Test 2 (chp 9-15)

Question 1

Mitosis

Answer 1

A process of nuclear division in eukaryotic cells conventionally divided into five stages: prophase, prometaphase, metaphase,anaphase, and telophase. Mitosis conserves chromosome number by allocating replicated chromosomes equally to each of the daughter nuclei.

Question 2

Meiosis

Answer 2

A modified type of cell division in sexually reproducing organisms consisting of two rounds of cell division but only one round of DNA replication. It results in cells with half the number of chromosome sets as the original cell.

Question 3

DNA replication

Answer 3

The process by which a DNA molecule is copied; also called DNA synthesis.

Question 4

Protein synthesis

Answer 4

The process by which amino acids are linearly arranged into proteins through the involvement of ribosomal RNA, transfer RNA, messenger RNA, and various enzymes during transcription and translation to link together into polypeptide chains.

Question 5

Operons

Answer 5

A unit of genetic function found in bacteria and phages, consisting of a promoter, an operator, and a coordinately regulated cluster of genes whose products function in a common pathway.

Question 6

Cell cycle

Answer 6

interphase
prophase
pro metaphase
metaphase
anaphase
telophase
cytokinesis

Question 7

Interphase

Answer 7

Can be divided into the G1 phase (“first gap”) where the cell grows, the S phase (“synthesis”) where the cell continues to grow and copies its chromosomes, and the G2 phase (second gap”) where the cell grows more as it completes preparations for cell division.
During all three subphases, a cell that will eventually divide grows by producing proteins and cytoplasmic organelles such as mitochondria and endoplasmic reticulum.
G0 phase is a nondividing stage the cell goes in when the cell does not receive the go-ahead signal.

Question 8

Prophase:

Answer 8

The mitotic spindle forms from the centrioles in the centrosome area starting to create spindle fibers made out of microtubules
Chromosomes start to coil and condense

Question 9

Prometaphase:

Answer 9

Nuclear envelope disappears
Nucleolus disappears
Centrosomes are on opposite poles of the cells
Spindle fibers attach to the sister chromatids

Question 10

Metaphase:

Answer 10

The sister chromatids line up on the metaphase plate (invisible line in the center of the cell)

Question 11

Anaphase:

Answer 11

Sister chromatids are pulled apart towards opposite ends of the cell using the shortening of the spindle fibers
The cell is starting to elongate

Question 12

Telophase:

Answer 12

(opposite of prophase)
Nuclear envelope returns
Nucleolus returns
Spindle fibers are broken down

Question 13

Cytokinesis:

Answer 13

Dividing of the cytoplasm to form two new cells
Cleavage furrow in animal cell
Golgi apparatus puts down vesicles that build the plate in plant cell

Question 14

Make sure you are able to recognize crossing over. What should you see happening?

Answer 14

In Meiosis, Prophase 1
A genetic rearrangement between nonsister chromatids involving the exchange of corresponding segments of DNA molecules, begins during pairing and synaptonemal complex formation and is completed while homologs are in synapsis.
Potentially can switch ends of their chromosomes of the non-sister chromatids (in the middle)

Question 15

Know the process of DNA replication: molecules involved, how leading strand/lagging strand copied.

Answer 15

Begins at the origin of replication which are short stretches of DNA having a specific sequence of nucleotides. Proteins that initiate DNA replication recognize this specific sequence and attach to the DNA, causing the two strands to separate and open a replication “bubble”. At the end of the bubble is the replication fork, a Y-shaped region where the parental strands of DNA are being unwound. To unwind the DNA, several kinds of proteins help. Helicase is an enzyme that untwists the double helix at the replication fork by breaking the hydrogen bonds, separating the two parental strands and making them available as template strands. After the parental strands separate, single-strand binding proteins bind to the unpaired DNA strands, keeping them open and preventing them from re-pairing. Topoisomerase helps relieve the strain of the tighter twisting ahead of the replication fork by breaking, swiveling, and rejoining DNA strands (aka reliefs tension of the supercoiling of the DNA behind replicating fork due to the unwinding). Replication then proceeds on the leading and lagging strands, but only in the 5’3’ direction. The enzyme primase puts an RNA nucleotide (5-10 nucleotides long) primer down on the strands for the DNA polymerase to start synthesizing a new DNA strand by adding nucleotides. DNA polymerase 3 synthesises the complementary strand in the 5’3’ direction by adding nucleotides to the new complementary strand as the fork progresses. DNA polymerase 3 is also working on the lagging strand in the 5’3’ direction away from the replication fork with multiple primers that were put down by primase. The lagging strand is synthesized discontinuously with those primers (put down by primase), DNA polymerase 3 attaches and lays down opposite nucleotides until it reaches a primer, then it jumps backwards to the next primer. These series of segments called Okazaki fragments. Once the one long leading strand is synthesized and the all of the Okazaki fragments are synthesized DNA polymerase 1 replaces the RNA primer with DNA nucleotides. Then, an enzymes called DNA ligase joins the sugar-phosphate backbones of all the Okazaki fragments into a continuous strand by joining the 3’ ends to the 5’ ends.
If no DNA ligase those fragments will stay as fragments so the next replication will be problematic

Question 16

When is DNA replication needed?

Answer 16

In a favorable environment the cell can copy all of the DNA and divide to form two genetically identical daughter cells.
This DNA Replication is needed to replace cells(mitosis), growth and repair (mitosis) and for production of gametes (meiosis) (replicate own cells) and meiosis (reproduction)

Question 17

Know the chromosome numbers throughout the mitotic process.

Answer 17

Mitosis: Interphase G1 is c, Interphase G2 through Telophase is 2c, Cytokinesis c.
Meiosis 1: Interphase G1 is c, Interphase G2 through Telophase is 2c, Cytokinesis c.
Meiosis 2: (no interphase in M2) All phases c, Cytokinesis 1/2c.

Question 18

How does the replicated DNA stay protected?

Answer 18

Telomerase puts down Telomeres that do not contain genes; instead, the DNA typically consists of multiple repetitions of one short nucleotide sequence. Telomeric DNA acts as a buffer zone that protects the organism’s genes.
Prevents the DNA from getting shorter and shorter because of its buffer area
Extra strand of unneeded DNA that protects the needed genes of the cell

Question 19

Be able to calculate the A/T, C/G percentages.

Answer 19

Chargaff’s Rule (Out of 100%) Only with double stranded DNA-not single strand
Number of T=Number of A
Number of C=Number of G
EX: if you have 15% A then you have 15% T. Then you add those percentages (you get 30%) and then subtract from 100% (70%)

Question 20

Why are we able to insert genes of one organism into another organism?

Answer 20

Using plasmids (a plasmid is a small circular DNA molecules that replicate separately from the bacterial chromosome).
A gene of interest is inserted into a plasmid which is now recombinant DNA (a DNA molecules formed when segments of DNA from two different sources are combined in vitro-in a test tube). That plasmid is now put into a bacterial cell, producing a recombinant bacterium. This single cell reproduces through repeated cell divisions to form a clone of cells,a population of genetically identical cells. The production of multiple copies of a signal gene is gene cloning.
Gene cloning is used for two basic purposes:
To make many copies of a particular gene
To produce a protein product
the second reason we are able to insert a gene of one organism into another organism is because DNA is the universal code. All organisms have the same nitrogen bases-A,T,C,G

Question 21

Understand how restriction enzymes work to cut DNA

Answer 21

Restriction enzymes protect the bacterial cell by cutting up foreign DNA from other organisms or phages. Each restriction enzyme is very specific, recognizing a particular short DNA sequence-or restriction site- and cutting both DNA strands at precise points within this restriction site.
EX: The sequence of nucleotides is the same on both strands when read in the 5’3’ direction. A restriction enzyme will make cuts in a DNA molecule, yielding a set of restriction fragments. One of these fragments has a sticky end (one single-stranded end) that can form hydrogen -bonded base pairs (hybridize) with complementary sticky ends on any other DNA molecules cut with the same enzyme. This is temporary but can be made permanent by DNA ligase by joining the sugar-phosphate backbones. The ligase-catalyzed joining of DNA from two different sources produces a stable recombinant DNA molecule.
There are some restriction enzymes that cut straight through the DNA molecule not producing sticky ends. When doing recombinant DNA it is preferred to use the ones that cut into sticky ends.
The DNA of a bacterial cell is protected from the cell’s own restriction enzymes by the addition of methyl groups (-CH3) to adenines or cytosines within the sequences recognized by the enzymes

Question 22

Be able to interpret a karyotype

Answer 22

A karyotype is a display of the chromosome pairs of a cell arranged by size and shape.
Shows the two chromosomes of each of the 23 types. When looking at it you can see if nondisjunction has occurred causing there to be extra chromosomes in a pair or causing missing chromosomes in a pair. It is also looked at to see the sex of the child

Question 23

understand the steps of protein synthesis

Answer 23

protein synthesis consists of transcription and translation

Aka: Gene expression the process by which DNA directs the synthesis of proteins

Question 24

transcription

Answer 24

The synthesis of RNA using information in the DNA
The two nucleic acids are written in different forms and the information is simply transcribed (“rewritten”) from DNA to RNA. A DNA strand provides a template for making a new complementary strand during RNA replication, it also may serve as a template for assembling a complementary sequence of RNA nucleotides. This starts at the promoter’s TATA box region that helps to bind RNA polymerase along with the help of transcription factors. Then replication begins by then unwinding the DNA and reading the template strand DNA and adding corresponding complementary RNA nucleotides. It stops once it reaches the terminator sequence and will stop right after this sequence. The pre-mRNA is the result of this replication. For the pre-mRNA to become mRNA a process called RNA splicing occurs. During RNA splicing it cuts out the unneeded noncoding stretches of DNA called the introns so only the needed introns (which will vary) and exons are left in the RNA strand and are spliced together. The resulting RNA molecule is a transcript of the gene’s protein-building instructions called a messenger RNA (mRNA) because it carries a genetic message from the DNA to the protein-synthesizing “machinery” of the cell. The 5’ cap (a bunch of Guanisines) and the poly-A tail (a bunch of Adenines) are added and then it is ready to leave the nucleus for translation to start in the cytoplasm.
Occurs inside the nucleus

Question 25

translation

Answer 25

The synthesis of a polypeptide using the information in the mRNA.
(in the cytoplasm) The cell translates the nucleotide sequence of an mRNA molecule into the amino acid sequence of a polypeptide. The sites of translation are ribosomes, complex particles that facilitate the orderly linking of amino acids into polypeptide chains. tRNA is a translator that transfers amino acids from the cytoplasmic pool to a growing polypeptide in a ribosome. At the ribosome, the small subunit attaches first, the first tRNA comes and attaches to the start codon (the first is always methionine) and then the large subunit attaches. The codons are translated into amino acids, one by one. The tRNA molecules interpret each nucleotide triplet (called an anticodon which is the complementary RNA) and transfer the specific amino acid to the end, creating a polypeptide chain. This works smoothly because the ribosome has three binding site, an E site (exit site), P site (peptidyl-tRNA binding site also growing polypeptide), and A site (aminoacyl-tRNA binding site also arrival site). tRNA arrives in the A first, then it moves into the E while another one replaces it in the A site. Then it moves into the E and leaves once the amino acid is bonded to the polypeptide chain and goes to get a new amino acid. The one in the A site moves into the P and a new one is added to the A site. A release factor comes in where the stop codon is and the ribosome comes apart.
There has to be a signal peptide on the beginning of the RNA is it is going to attach to the ER after
Occurs outside the nucleus

Question 26

What are operons and where are they found?

Answer 26

A unit of genetic function found in bacteria and some phages, consisting of a promoter, an operator, and a coordinately regulated cluster of genes whose products function in a common pathway.

Question 27

what are two common types of operons

Answer 27

the lac operon and the trp operon

Question 28

trp operon

Answer 28

(repressible operon) (usually always creating something)
The trp operon is always on (meaning tryptophan is absent); therefore, the repressor is normally inactive and not bond to the operator. RNA polymerase can bind to the promoter and transcribe the genes of the operon. The trp operon can be switched off by a protein (that is turned off when there is too much tryptophan) called the trp repressor. The inactive repressor receives a tryptophan that acts as a corepressor, turning the repressor on and causing it to bind to the operator and blocks the attachment of RNA polymerase to the promoter, preventing transcription of the genes

Question 29

lac operon

Answer 29

The lac operon is always off (meaning lactose is absent); therefore, the lac repressor is normally active and bound to the operon. RNA polymerase is unable to bind to the promoter and unable to transcribe the genes of the operon. The lac inducer in the presence of lactose, allolactose, attaches to the repressor causing it to inactivate the repressor and turns the operon on to break down the present lactose

Question 30

Understand how gel electrophoresis works to separate DNA fragments

Answer 30

Gel Electrophoresis is a technique for separating nucleic acids or proteins on the basis of their size and electrical charge, both of which affect their rate of movement through an electric field in a gel made of agarose or another polymer.
DNA is put into wells in a gel made of agarose near the negative side of the electrophoresis machine. The machine is turned on and electrical waves are sent through the gel. The electric current causes the different sized fragments of DNA to move through the gel and separate, smaller ones move farther than larger ones because it can move through the gel easier

Question 31

What other molecules can gel electrophoresis be used for

Answer 31

Any nucleic acid fragments (DNA or RNA) or proteins

Question 32

What is the purpose of radioactive labeling molecules? What would you use to label nucleic acids? Proteins?

Answer 32

The purpose would be able to track the specific molecules that are labelled and see where they end up.
Nucleic acids are labelled with phosphorus.
Proteins are labelled with sulfur.
This was Hershey and Chase’s experiment.

Question 33

What is the purpose of the pre-mRNA transcript?

Answer 33

To be able to use one gene to code for multiple proteins depending on the introns that are cut out. To be able to create different messenger RNAs from the same gene by the introns that are taken out.
Depending on what I want to make certain introns are taken out.
Depending on what wants to be made there are different intron regions
The 5’ cap end is synthesized of a modified form of a guanine (G) nucleotide added onto the 5’ end after transcription of the first 20-40 nucleotides. The poly-A tail is also added to the 3’ end where an enzyme adds 50-250 more adenine (A) nucleotides.
These are important because…
They seem to facilitate the export of the mature mRNA from the nucleus.
They help protect the mRNA from degradation by hydrolytic enzymes.
They help ribosomes attach to the 5’ end of the mRNA once the mRNA reaches the cytoplasm.

Question 34

Where are TATA boxes found? What types of organisms?

Answer 34

In the promoter

It is a DNA sequence in eukaryotic promoters that are crucial in forming the transcription initiation complex.

Question 35

What is non disjunction? Be able to recognize the gametes if it happens in meiosis I versus meiosis II.

Answer 35

Nondisjunction is an error in meiosis or mitosis in which members of a pair of homologous chromosomes or sister chromatids fail to separate properly from each other during anaphase in Meiosis 1 or Meiosis 2.
The homologous chromosomes do not move apart properly or do not move apart at all.
This can be recognized when cells have an inadequate amount of chromosomes.
EX: gametes should have 2 chromosomes but if they have anything other than 2 then nondisjunction occurred.