Topic 3.5 & Some Topic 7: Biotechnology and Genetic Modification/Nucleic Acids Flashcards

DP2- U2- Molecular Genetics

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1
Q

A. Intro to Molecular Genetics

  1. State that DNA and RNA are nucleic acids and are polymers of nucleotides.

Define nucleotide/consists and nucelic acids

A

A nucleotide: a single unit of a nucleic acid

Nucleic acids: very large molecules– linking together nucleotides to form a polymer

Nucleotide consists of a phosphate group (acidic, negatively charged), sugar (5-C/pentose), nitrogenous base (ATGC or U). All lined by covalent bonds.

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2
Q

A. Intro to Molecular Genetics

2.Compare and contrast the structures of DNA and RNA

plus purine and pyrimidine

A

Purine= Adenine and Guanine
Pyrimidine= Thymine/Uracil and Cytosine

DNA:
* Adenine, Guanine, Thymine, Cytosine
* Deoxyribose sugar
* Two antiparallel complementary strands that form a double helix
RNA:
* Adenine, Guanine, Uracil, Cytosine
* Ribose sugar
* Single stranded and often linear in shape

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3
Q

A. Intro to Molecular Genetics

3.Outline the detailed structure of the DNA double helix including the antiparallel strands and the complimentary base pairs.

phopic&phylic regions/h-bond/complementary base pairing/antiparallel

A
  • Each polynucleotide chain (strand) consists of a chain of nucleotides bonded covalently The sugar-phosphate is on the outside (hydrophilic) and the reactive nitrogenous bases are on the inside (hydrophobic)
  • Two polynucleotide chains of DNA are held together by hydrogen bonds between complementary base pairs:
    -(A=T) via 2 hydrogen bonds
    -(G=C) via 3 hydrogen bonds
  • Complementary base pairing reduces the chance of error during DNA replication or transcription as only T goes with A and C with G.
  • Antiparallel: In order for bases to be orientated correctly and thus able to pair, the two strands must run in opposite directions
    -One strand at C5 at the top and the other has C3 at the top. 5’ and 3’ directions
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4
Q

A. Intro to Molecular Genetics

4.Draw a simple diagram of the structure of single nucleotides of DNA and RNA, using circles, pentagons and rectangles to represent phosphates, pentoses and bases.

how to draw one and 4 nucleotides

A

view sheet

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5
Q

A. Intro to Molecular Genetics

5.Analyze the results of the Hershey and Chase experiment that provided evidence that DNA is the genetic material.

aim/context/procedure/method/conclusion

A

Aim:
wanted to see if proteins or chromosome were the genetic material of cells

Context:
* Took T2 bacteriophage virus which can infect E. Coli bacterium. It attaches to the outside of E coli. and injects a substance into the bacteria to create many new copies of the virus.
* These newly cultured copies of the virus are then expelled when the bacteria bursts.
* Therefore, the substance that is initially injected into the bacteria, whether it is DNA or proteins, will have the instructions on how to make the new viruses. They wanted to discover if it was the DNA core or protein coat that was injected into the bacteria.

Procedure:
* They ran two experiments to test their aim
1. Radioactive sulfur 35 was labeled for the protein
2. Radioactive phosphorus 35 was labeled for the dna
-Elegant method because phosphorus is only found in the nucleotides that make up dna and not in proteins. Vise versa for phosphorus.
* After the viruses infected the bacteria they wanted to see if the radioactive signal was found in the bacteria or outside of it.
* They separated the viruses attached to the bacteria by using a waring blender to do so
-They then centrifuged the samples- since bacteria is bigger it is centrifuged to the bottom into a pellet and the virus was at the top of the tube in the supernatant
-They found that the sulfur labeled virus remained outside of the bacteria and none of the replicated copies had the radioactive virus included. Therefore the protein wasn’t being inherited
-On the other hand, the phosphorus labeled virus was injected into the bacteria and the cultured viruses all had the radioactivity from it. Therefore, DNA was in fact inherited.

Conclusion:
This experiment showed that DNA is in fact the genetic material and protein was not

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6
Q

A. Intro to Molecular Genetics

7.Outline what Crick and Watson were able to deduce regarding the structure of DNA using their model.

method/why/first model-faults/ what they realized/second model suggests

A
  • Watson and Crick used a stick and ball modeling method to test their hypotheses on the possible structures of DNA.
  • This allowed them to visualize the molecule and to quickly see how well it fitted the available evidence.
  • Their first model, a triple helix, was rejected:
    -The ratio of Adenine to Thymine was not 1:1 (as discovered by Chargaff)
    -It required too much magnesium to be true (identified by Franklin)
  • Through this they realized:
    -DNA must be a double helix.
    -The complementary relationship between the bases and base pairing
    -The strands must be anti-parallel to allow base pairing to happen
  • Because of the visual nature of their work the second and the correct model quickly suggested:
    -Possible mechanisms for DNA replication
    -That information was encoded in triplets of bases
  • Roslalind franklin whose x-ray diffraction of DNA was crucial to their discovery
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7
Q

B. DNA Replication

1.Outline Meselson and Stahl’s evidence for the semi-conservative replication of DNA.

state the three theories of dna replication/results from procedure

A

There were three theories to how DNA could have been replicated:
* Semi-conservative replication: each new DNA molecule is composed of one parent strand and one newly synthesized strand
* Conservative replication: the new DNA molecule consists entirely of newly synthesized DNA
* Dispersive replication: each new DNA molecule consists of segments of new and old DNA interspersed

Meselson and Stahl’s work shows that only semi-conservative replication was biologically significant.
Procedure:
* Gen 0: Grew e-coli in heavy nitrogen= N15 100% of the bacteria contained N15.
* Then they transferred them to e coli living in the lighter normal nitrogen N14.
* They spun the e coli in an ultracentrifuge, and DNA with the heavy nitrogen will be moved to the bottom of a test tube whereas those that contain lighter nitrogen (N14) will be moved to the top due to densities.
* Gen 1: The N15 DNA copied again but with the N14 added. This made one half of the DNA be heavy and the other half be lighter (hybrid strands)- due to the different densities of nitrogen- showed up in the middle of the test tube. Debunks conservative replication as it’s a hybrid.
* Gen 2: the strand from gen 1 stayed the same but another DNA molecule appeared above the hybrid location signifying that it consisted of 100% N14 DNA. Dispersive replication is falsified as this shows for there to be one 100% newly synthesized strand and one hybrid while dispersive would just show two that are partially segmented.

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8
Q

B. DNA Replication

2.Explain DNA replication.

both/leading/lagging– reference all 6 enzymes

A

REMEMBER: DNA replication occurs during (S phase of ) interphase, in preparation for mitosis and cell division
BOTH STRANDS:
* Helicase unwinds the double helix and separates the two polynucleotide strands of DNA. Binding proteins keep the strands apart.
* It uses ATP to break the hydrogen bonds between the two strands and move along the DNA strands (single stranded binding proteins keep the separated strands apart so that nucleotides can bind)
* DNA topoisomerase moves in advance of helicase to prevents the DNA from supercoiling.
* Each strand of parent DNA is used as template for the synthesis of the new strands
* synthesis always occurs in 5´ → 3´ direction on each new strand
* To synthesize a new strand an RNA primer is synthesized on the parent DNA using DNA primase.

LEADING STRAND:
* DNA primase synthesizes an RNA primer on the leading parent DNA for attachment and initiation of DNA polymerase III.
* DNA polymerase III adds the nucleotides (only in a 5’ to 3’ direction on the complementary strand- parent strand would be opposite for each side- refer to drawing on sheet) according to the complementary base pairing rules; adenine pairs with thymine and cytosine pairs with guanine; (names needed, letters alone not accepted).
* The free nucleotides which are added by the DNA polymerase III for complementary base pairing are in the form of deoxynucleoside triphosphates. Two phosphate groups are released from each nucleotide and the energy is used to form the covalent phosphodiester bonds between sugars and phosphate groups on the growing chain.
* Once the DNA polymerase III begins it work of adding the nucleotides, DNA polymerase I then removes the RNA primer and replaces them with DNA
* The new strand is made continuously towards the replication fork and as mentioned the bonds being made are phosphodiester covalent bonds.

LAGGING STRAND:
* DNA primase synthesizes a series of RNA primers at intervals leading away from the replication fork of the lagging parent DNA.
* Next the DNA polymerase III adds the nucleotides (to the 3´ end) added according to the complementary base pairing rules; between the primers.
* The free nucleotides which are added by the DNA polymerase III for complementary base pairing are in the form of deoxynucleoside triphosphates. Two phosphate groups are released from each nucleotide and the energy is used to form the covalent phosphodiester bonds between sugars and phosphate groups on the growing chain.
* This leads to the formation of Okazaki fragments on the lagging strand, in order to elongate it away from the replication fork.
* Once the DNA polymerase III begins it work of adding the nucleotides, DNA polymerase I then removes the RNA primer and replaces them with DNA
* DNA ligase next joins Okazaki fragments on the lagging strand to form a continuous strand

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9
Q

B. DNA Replication

3.Outline the important functions of non-coding regions of DNA.

define/role- 6 key non-coding +role

A

Non-coding DNA: all organisms have regions of DNA which are not expressed as polypeptides. They still do however have an important role to organisms, for example:
* Produces tRNA (t=transfer) and rRNA (r=ribosomal)
* Important role in regulating (controlling) gene expression.

Introns:
* Most eukaryotic DNA have introns and exons.
* Introns do not code for a protein so are removed when the mature mRNA is made (the introns and exons are both in DNA and translated to mRNA but spilced out for mature mRNA)
* Since mature mRNA is translated by ribosomes into polypeptides only exons code for polypeptides
* Introns however are needed because it enables multiple proteins to be made from a single gene.

Promoters:
* attachment point for RNA polymerase enabling transcription

Enhancers:
* sequences that increase the rate of transcription when bound to an activator protein

Silencers:
* inhibit transcription when bound to a repressor protein. Block or reduce transcription.

Telomeres:
* There are two types of repetitive sequences that are non-coding: moderately and highly repetitive sequences AKA satellite DNA
* Telomeres are non-coding highly repetitive sequences and they protect the DNA molecule from degradation during replication. They are located towards the end of chromosomes.

Tandem repeats (more below):
* used in DNA profiling because there is more variation in the repeats than in coding sections of genes.

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10
Q

B. DNA Replication

4.Outline how tandem repeats are used in DNA profiling.

define/how its used in profiling

5 is skipped- HL

A
  • Tandem repeat sequences are short sequences of (non-coding) DNA, normally of length 2-5 base pairs, that are repeated numerous times.
  • In a population, the repeating sequence may stay the same but varies greatly in the number of times it may repeat.

For profiling:
* dyes/markers are attached to the tandem repeats during PCR (DNA Replication).
* Restriction enzymes are used to cut DNA between the tandem repeats.
* Gel Electrophoresis enables scientists therefore to calculate the length of the tandem repeat sequence of an individual.
* If different tandem repeats at different loci are used, then a unique profile for an individual can be identified.

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11
Q

C. Transcription & Translation

1.Explain transcription.

4-1st, 4-2nd, 2-3rd

A

Occurs in the nucleus then moves to the cytoplasm for translation.
Initiation
* Transcription begins at a region upstream of the gene called a promoter (eukaryotes call it the TATA box because of the appearance before every gene)
* Proteins called transcription factors first bind to the promoter.
* The transcription factors then assist the enzyme RNA polymerase II to bind. (transcription factors + RNA polymerase II + promoter DNA = transcription initiation complex)
* RNA polymerase then binds to unwind the DNA.

Elongation
* RNA polymerase moves along the DNA and adds ribonucleoside triphosphates that are complementary to the exposed DNA nucleotides to the 3’ end of the growing mRNA molecule -> Transcription occurs in a 5’ to 3’ direction!
* RNA polymerase moves along the anti-sense (template) strand. Like in DNA replication, this complementary strand runs the opposite way ( 3’ to 5’)
* The energy from the cleavage of the additional phosphate groups is utilized to make the covalent bond between nucleotides
* The DNA double helix re-forms behind RNA polymerase and the new mRNA molecules peels away from the DNA template

Termination
* RNA polymerase continues to move along the DNA and synthesize mRNA until it reaches a terminator sequence. This sequence will be located downstream of a stop codon.
* It then detaches from the DNA strand and the newly formed pre-mRNA strand is released.

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12
Q

C. Transcription & Translation

2.Outline the role of the promoter.

A

Promoter: a region of DNA located close to the specific gene and the attachment point for RNA polymerase enabling transcription during initiation. This is where the majority of gene expression is controlled, by either accessing or blocking regions to this site from the RNA polymerase. Once the RNA polymerase is attached it then begins to create the mRNA strand by transcribing the gene.

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13
Q

C. Transcription & Translation

3.Outline how gene expression is regulated by proteins that bind to a specific base sequence in DNA.

two components and their functions

A
  • Transcription is regulated by both transcription factors and regulatory proteins which mediate the binding of RNA polymerase to the promoter
  • Transcription factors:
    -Form a complex with the RNA polymerase at the promoter
    -Since RNA polymerase cannot initiate transcription without these factors– it is crucial part of gene expression
  • Regulatory proteins
    -Bind to DNA sequences outside of the promoter and interact with the transcription factors
    -Activator proteins bind to enhancer sites and increase the rate of transcription (by mediating complex formation)
    -Repressor proteins bind to silencer sequences and decrease the rate of transcription (by preventing complex formation)
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14
Q

C. Transcription & Translation

4.Use examples to outline how the environment has an impact on gene expression.

A

Human hair and skin color are impacted by sunlight and high temperatures.

The Himalayan rabbit produces a different fur pigment depending on the temperature (>35ºC = white fur ; <30ºC = black fur)

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15
Q

C. Transcription & Translation

6.Outline nucleosome role in DNA supercoiling and eukaryotic gene expression.

structure and why its done

A

Structure- formed by wrapping dna around histone proteins
* A nucleosome consists of an octamer of histones (2 copies of 4 different types of histone proteins)
* Histone #1 binds dna in a way to form a 30 nm fiber structure which facilitates further packing
* Linker DNA which holds it all together

Why
* Nucleosomes protect DNA and allow it to be packaged through supercoiling
* It’s essential to supercoil DNA because it helps with…
-organizing DNA in prep for cell division
-controlling DNA expression as supercoiled DNA cannot be transcribed
-allows for the specialization of cells by permanently supercoiling DNA like heterochromatin
-promoting or inhibiting transcription of active chromatin through the use of associated histones

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16
Q

C. Transcription & Translation

9.Outline how the environment can trigger heritable changes in epigenetic factors.

use an example

A

Highly nurtured rat pups tend to grow up to become adults while rat pups who receive little nurturing tend to grow up to be anxious
* The difference is due to epigenetics
* Nurturing behavior of the mother the right during the first week of life affects the pups epigenomes

17
Q

C. Transcription & Translation

10.Outline how eukaryote cells modify mRNA after transcription.

process has a name

A

Within eukaryotic genes, the non-coding sequences, introns, must be removed prior to forming mature mRNA
The coding regions, exons, are fused together when introns are removed to form a continuous sequence
The process by which introns are removed is called splicing

18
Q

C. Transcription & Translation

11.Outline how mRNA splicing increases the number of different proteins an organism can produce.

A

Splicing can also result in the removal of exons – a process known as alternative splicing
The selective removal of specific exons will result in the formation of different polypeptides from a single gene sequence

19
Q

C. Transcription & Translation

12.Explain translation.

define and three stages 4,5,1 steps for each. subunits in 2nd and 3rd

A

Translation is the process of protein synthesis. Genetic information encoded in mRNA is translated into a sequence of amino acids in a polypeptide chain.

Initiation:
* mRNA binds to the small subunit of the ribosome.
* The small subunit of the ribosome moves along the mRNA molecule in a 5’ - 3’ direction until it reaches a start codon (AUG)
* A molecule of tRNA (with it’s amino acid, Methionine attached) with complementary anticodon (UAC) binds to the P site of the ribosome
* The large subunit of the ribosome binds to the tRNA and small subunit

Elongation:
* A second tRNA (with amino acid attached) complementary to the second codon on the mRNA then binds to the A site of the ribosome
* The amino acid carried by the tRNA in the P site is transferred to the amino acid in the A site as a consequence of the ribosome catalyzing a new peptide bond (condensation reaction). The growing polypeptide increases in length.
* The ribosome moves one codon along the mRNA (in a 5’ – 3’ direction):
-The tRNA in the P site is moved to the E site and then released
-The tRNA in the A site is moved into P site
* Another tRNA binds, complementary to the next codon on the mRNA, binds to the A site.
* Steps 6, 7, and 8 are repeated until a stop codon is reached.

Termination:
* When a stop codon is reached translation is stopped:
-a release factor attaches to the A site
-the polypeptide chain is released
-the ribosome complex dissembles ready for reuse translating another mRNA molecule

20
Q

C. Transcription & Translation

13.Use a table of mRNA codons and their corresponding amino acids to deduce the sequence of amino acids coded by a short mRNA strand known base sequence.

A

practice and use table

21
Q

C. Transcription & Translation

14.Outline the production of human insulin in bacteria as an example of the universality of the genetic code.

what is the genetic code and how is human insulin an ex

A
  • All living things use the same bases and the same genetic code
  • Each codon codes for the same amino acid in transcription and translation-regardless of the species
  • The sequence of amino acids in a polypeptide chain remains unchanged
  • Therefore we can take genes from one species and insert them into the genome of another.
  • Human insulin for instance was produced in 1982 through genetically modified E. Coli bacteria. Since then the method of production has been developed using yeast cells and more recently safflower plants.
22
Q

C. Transcription & Translation

17.Contrast protein synthesis by free ribosomes and ER bound ribosomes.

with exception

A
  • Proteins by free ribosomes are produced within the cell
  • Proteins by ER bound ribosomes transport to the golgi apparatus and then export out the cell through a vesicle. Or Proteins by ER bound ribosomes transport are for lysosomes.
23
Q

C. Transcription & Translation

18.Contrast transcription and translation in eukaryotes and prokaryotes.

synchronous vs asynchronous

A

Prokaryotes lack compartmentalized structures (like the nucleus) and so transcription and translation are a synchronous process.
* Ribosomes may begin translating the mRNA molecule while it is still being transcribed from the DNA template
* This is possible because both transcription and translation occur in a 5’ → 3’ direction

The ribosomes in the Eukaryotes, on the other hand, are separated by genetic material (RNA and DNA in nucleus) and thus go through an asynchronous process.
* After transcription, the mRNA must be transported from the nucleus (via nuclear pores) prior to translation by the ribosome.
* This transport requires modification to the RNA construct (e.g. splicing)

24
Q

C. Transcription & Translation

21.Deduce the DNA base sequence for the mRNA strand.

A

practice

25
Q

D.Biotechnology

1.Outline how gel electrophoresis separates proteins or fragments of DNA according to size.

A

Through gel electrophoresis, fragments of DNA are moved through an electric field and separated based on their size.
1.DNA samples are taken and amplified with PCR.
2.Restriction enzymes cut DNA into fragments at specific base sequences in each sample.
3.A fluorescent marker is bound to the DNA fragments or a dye is used, so that results can be seen.
4.Samples are added to a gel electrophoresis chamber. Electric current is passed through, pushing the fragments along.
5.Heavier fragments stay closer to the origin and smaller fragments go further. Negative fragments are moved to the right.
6.A banding pattern shows up for each DNA sample and can be compared.

26
Q

D.Biotechnology

2.Outline how PCR amplifies small amounts of DNA

A
  • The polymerase chain reaction is a direct method of making multiple copies of a desired DNA sequence.
  • It is useful when only a small amount of DNA is available for testing as most DNA analysis tactics are impossible unless you have an ample amount of DNA. e.g. crime scene samples of blood, semen, tissue, hair, etc.
  • PCR occurs in a thermal cycler and involves a repeat procedure of 3 steps:
    1.Denaturation: DNA sample is heated to separate it into two strands (94-96ºC)
    2.Annealing: DNA primers attach to opposite ends of the target sequence (50-65ºC)
    3. Extension: A heat-tolerant DNA polymerase (Taq) copies the strands (72ºC)
  • One cycle of PCR yields two identical copies of the DNA sequence
27
Q

D.Biotechnology

3.Outline the use DNA profiling in paternity and forensic investigations.

A
  • DNA is often left behind at crime scenes and can provide loads of evidence e.g. crime scene samples of blood, semen, tissue, hair, etc.
  • 1986 DNA analysis was first called DNA fingerprinting and is now called DNA profiling or DNA testing for a clear distinction between traditional skin fingerprinting
  • it is easier to exclude a suspect than to convict someone based on a DNA match. The FBI estimates that one-third of initial rape suspects are excluded because DNA samples fail to match.
  • Precautions are taken to limit human mistakes however it is inevitable
  • DNA profiling can be used to determine:
    Inheritance of property, savings etc.
    Paternity investigation (samples are needed from the mother and father in question): Father is required to pay maintenance to support his biological child
28
Q

D.Biotechnology

4.Analyze examples of DNA profiles.

A

practice

29
Q

D.Biotechnology

5.Explain genetic modification using the gene transfer technique with bacteria plasmids, restriction endonucleases and DNA ligase as an example.

A

Gene transfer: the introduction of new DNA into an existing organisms cell usually by vectors such as plasmids or modified viruses.

Genetic modification using gene transfer:
1. The DNA plasmids in E. coli bacteria can be removed and used as a vector; what is used to transfer the gene from one species to another.
2. Restriction enzymes (endonucleases) ‘cut’ the desired gene from the genome. Or an easier way to obtain messenger RNA transcript is to use reverse transcriptase enzymes which can be used to make DNA copies of the RNA molecules called cDNA (complementary DNA). These copies can then be inserted into the plasmids for genetic modification.
3. The same restriction enzyme cuts into the plasmid
4. Because it is the same restriction enzyme the same bases are left exposed, creating ‘sticky ends’
5. Ligase joins the sticky ends, fixing the gene into the E. coli plasmid
6. This recombinant plasmid is inserted into the host cell. It now expresses the new gene. An example of this is human insulin production.

Other examples: golden rice with beta carotene to help with vitamin A production. Salt tolerant tomatoes to improve the yield and uniformity of crops in naturally salt affected areas.

30
Q

D.Biotechnology

6.Assess the risks and benefits of GM crops.

A

Bt corn: Genes from Bt have been inserted into maize so GM plants can produce an insecticidal toxin and therefore be resistant to pests, e.g. European Corn Borer.

Benefits:
* Greater resistance to pests– Bt gene increases resistance to pests such as the European Corn Borer
Bt was successfully introduced and Bt Corn is significantly more resistant to pests
* Increased productivity– less land used / greater yield / less crop damage
Maximum productivity has not increased, but losses in ‘bad’ years have been reduced.
* Less use of chemical pesticides – reduced cost / reduced ecological damage
Bt toxins are much more safer for humans and non-target organisms than most conventional insecticides

Possible risks:
* Could be toxic to or cause allergic reactions in humans
Bt toxin has been used for 30 years without toxicity in humans being detected, however some experimental transgenic plants have caused allergic responses.
* Transferred genes could mutate after testing
* Non-target organisms affected by toxins & Increases resistance to toxin evolves in pests
Bt is a very common soil bacterium – therefore organisms have regularly been exposed to Bt toxins previously. Negative effects on many non-target organisms thought to be minor …
… However many species of caterpillars occur in and around cornfields during the growing season, and might be affected by Bt corn.

Found risks:
* Accidental release may result in competition with native plant species
Unknown but there is a risk
* Super weeds - through cross-breeding the introduced gene could be transferred to wild varieties
known and there is evidence of superweeds evolved from other transgenic crops
* Biodiversity reduced – both plant populations by direct competition and animal populations directly and indirectly could be affected
Unknown ecosystem-wide risks, specific examples have been examined

31
Q

D.Biotechnology

7.Analyze the risks to monarch butterflies of BT corn.

A
  • Caterpillar stage of the monarch butterfly exclusively feeds on milkweed, which grows on the edge of corn fields
  • Studies have shown that some mortality in monarch caterpillars fed on the milkweed leaves covered with Bt corn pollen

First study, Laboratory condition:
* To test this in 1999 A study was conducted where Monarch caterpillars were fed milkweed leaves that had been dusted with pollen from Bt corn
* Growth and mortality rates were compared against caterpillars fed on non-dusted leaves or leaves dusted with non-GM pollen
* Caterpillars exposed to Bt pollen were found to have eaten less, grew more slowly and exhibited higher mortality rates
* This study may lack ecological validity
-There were higher amounts of Bt pollen on the leaves than would be found naturally (e.g. rain would diminish build up)
-Larva were restricted in their diet (in the field, larva could feasibly avoid eating pollen dusted leaves)

Second study, Field condition:
* Compared the survival rate of monarch butterflies based on proximity to Bt corn fields
* There was no significant increase in mortality when monarch larva were placed in or near an actual Bt corn field
* From this it was concluded that exposure to Bt pollen poses no significant risk to monarch butterfly populations

32
Q

D.Biotechnology

8.Define clone.

A

A group of genetically identical organisms. A group of cells derived from a single parent cell.

33
Q

D.Biotechnology

9.Outline natural methods of cloning with examples from plant and animals.

A

Plants:
* Tubers - the swollen tips of underground stems are storage organs in plants (potatoes!)
* Plant stem Cuttings – many but not all plants can be cloned by taking a cutting and immersing it in water.
* Runners like strawberry plants grow horizontal stems with plantlets that can grow roots and become independent clones.
* Bulbs - a single bulb of garlic, when planted will grow leaves that will result in groups of clone bulbs

Animals:
* Monozygotic twins are naturally-occuring clones, epigenetics make it look different.
* Starfish can regenerate a whole body from a single leg.
* Asexual reproduction such as binary fission.

34
Q

D.Biotechnology

10.Design an experiment to assess one factor affecting the rooting of stem-cuttings.

A

Many common plants root easily from stem cuttings producing full-grown (clone) plants quickly.
* Examples of factors that can be investigated:
* Substrate - water, type of soil
* Number of leaves on the stem
* Length of stem cut
* Effectiveness of commercial rooting powders
* Covering with a plastic bag (to reduce transpiration)
* How the stem is cut
* Proximity of a node (point of branching) to the base of the stem

35
Q

D.Biotechnology

11.Outline the method of cloning animals using embryos.

A

Embryos can split and then continue to develop separately to form identical twins.
This is possible because in embryonic development the cells are still unspecialized and can become any type of cell.
Below shows this process being done artificially: using a pipettes cells are extracted from the embryo (A) and implanted into an embryo (to be implanted into a surrogate mother). (B).

see doc for photo if need be

36
Q

D.Biotechnology

12.Outline the cloning method that uses differentiated cells. Use the creation of Dolly the sheep as an example.

A
  • Cloning an organism with desired traits is problematic as a developed organism consists of specialized cells which are multipotent, unipotent, or cannot divide at all.
  • For a clone to develop from somatic (diploid body) cells these cells need to be induced to become pluripotent (cells capable of dividing to become any type of cell).
  • John Gurdon, in 1958, transplanted the nucleus of a (specialized diploid) tadpole intestinal cell into an enucleated (nucleus removed) frog egg.This created tadpoles that were genetically identical to the original tadpole.

This is called somatic-cell nuclear transfer (SCNT)
Creating a genetically identical organism through transfer of a differentiated diploid nucleus.

Dolly the sheep was the first successful cloning of a mammal from a differentiated somatic cell.
Dolly came from Finn Dorset male sheep and Scottish blackface female sheep.

How was dolly created (and any other SCNT process):
* Remove and culture the desired somatic cell (non-gamete).
* Remove a differentiated diploid nucleus from the individual’s cultured cells.
* Collect and enucleate a donor unfertilized egg cell.
* Fuse together the diploid nucleus into the enucleated egg cell using a zap of electrical current.
* The egg is then stimulated to divide and develop into an embryo by the electrical current.
* Implant the egg into the endometrium of a surrogate mother for gestation.
* The newborn will be genetically identical to the donor nucleus parent.