Can attach to mRNA via their anticodon (a complementary codon to mRNA). Allow information to be translated into a peptide sequence.

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31
Q

Ribosomal RNA

A

rRNA helps form ribosomes. Helps link amino acids together.

32
Q

The Genetic Code

A

DNA contains the sequences of nucleotides that codes for proteins. The sequence is read in groups of three called the triplet code. During transcription, only one DNA strand is being transcribed. Known as the template strand, (also known as the noncoding strand, minus strand, or antisense strand). mRNA molecules formed are antiparallel and complementary to the DNA nucleotides. Base pairing A=U & C=G. The mRNA nucleotide triples are called codons. Codons code for amino acids.

33
Q

The Genetic Code (Continued)

A

64 different codon combinations. 61 code for amino acids, 3 code for stop codons. Universal to all life. Redundancy more than one codon codes for each amino acid.

34
Q

Reading Frame

A

The codons on the mRNA must be read in the correct grouping during translation to synthesize the correct proteins. Example: the fat cat ate the rat. If the frame shifts even by one letter, it will produce a completely different outcome: Hef atc ata tet her at.

35
Q

Steps of Transcription

A
  1. Initiation
  2. Elongation
  3. Termination
36
Q

Step 1: Initiation

A

Transcription begins when RNA polymerase molecules attach to a promoter region of DNA. Do not need a primer to attach. Promoter regions are upstream of the desired gene to transcribe.
Eukaryotes: The promoter region is called a TATA box. Transcription factors help RNA polymerase bind.
Prokaryotes: RNA polymerase can bind directly to the promoter. (*The promoter is always upstream of the gene of interest to be transcribed).

37
Q

Step 2: Elongation

A

RNA polymerase opens the DNA and reads the triplet code of the template strand. Moves in the 3’ to 5’ direction. The mRNA transcript elongates 5’ to 3’. RNA polymerase moves downstream. Only opens small sections of DNA at a time. Pairs complementary RNA nucleotides. The growing mRNA strand peels away from the DNA template strand. DNA double helix then reforms. A single gene can be transcribed simultaneously by several RNA polymerase molecules. Helps increase the amount of mRNA synthesized. Increases protein production.

38
Q

Step 3: Termination

A

Prokaryotes: Transcription proceeds through a termination sequence. Causes a termination signal. RNA polymerase detaches. mRNA transcript is released and proceeds to translation. mRNA does NOT need modifications.
Eukaryotes: RNA polymerase transcribes a sequence of DNA called the polyadenylation signal sequence. Codes for a polyadenylation signal (AAUAAA). Releases the pre-mRNA from the DNA. Must undergo modification before translation.

39
Q

Pre-mRNA Modifications

A

There are three modifications that must occur to eukaryotic pre-mRNA before it is ready for translation.
1. 5’ Cap GTP
2. Poly-A-Tail
3. RNA Splicing
Once all modifications have occurred, the pre-mRNA is now considered mature mRNA and can leave the nucleus and proceed to the cytoplasm for translation at the ribosomes.

40
Q

5’ Cap GTP

A

The 5’ end of the pre-mRNA receives a modified guanine nucleotide “cap.”

41
Q

Poly-A-Tail

A

The 3’ end of the pre-mRNA receives 50-250 adenine nucleotides. Both the 5’ Cap and the Poly-A-Tail function to help the mature mRNA leave the nucleus, help protect the mRNA from degradation, and help ribosomes attach to the 5’ end of the mRNA when it reaches the cytoplasm.

42
Q

RNA Splicing

A

Sections of the pre-mRNA called introns, are removed and then exons are joined together.

43
Q

Introns

A

intervening sequence, do not code for amino acids

44
Q

Exons

A

Expressed sections code for amino acids

45
Q

Why does splicing occur?

A

A single gene can code for more than one kind of polypeptide. Known as alternative splicing.

46
Q

Translation

A

The synthesis of a polypeptide using information from the mRNA. Occurs at the ribosome. A nucleotide sequence becomes an amino acid sequence. tRNA is a key player in translating mRNA to an amino acid sequence.

47
Q

Transfer RNA in Translation

A

tRNA has an anticodon region which is complementary and antiparallel to mRNA. tRNA carries the amino acid that the mRNA codon codes for. The enzyme aminoacyl-tRNA synthetase is responsible for attaching amino acids to tRNA. When tRNA carries an amino acid it is “charged.”

48
Q

Ribosomes in Translation

A

Translation occurs at the ribosome. Ribosomes have 2 subunits: small and large. Prokaryotic and eukaryotic ribosomal subunits differ in size. Prokaryotes: small subunits (30s) large subunits (40s)
Eukaryotes: small subunits (40s) large subunits (60s)

49
Q

Large Subunit

A

The large subunit has three sites: A, P, and E

50
Q

A Site

A

Amino acid site. Holds the next tRNA carrying an amino acid.

51
Q

P Site

A

Polypeptide site. Holds the tRNA carrying the growing polypeptide chain.

52
Q

E Site

A

Exit site

53
Q

Translation Stages

A
  1. Initiation
  2. Elongation
  3. Termination
54
Q

Step 1: Initiation (Translation)

A

Translation begins when the small ribosomal subunit binds to the mRNA and a charged tRNA binds to the start codon, AUG, on the mRNA. The tRNA carries methionine. Next, the large subunit binds. (*The first tRNA carrying Met will go to the P site, every other tRNA will go to the A site first).

55
Q

Step 2: Elongation (Translation)

A

Elongation starts when the next tRNA comes into the A site. mRNA is moved through the ribosome and its codons are read. Each mRNA codon codes for a specific amino acid. Codon charts are used to determine the amino acid. Since all organisms use the same genetic code, it supports the idea of common ancestry.

56
Q

Elongation Steps

A
  1. Codon Recognition: the appropriate anticodon of the next tRNA goes to the A-site.
  2. Peptide Bond Formation: Peptide bonds are formed that transfer the polypeptide to the A-site tRNA.
  3. Translocation: the tRNA in the A-site moves to the P-site. The tRNA in the P-site goes to the E-site. The A-site is open for the next tRNA.
57
Q

Step 3: Termination (Translation)

A

Termination occurs when a stop codon in the mRNA reaches the A-site of the ribosome. Stop codons do not code for amino acids. The stop codon signals for a release factor. Hydrolyzes the bond that holds the polypeptide to the P-site. Polypeptide releases. All translational units disassemble.

58
Q

Protein Folding

A

As translation takes place, the growing polypeptide chain begins to coil and fold. Genes determine the primary structure. The primary structure determines the final shape. Some polypeptides require chaperone proteins to fold correctly and some require modification before they can be functional in the cell.

59
Q

Retroviruses

A

Retroviruses, like HIV, are an exception to the standard flow of genetic information. Information flows from RNA to DNA. Uses an enzyme known as reverse transcriptase. Couples viral RNA to DNA. DNA then becomes part of the RNA.

60
Q

Gene Expression

A

Prokaryotes and eukaryotes must be able to regulate which genes are expressed at any given time. Genes can be turned “on” or “off” based on environmental and internal cues. On/off refers to whether or not transcription will take place. Allows for cell specialization.

61
Q

Bacterial Gene Expression

A

Operons: a group of genes that can be turned on or off. Operons have three parts:
1. Promoter: where RNA polymerase can attach
2. Operator: the on/off switch
3. Genes: code for related enzymes in the pathway.
Operons can be repressible or inducible. Repressible (on to off): transcription is usually on, but can be repressed (stopped). Inducible (off to on): transcription is usually off but can be induced (started).

62
Q

Regulatory Gene

A

Produces a repressor protein that binds to the operator to block RNA polymerase from transcribing the gene. Always expressed, but at low levels. The binding of a repressor to an operator is reversible.

63
Q

Allosteric Regulation: Activator

A

Substrate binds to the allosteric site and stabilizes the shape of the enzyme so that the active sites remain open.

64
Q

Allosteric Regulation: Inhibitor

A

Substrate binds to allosteric site and stabilizes the enzyme shape so that the active sites are closed (inactive form).

65
Q

Repressible Operons

A

Example: the trp operon. The trp operon in bacteria controls the synthesis of tryptophan. Since it is irrepressible, transcription is active. It can be switched off by a trp repressor. Allosteric enzyme that is only active when tryptophan binds to it. When too much tryptophan builds up in bacteria, tryptophan is more likely to bind the repressor turning it active, which will then temporarily shut off transcription for tryptophan.

66
Q

Inducible Operon

A

Example: the lac operon. The lac operon controls the synthesis of lactase, an enzyme that digests lactose (milk sugar). SInce it s inducible, transcription is off. A lac repressor is bound to the operator (allosterically active). The inducer for the lac repressor is allolactose. When present, it will bind to the lac repressor and turn the lac repressor off (allosterically inactive). The genes can now be transcribed.

67
Q

Eukaryotic Gene Expression

A

The phenotype of a cell of an organism is determined by a combination of genes that are expressed and the levels at which they are expressed. The difference between cell types is known as differential gene expression. Chromatin structure: If DNA is tightly bound it is less accessible for transcription. How can it be modified? Histone acetylation adds acetyl groups to histones, which loosens the DNA. DNA methylation adds methyl groups to DNA, which causes the chromatin to condense. Epigenetic Inheritance: chromatin modifications do not alter the nucleotide sequence of the DNA, but they can be heritable to future generations. modification can be reversed, unlike mutations. Explains why one identical twin may inherit a disease while the other does not.
Transcription Initiation: once chromatin modifications allow the DNA to be more accessible, specific transcription factors bind to control elements. Sections of noncoding DNA that serve as binding sites. Gene expression can be increased or decreased by binding of activators or repressors to control elements.
RNA Processing: Alternative splicing of pre-mRNA.
Translation Initiation: Translation can be activated or repressed by initiation factors. MicroRNAs and small interfering RNAs can bind to mRNA and degrade it or block translation.

68
Q

Eukaryotic Development

A

During embryonic development, cell division, and cell differentiation occurs. Cells become specialized in their structure and function. Morphogenesis: the physical process that gives an organism its shape. How do cells differentiate during early development? Cytoplasmic determinants: substances in the maternal eggs that influence cells. Induction: cell-to-cell signals that can cause a change in gene expression. Both cytoplasmic determinants and induction influence pattern formation: a “body plan” for the organism. Homeotic genes map out the body structures. As cells differentiate, apoptosis plays a critical role. Apoptosis: programmed cell death. Allows structures to take their form. Example: if apoptosis did not occur during the development of human hands and feet, we would be born with webbed fingers and toes.

69
Q

Mutations

A

Changes in the genetic material of cells can alter phenotypes. The primary source of genetic variation. Normal function and production of cellular products are essential. Any disruption can cause new phenotypes. Changes can be large-scale or small-scale. Large scale: chromosomal changes. Small scale: nucleotide substitutions, insertions, or deletions.

70
Q

Small Scale Mutations

A

Point Mutations: changes a single nucleotide pair of a gene.
Substitution: the replacement of one nucleotide and its partner with another pair of nucleotides.
Silent Substitution: change still codes for the same amino acid (Remember: redundancy in the genetic code).
Missense Substitution: changes results in a different amino acid
Nonsense Substitution: Change results in a stop codon.
Frameshift mutation: when the reading frame of genetic information is altered. Disastrous effects on resulting proteins. Insertion: a nucleotide is inserted. Deletion: a nucleotide is removed.

71
Q

Large Scale Mutations

A

Mutations that affect chromosomes.
Nondisjunction: when chromosomes don’t separate properly in meiosis. Results in the incorrect number of chromosomes (Example: Down Syndrome (Trisomy 21)).
Translocation: a segment of one chromosome moves to another.
Inversions: a segment is reversed.
Duplications: a segment is repeated.
Deletions: a segment is lost.

72
Q

Natural Selection

A

Any time mutations occur, they are subject to natural selection. Genetic changes can sometimes enhance the survival and reproduction of an organism.

73
Q

Biotechnology

A

Both RNA and DNA can be manipulated through genetic engineering.

74
Q

Gel Electrophoresis

A

A technique used to separate DNA fragments by size. DNA is loaded into wells on one end of a gel and an electric current is applied. DNA fragments are negatively charged so they move towards the positive electrode.

75
Q

PCR

A

Polymerase Chain Reaction: a method used in molecular biology to make several copies of a specific DNA segment. Segments of DNA are amplified. Results can be analyzed using gel electrophoresis.

76
Q

DNA Sequencing

A

The process of determining the order of nucleotides in DNA.

AP BIOLOGY (5 decks)

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