HMB265H1 Lecture Notes - Lecture 16: Gene Knockout, Cloning, Pigment

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emeraldcrow840

20 May 2018

School

UTSG

Department

Human Biology

Course

HMB265H1

Professor

Maria Papaconstantinou

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Lecture 16: Transposable Elements

Positional Cloning

- Tomato plants

- Metaphoric description using colour that gives us a description of how much of a certain

DNA is in a certain plant

- First plant is all grey (the plant has the DNA of that particular strain)

- Second plant is black (the pants has DNA from a different strain)

o Can distinguish between the strains

- One plant has a particular trait that we want to isolate for

- One plant has the trait for a big tomato

o Cross this plant with a strain for tiny fruit

- In the F1 generation

o 50% of the DNA from each parent

▪ can have big fruit, small fruit and everything in between

- we only want the big fruit gene isolated

o only pick the big fruit plant (black DNA)

- cross the big fruit plant with the tiny fruit again (back cross)

o more tiny fruit DNA

o 75% of tiny fruit DNA and 25% big fruit DNA

▪ narrowing down the region of where that gene resides for big fruit (25%)

- continue to back cross the big fruit plants until 99% of the DNA is for the tiny plant

o everything BUT the fruit size will come from the tiny fruit plant DNA

o the fruit size will come from the parental black DNA for the big fruit

Transposition

- Movement of small segments of DNA called transposable elements

from one position to another in the genome

- Have evolved the ability to move in the genome

o Jumping genes

- Can insert themselves into open reading frames and knockout genes

- Can excise other genes

- Make certain genes more mutable than others

- Disoered  Barara MClitok i late ’s aarded Noel Prize i 

o Scientists did not agree with her work as it did not align with the chromosomal

theory of inheritance

- Corn: transposable element of pigment genes – when it is inserted into the gene the

colour can be inserted or knocked out

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Trasposale eleets TE’s are ovale geeti eleets

- TE’s are a seget of DNA that eoles the ailit to oe fro place to place within

a genome

- Marus Rhoades ’s) and Barbara McCClintock (1940-’s iferred eistee of

TE’s fro geeti studies of or

o Barbara performed genetic experiments

- TE’s hae o ee foud i all orgaiss

o Scientists were able to clone the transposable elemetns and then realize that

they were in the genomes of all organisms

o Human genome contains 44% of transposable elements

- Previously considered to be selfish DNA – carried no genetic information useful to host

- No ko that soe TE’s hae eoled futios that are eefiial to host

o Drosophila to not have repeats in their telomeres

o Transposable elements insert themselves into the chromosomes and maintains

the size of the chromosomes (beneficial)

- TE length ranges from 50 bp to 10 kb

- TE’s a preset i hudred of thousads of opies per genome

MClitok’s experiets

- One strain of corn frequently had break in chromosome 9

- Ds element helps cause break

- Another element Ac required to activate break at Ds locus

- Dissociation locus

o In order for this to happen another element

(activator) must be present for the break to happen

- Used visual markers to test her theories

- Chromosome 9 had 3 different markers (all dominant)

o Pigment marker

o Plump

o Starchy

▪ This was the wild type phenotype

- If chromosome 9 had all recessive phenotypes

o Plant was colourless, shrunken, and waxy

- Noticed some plants that were colourless plump and starchy

- Dissociation Locus (Ds): nonautonomous element

o Completely dependent on the Ac

- Activator Locus (Ac) autonomous element

- Autonomous elements encode information required for own movement and for

movement of nonautonomous elements

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HMB265H1 Full Course Notes

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Document Summary

Metaphoric description using colour that gives us a description of how much of a certain. First plant is all grey (the plant has the dna of that particular strain) Second plant is black (the pants has dna from a different strain: can distinguish between the strains. One plant has a particular trait that we want to isolate for. One plant has the trait for a big tomato: cross this plant with a strain for tiny fruit. In the f1 generation: 50% of the dna from each parent, can have big fruit, small fruit and everything in between. Movement of small segments of dna called transposable elements from one position to another in the genome. Have evolved the ability to move in the genome: jumping genes. Can insert themselves into open reading frames and knockout genes. Make certain genes more mutable than others.

Related Questions

1. A person is described as a carrier for a particular genetic disease. What does this tell you about their genotype?

2. For any given gene, the principle of independent segregation predicts that each gamete produced as a result of meiosis will contain ____________________

3. A pair of alleles is said to be heterozygous when:

4. In a diploid species, cells can have up to ___ alleles of a particular gene.

5. In a cross between a homozygous dominant parent and a homozygous recessive parent, the number of phenotypic classes observed in the offspring is ______

6. For a person of genotype Xx, the genotype of their gametes may be ______________

7. In a "test cross", an individual of unknown genotype is crossed with a ___ individual to determine the unknown genotype.

8. An albino woman has a child with a phenotypically normal man with no family history of albinism (which is a rare recessive disorder). What percentage of their children are likely to be albino?

9. An individual with the dominant phenotype, but an unknown genotype, is test-crossed, and all of the offspring exhibit the dominant phenotype. The genotype of the unknown individual is ______________

10. If two heterozygous individuals for a recessive disease (carriers) have a child, what is the chance that their child will develop the disease?

11. In a monohybrid cross between two true-breeding parents that both exhibit the dominant phenotype, ___ percent of the offspring will exhibit the recessive phenotype.

12. In a particular type of tomato plant, red fruit is dominant over yellow fruit, and dwarf height is a recessive trait. If two heterozygous plants were crossed, what proportion of the offspring would show the dominant phenotype?

13. In a cross between a BbRr individual and a BBRR one, what do the following represent: BR, Br, bR, br?

14. In humans, the gene that determines whether an embryo develops as a male or a female is carried on _____________

amaranthhare726

1- A new beetle species has arrived in Ohio and ate the wheat, destroying much of the harvest. Research in lab has found that a toxin from the bacterial species Bacillius subtilius can kill the beetle when ingested. You are able to make recombinant DNA containing a plant expression vector and the toxin gene from Bacillius subtiliu. This toxin gene is transcribed and translated in plant cells in the lab. Before introducing this recombinant DNA into corn plants, what main concern do you need to first address?

A) Do we have the technology to express the toxin in this plant cell?

B) Will this gene escape the corn plant and the DNA inserts itself into other plant species?

C) Will the toxin harm humans/livestock when they eat the corn?

D) Will introfuction of the recombinant DNA select for naturally resistent weeds?

E) other

2- You are working on influenza virus infection and want to know if the isg15 gene in mice protects against a particular iinfluenza viru strain? You ask if the isg15 decreases influenza viral replication. You want to silence the expression of the isg15 gene in cells, and you only need the gene silence for a short time (24-48hours). What is the fastest method you would use to silence the isg15 gene?

A) Polymerase chain reaction (PCR) for the isg15 gene

B) Make a ecombinant DNA molecule conyaining the isg15 gene that can be expressed in mouse cells

C) Use homologous recombination to knock out the isg15 gene and then grow up those cells

D) Use a short interfering (si)RNA to the isg15 gene

E) Use a microarray to look at the isg15 levels

3- A mutation in the TATA box would:

A) allow more transcription factors to bind to the DNA

B) prevent RNA polymerase from bindng to the DNA

C) prevent alternative spilcing

D) Mutate the silencer sequence of the DNA

E) Allow RNA polymerase to start transcription

4- If a genes template strand's DNA sequence is 3'-TGA-5', the codon sequence is ___ and the anticodon sequence is ___.

A) 5'-ACU-3' and 3'-UGA-5'

B) 5'-ACT-3' and 3'-TGA-5'

C)5'-UGA-3' and 3'-ACU-5'

D)5'-ACU-3' and 3'-ACU-5'

E)5'-UGA-3' and 3'-UGA-5'

Please help and explain why! Thanks!!

1. Characters that show a continuous range of variation, such as height and eye color, usually are controlled:

a.	by a single gene with two alleles that are codominant.

b.	by many genes with an additive effect.

c.	by epistatic interactions between two genes.

d.	mainly by the environment, with only a small genetic component.

2. In humans, red-green colorblindness is inherited as a sex-linked recessive trait. In order for a woman to be red-green colorblind, which of the following statements must be true.

a.	Her mother must be red-green colorblind.

b.	All of her brothers must be red-green colorblind.

c.	Her father must be red-green colorblind.

d.	All of the above statements must be true if a woman is red-green colorblind.

3. The x-ray crystallography data collected by Rosalind Franklin suggested to Watson and Crick that the:

a.	structure of DNA is a double helix.

b.	two strands of the DNA molecule are joined by hydrogen bonds between the bases.

c.	four bases within DNA pair in a specific way.

d.	two strands of the DNA molecule are joined by covalent bonds between the bases.

4. In the genetic code, _________ one amino acid.

a.	one nucleotide specifies

b.	two nucleotides specify

c.	three nucleotides specify

d.	four nucleotides specify

5. During Meiosis I, a homologous pair of chromosomes may not separate, resulting in daughter cells that have extra chromosomes or are missing chromosomes. This can lead to genetic disorders, including Down Syndrome. This phenomenon is called:

a.	independent assortment.

b.	nondisjunction.

c.	segregation.

d.	crossing over.

6. You are a human geneticist studying the incidence of retinitis pigmentosa in the residents of Tristan de Cunha, a group of small islands in the middle of the southern Atlantic Ocean. The allele for retinitis pigmentosa, which causes a form of blindness, is inherited as an autosomal recessive. You have determined that the frequency of this allele (r) in the population is 0.4 (40%). Using the principles of the Hardy-Weinberg rule, you would estimate the frequency of individuals who are heterozygous for this allele (Rr) in the population to be:

a.	0.16 (16%)

b.	0.24 (24%)

c.	0.36 (36%)

d.	0.48 (48%)

7. Natural selection acts at the level of the:

a.	phenotype.

gene.

c.	population.

d.	nucleotide.

8. You are working with pea plants, trying to recreate the experiments that Mendel performed. You are doing a dihybrid cross with a plant that is heterozygous for both seed shape and seed color, with the genotype RrYy. Which allelic combinations would you expect to find in the gametes produced by this plant?

a.	This plant would produce only RY and ry gametes.

b.	This plant would produce only RrYy gametes.

c.	This plant would produce RY, Ry, rY, and ry gametes.

d.	You cannot determine which gametes this plant can produce without knowing the genotypes of its parents.

9. Biochemist Erwin Chargaff found that in DNA there is a special relationship between the four bases that we now call Chargaff's rule. His observation was that, in an organism's genome the:

a.	percentage of A nucleotides = the percentage of T nucleotides, and the percentage of C nucleotides = the percentage of G nucleotides.

b.	four bases all occur in an equal frequency (25%) within each organism.

c.	percentage of A nucleotides = the percentage of G nucleotides, and the percentage of C nucleotides = the percentage of T nucleotides.

d.	genetic material is composed of proteins, not DNA.

10. During DNA replication:

a.	each strand of the double helix acts as a template for the synthesis of a new strand.

b.	the enzyme DNA polymerase adds nucleotides to the strand being synthesized.

c.	the bases A,C,G and T are required.

d.	All of the above are true of DNA replication.

11. During translation, amino acids are joined by peptide bonds to make polypeptides. The formation of these peptide bonds is catalyzed by:

12. If an allele (R) at a gene with two alleles shows complete dominance, individuals with the genotypes ______ will have the same phenotype.

a.	RR and rr.

RR and Rr

Rr and rr

d.	Each of the three possible genotypes will have a different phenotype.

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HMB265H1 Lecture Notes - Lecture 16: Gene Knockout, Cloning, Pigment

HMB265H1 Full Course Notes

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