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Exam 3 Flashcards

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1
Q

3 components of a cloning vector

A
  1. Origen of Replication
  2. Selectable makers: genes or traits that enable cells containing the vector to be identified
  3. Cleavage or restriction sites for each of one or more restriction enzymes used
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2
Q

What kind of DNA cut gives sticky or cohesive ends?

A

Staggered cut

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3
Q

In electrophoresis what type of fragments move faster to the positive pole?

A

small fragments

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4
Q

What is the direction of DNA in electrophoresis?

A

From negative pole to positive pole

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5
Q

When a restriction enzyme cuts in the middle of its recognition site it produced ______ ______ fragments

A

blunt ends

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6
Q

what is a cloning vector?

A

Is a stable, replicating DNA molecule to which a foreign DNA fragment can be attached for introduction into a cell.

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7
Q

Expression vectors are used to ___________ ______ ________

A

express the gene

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8
Q

Is there germ-line gene therapy available yet?

A

No.

Only gene therapy has been used to treat somatic cells “Somatic gene therapy”

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9
Q

What is gene therapy?

A

It delivers genes to cells via human virus vectors

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10
Q

Human virus vectors have had viral genes removed to cause them to be _______ _________

A

non-replicative

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11
Q

Types of virus vectors

A
  1. Adenovirus (herpes-simplex virus) - virus remains an episome (not integrated onto chromosome) –> gene expression is transient –> re-dose is necesary because it gets out of the nucleus. –>only for normal dominant genes
  2. Retrovirus - virus integrates into the chromosome. –> long term expression, but can only integrate in dividing cells because cannot cross nuclear membrane.
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12
Q

Adenovirus (4 Characteristics)

A
  1. (herpes-simplex virus) - virus remains an episome (not integrated onto chromosome) –>
  2. gene expression is transient (lasting for a short time) –>
  3. re-dose is necesary because it gets out of the nucleus. –>
  4. only for normal dominant genes
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13
Q

Retrovirus - (4 characteristics)

A
  1. viral DNA integrates into the chromosome. –>
  2. Long term expression,
  3. but can only integrate in dividing cells because cannot cross nuclear membrane.
  4. works on **cells that divide ÷ a lot **
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14
Q

transgenic mouse

  1. Definition
  2. Uses
A
  1. a mouse that has been altered permanently by the addition of a DNA sequence to its genome.
  2. Used to study the function of human genes
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15
Q
A
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16
Q

DNA fragments that are 500 bp, 1000 bp, and 2000 bp in length are separated by gel electrophoresis. Which fragment will migrate farthest in the gel?

A

500 bp

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17
Q

How is a gene inserted into a plasmid cloning vector?

A

The gene and plasmid are cut with the same restriction enzyme and mixed together. DNA ligase is used to seal nicks in the sugar-phosphate bonds.

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18
Q

How gene therapy delivers genes?

A

via human virus vectors

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19
Q

How gene therapy can cause leukemia in patients that were treated for SCID (severe combined immunodeficiency)?

A
  • The virus integrated next to an oncogene in some cells and became cancerous.
  • This promoted the over expression of a normal oncogene –> promoting cell division
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20
Q

Why is SCID (severe combined immunodeficiency) ideal for gene therapy?

A
  1. Easy to get T-cells from the body
  2. T-cells grow outside the body easily
  3. Retrovirous is integrated into T-cells
  4. T-cells with retrovirous is culture
  5. Then T-cells are injected back to the donor
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21
Q

Name one problem with viruses in gene therapy.

A
  • viruses have very active promoters and attract polymerase very quickly.
  • If the gene is inserted next to an oncogene cancer starts as in SCID patients that got leukemia.
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22
Q

Ups and downs of gene therapy

1990

1999

A
  • 1990 Ashanti dSilva was cured of her SCID (severe combined immunodeficiency)
  • 1999 Jesse Gelsinger died of multiple organ failure following injection of a trial dosage of engineered ADENOVIRUS with a functional gene for OTC (ornithine transcarbamylase deficiency)
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23
Q

ups and downs of gene Therapy

2000/2001

2002

A
  • 10 out of 12 patients were cured of SCID (severe combined immunodeficiency)
  • 2 of the 10 cured SCID patients developed leukemia as a result of gene therapy
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24
Q

Explain how an antibiotic-resistance gene and the lacZ gene can be used as markers to determine which cells contain a particular plasmid.

A
  1. Foreign DNAs are inserted into one of the unique restriction sites in the lacZ gene and transformed into E. coli cells.
  2. Transformed cells are plated on a medium containing the appropriate antibiotic to select for cells that carry the plasmid, an inducer of the lac operon, and X-gal, a substrate for B-galactosidase that turns blue when cleaved.
  3. Colonies that carry the plasmid without foreign DNA inserts will have lacZ genes, make functional B-galactosidase, cleave X-gal, and turn blue.
  4. Colonies that carry the plasmid with foreign DNA inserts will not make functional B-galactosidase and will remain white.
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25
Q

How does a genomic library differ from a cDNA libray?

A
  • a genomic library is created by inserting fragments of chromosomal DNA into a cloning vector. Chromosomal DNA is randomly fragmented by shearing or by partial digestionwith a restriction enzyme.
  • a cDNA library is made from mRNA sequences. Cellular mRNAs are isolated and then reverse transcriptase is used to copy the mRNA sequences to cDNA, which are cloned into plasmid or phage vectors.
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26
Q

What is a library?

A

is a set of clones (i.e. bacterial or yeast cells) that collectively contain an entire “set” of DNA

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27
Q

2 Types of library

A
  1. Genomic libraries (contains all of the DNA sequences found in an organism’s genome)
  2. cDNA libraries (from mRNA that is first converted into DNA and then cloned into bacteria.)
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28
Q

Genomic libraries

A
  1. the set of bacterial colonies or phages containing all the DNA fragments from human cells
  2. Total genome of an organism.
  3. a lot of gene material is needed e.g. white blood cells
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29
Q

cDNA library

A

is created from mRNA that is first converted into DNA and then cloned into _bacteria. _

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30
Q

How to create a genomic library

A
  1. cells are collected and disrupted which causes them to release their DNA
  2. DNA is extracted from solution
  3. DNA is cut into fragments by restriction enzymes for LIMITED AMOUNT OF TIME (partial digestion)
  4. DNA molecules will be cut in ¥ (different) places
  5. set of overlapping fragments are produced
  6. fragments are put into vectors
  7. vectors are transfered to bacteria.
  8. The set of clones contain overlapping genomic fragments, some of which may include segments of the gene of interest.
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31
Q

CONCLUSION of construction of genomic libraries

A

some clones conain the entire gene of interest, others include part of the gene, and most contain none of the gene of interest.

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32
Q

Where do restriction enzymes come from?

A

Restriction enzymes exist naturally in bacteria, which use them to prevent the entry of viral DNA

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33
Q

Structural genomics (def.)

A
  1. Organization and
  2. sequence of genetic information contained within a genome.
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34
Q

Basic approach of STRUCTURAL GENOMICS (4 steps)

A
  1. Map a genome –> physical and genetic maps
  2. Break the chromosomes into fragments
  3. sequence the fragments (LIBRARY)
  4. use a computer to resassemble the fragment sequences into whole chromosomes sequences.
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35
Q

Genetic maps are based on rates of _____________ and are measured in __________ _____________, or _____________

A
  • recombination
  • percent recombination
  • centimorgans
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36
Q

True or False?

Withing a year, the publicly-funded HGP (human genomic project) closed down all but 3 largest sequencing centers and began a “quick and dirty” sequence of their own

A

True

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37
Q

The HGP sequence data had always been made available free to the public as it was obtained (GenBank database).

True or false

A

True

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38
Q

Functional genomics

A

Includes identification of:

  1. all RNAs that are transcribed from a genome
  2. all the proteins that are encoded in the genome,
  3. detemination of what those RNAs and proteins do in the cell.
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39
Q

SNP stands for?

A

Single nucleotide polymorphism

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40
Q

Haplotype

A

Specific set of SNPs (single nucleotide polymorphism) observed on a single chromosome.

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41
Q

Many SNPs can be _________ to certain ______________ genes and led to the _________ project

A
  • linked
  • disease
  • HapMap (haplotype map)
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42
Q

Can variation in gene expression, detected by microarrays, be uesed to predict the recurrence of breast cancer?

A

Seventy genes were identify whose expression patterns accurately predicted the recurrence of breast cancer within 5 years of treatment.

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43
Q

microarrays

A
  • rely on nucleic acid hybridization**, in which a known **DNA fragment is used as a probe to find complementary sequences.
  • can provide information about the expression of thousands of genes –> this enables us to study WHICH GENES ARE ACTIVE in particular tissues.
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44
Q

Comparative genomics

A

comparisions fo data from different species helps scientists understand how genomes function and evolve.

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45
Q

Prokaryotic genes:

  1. Proportion of unknown genes
  2. Most known genes
A
  1. 1/3
  2. metabolism
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46
Q

Percentage of human genome consisting of intersperesed repeats derived from transposable elements.

A

44.4%

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47
Q

The percentage of repetitive sequences is usually higher in those species with

A

larger genomes.

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48
Q

Percentage of human genes that is still unknown?

A

40%

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49
Q

Name the 2 most numerus genes known that are functional

A
  1. Transcription factors
  2. Nucleic acid enzymes.
50
Q

Human proteome project

A
  1. The goal of the HPP would be to assign functions to all of the proteins in the human body, and to undestand how the proteins interact with each other in the cell.
  2. Current technology is too SLOW and tedious for protein analysis
51
Q

New paradigm of the Central dogma

A
  1. RNAs are not always just the “middlemen” between DNA and protein.
  2. many RNAs are the “gene product” and are never translated to make protein at all. They do their own function in the cell
52
Q

genes for _________ may make up a good portion of the 40% of genes with “unknown molecular fucntion”

A

RNAs

53
Q

The human _______________ project, will include anaysis of mRNAs and the untranslated RNAs

A

transcriptome

54
Q

Genetics is said to be both a very old science and a young science. Explain, with specific examples what is meant by this statement

A

Genetics is old in the sense that humans have been aware of hereditary principles for thousands of years and have applied them since the beginning of agriculture. It is very young in the sense that the fundamental principles were not uncovered until Mendel’s time, and that the structure of DNA and the principles of recombinant DNA were discovered within the past 60 years.
The use of genetics by humans began with the domestication of plants and animals. The ancient Greeks developed the concepts of pangenesis and the inheritance of acquired characteristics. Preformationism suggested that a person inherits all of his or her traits from one parent. Blending inheritance proposed that offspring posses a mixture of the parental traits.

55
Q

Relate the concept of DNA methylation to the processes of embryonic development, from stem cells to fully differentiated cells. Then name and describe the physical symptoms of a disorder caused by the chemical disruption of this process.

A

DNA methylation is one way that genes are turned off (e.g. they will not be expressed in that cell). The methylation is kept throughout subsequent cell divisions. That means that the newly synthesized DNA (at the end of S phase in the cell cycle) will be methylated in the same places that the parental DNA was methylated.
Turning off genes is the main way that a totipotent stem cell changes into cell types that are increasingly determined to becoming a certain cell type, until the cell is fully differentiated.
_FASD (fetal alcohol spectrum disorder) or FAS (fetal alcohol syndrome) e_ncompasses three broad phenotypes: 1) prenatal and or postnatal growth retardation; 2) distinctive facial features (short palpebral fissure, smooth philtrum, thin vermillion border of the upper lip); and 3) brain damage. Others are cardiac septal defects and minor joint abnormalities.

56
Q

3) How are the events that take place in [mammalian] spermatogenesis and oogenesis similar? How are they different

A
  • Both spermatogenesis and oogenesis begin similarly in that the diploid primordial cells (spermatogonia and oogonia) can undergo multiple rounds of mitosis to produce more primordial cells, or both types of cells can enter into meiotic division.
  • Like spermatogenesis, oogenesis involves the creation of haploid sex cells through the process of meiosis
  • Spermatogenesis:
    • Meiosis I** product is **2 haploid secondary spermatocytes
    • Meiosis II product is 4 haploid spermatids
    • Final result of spermatogenesis is 4 haploid sperm cells for each spermatogonia.
    • Occurs in the male testes. Both maturation and divisions are completed in testes themselves.
    • Small amount of cytoplasm is present
  • Oogenesis:
    • Meiosis I product is 1 haploid secondary oocyte & 1 polar body.
    • Meiosis II** is **1 haploid ovum & 1 polar body.
    • Final result is one egg cell and 2 polar bodies for each oogonia but only if fertilization has occurred.
    • Occurs in female ovaries.
      First maturation division is completed in the ovaries but the second one is completed outside ovaries after fertilization begins.
    • Large amount of cytoplasm is present → providing most of the mitochondria and the cytoplasm for a new fertilized egg.
57
Q

4) Compare and contrast DNA replication and RNA transcription. Name 3 specific similarities and separately, 3 clear differences

A
  • Similarities:
  1. 5’→3’ (new strand) bidirectional from the origin in bacteria.
  2. It is synthesized in the nucleus.
  3. Involve specific complementary base pairing and unwinding of the double helix DNA.
  • Differences DNA replication
  1. The initiation point is at the origin of replication in DNA
  2. The final product is replicated chromosomes Dyads → sister chromatids
  3. Critical enzymes: initiator proteins, helicase, SSBPs, gyrase/topoisomerase, primase, DNA poly III, DNA poly I, ligase.
  • Differences RNA transcription:
  1. The initiation point is at the promoter in DNA
  2. The final product is RNAs (all types).
  3. Critical enzymes: RNA-polymerase (with a built in helicase)
58
Q

Cancer cells are characterized by (4 things)

A
  1. Uncontrolled growth
  2. Loss of “contact inhibition” (or “density-dependent inhibition”) –> “cells cannot tell it is getting too crowded so they keep replicating”
  3. Accumulation of gene mutations and large chromosomal abnormalities.
  4. Ability to metastasize (spread to other areas of the body)
59
Q

From birth to death, your chances of getting some type of cancer is about:

A

1 in 2

(or maybe as good as 1 in 3)

60
Q

3 Characteristics of cancer

A
  1. Most cancers are sporadic (85-95%) as opposed to those with and inhereted component (5-15%)
  2. Most cancers result from an accumulation** of several specific **gene mutations in the same cell, (i.e. the multi-step model of cancer)
  3. Tumors are clonal, so only one original cell needs to accumulate the mutations in order for a tumor to form
61
Q

What is Retinoblastoma?

What is the cause of Retinoblastoma?

A
  • Is a childhood cancer of the eye.
  • It is unusual because it is caused by a homozygous mutation in a single gene
62
Q

In 1971 what was the hypothesis that Alfred Knudsen proposed as the cause of Retinoblastoma?

A

Two-hit hypothesis

63
Q

Results if one Retinoblastoma RB-1 mutation is inhereted.

A

Just one cell of the retina needs to accumulate a mutation in the other RB copy, and tumor will form.

64
Q

Results if NO Retinoblast RB mutation is inhereted.

A

then a single cell must accumulate mutations in both copies of the gene to get the cancer –> have 2 sporadic mutations

65
Q

Familial retinoblastoma typically results in ___________ and ______________

A

early onset and bilateral tumors (both eyes)

66
Q

Sporadic retinoblastoma results in ______________ and _____________

A

later onset and unilateral tumor (one eye)

67
Q

Many cancers are not caused by mutations in just one gene. These types of cancer follow a ______ - ________ ___________

A

Multi-step progression

68
Q

What kinds of genes, when mutated cause cancer?

(the 2 most important)

A
  1. Genes which encode proteins that STOP the cell cycle (Tumor-suppressor genes)
  2. Genes whic encode proteins which PROMOTE the cell cycle (Proto-oncogenes)
69
Q

What kind of genes, when mutated cause cancer? (6 types)

A
  1. Genes which encode proteins important for DNA repair (DNA repair genes) Example: Xeroderma pigmentosum)
  2. Genes that regulate Telomerase
  3. Genes that promote angiogenesis
  4. Genes that are needed for proper _cytoskeleton (palladin gene) _or ECM structure
  5. Genes which encode proteins that STOP the cell cycle (Tumor-suppressor genes)
  6. Genes whic encode proteins which PROMOTE the cell cycle (Proto-oncogenes)(proto = first, before) (A proto-oncogene is a non-mutated form of normal cellular gene that control cell growth and it has the potential to become oncogene causing cancer usually after its overexpression. )
70
Q

Tumor-suppressor genes _______ cancer and are _______. so both alleles must be ________ before inhibition of cell division is removed.

A
  • inhibit
  • recessive
  • mutated
71
Q

Cancer can be the result of _______ of _________ of a _______ ________ gene

A
  • Loss of heterozygosity
  • tumor suppressor gene
72
Q

3 Tumor-suppressor genes and their function studied in class.

A
  1. p53: transcription factor, regulates apoptosis
  2. RB (retinoblastoma): transcription factor
  3. WT-1 Transcription factor (e.g. children’s kidney cancer)
73
Q

2 Facts about p53

A
  1. About ½ of all human cancers express a mutant p53 (Tumor suppressor gene)
  2. Li-Fraumeni Syndrome – an inhereted dominant p53 mutation with leads to high occurrences of a wide range of malignancies (risk developing several types of cancer)
74
Q

What does the tumor suppressor RB (retinoblastoma) do?

A

The RB protein helps control the progression through the G1/S check point by binding transcription factor E2F

  1. RB bind to EF2 keeps it inactive
  2. increasing concentration of cyclin-D-CDK and cyclin -E-CDK phosporilates RB
  3. phosporilation inactivates RB and it is released from E2F
  4. E2F binds to DNA and stimulates the transcription of genes required for DNA replication.
75
Q

Result of an RB (retinoblastoma) mutation

A
  • RB won’t be made
  • E2F will always be active —> which binds to DNA and stimulates the transcription of genes required for DNA replication.
76
Q

Oncogenes mode of action

A
  1. Proto-oncogenes normally produce factors that stimulate cell division
  2. Mutant alleles (oncogenes) tend to be dominant: one copy of the mutant allele is sufficient to induce excessive cell proliferation.
77
Q

About _____ of all tumors have a ras mutation

A

1/3

ras is a Proto-oncogene

78
Q

How to create a fusion protein that is overactive? –> In what type of cancer this is common

A
  1. With a reciprocal translocation** between chromosome 9 and 22 –> **philadelphia chromosome
  2. Chronic myelogenous leukemia (CML)
79
Q

Describe ras proto-oncogene

A
  • Its product is located in the cell membrane
  • Its function is GTP binding and GTPase
80
Q

Describe myc proto-oncogene

A
  1. Its product is located in the nucleus
  2. Its function is transcription factor
81
Q

What causes Burkitt’s lymphoma?

A
  1. Translocation of a potential cancer-causing gene to a new location, where it is activated by different regulatory sequences.
  2. The c-MYC protein stimulates the division fo the B cells and leads to Burkitt’s lymphoma
82
Q

How a virus causes cancer?

A
  1. They cause cancer if they carry a mutated oncogene into the cell –> mutating and rearranging proto-oncogenes, or
  2. if they integrate** onto a chromosome **next** to a **proto-oncogene (A normal gene that has the potential to become an oncogene. T) –> remember viruses carry strong promoters
83
Q

Colon cancer progresses in a _____-_____ manner

A

multi-step

84
Q

Colon cancer progresses in a multi-step manner.

Name the steps and genes involved

A
  1. Loss of normal tummor suppressor gene APC –> a polyp (a bening, precancerous tumor grows)
    * 2 copies need to be lost or mutated to cause cancer (homozygous recessive)
  2. Activation of oncogene ras —> an adenoma (a bening tumor grows)
    * only 1 copy needed to be mutated or lost to cause cancer.
  3. Loss of tumor-suppressor gene p53 –> a carcinoma (s malignant tumor develops)
  • 2 copies need to lost or mutated to cause cancer.
  • So far 4 mutations have accumulated
  1. Other changes; loss of antimetastasis gene –> the cancer metastasizes through the bloodstream
85
Q

2 inhereted forms of colon cancer

A
  1. FAP (familial adenomatous polyposis)–> APC mutation (loss of normal tumor-suppressor gene APC)
  2. HNPCC (hereditary non-polyposis colon cancer) –> DNA repair mutation.

Both result from the same collection of mutations in 1 colon cell, but the gene mutation which is inhereted is different.

[APC, p53, ras]

86
Q

Criteria of Hardy Weinberg Equilibrium

A
  • Large population
  • Random mating
  • No significant rate of mutation, migration, or natural selection
87
Q

CCR5 is a gene that encodes:

A

a cell surface receptor protein that is necessary for HIV-1 to bind to cells.

88
Q

What type of mutation confers resistance to HIV infection?

A

The 32-bp deletion in exon 4 of the CCR5 gene, if homozygous, confers resistance to HIV infection. The Greek letter Δ (delta) stands for “deletion.”

The CCR5 gene has two alleles of importance to us:

  • CCR5-1 (normal)
  • CCR5-Δ32 (mutant)
89
Q

Genetic drift

A

Loss of genetic variation

90
Q

Recurrent mutation eventually leads to

A

an equilibrium, with the allelic frequencies being determined by the relative rates of forward and reverse mutation.

91
Q

Definition of member of the same species

A

An individual belonging to a group of organisms (or the entire group itself) having common characteristics and (usually) are capable of mating with one another to produce fertile offspring

92
Q

Implications of mutation in the Hardy Weinberg equilibrium:

A

New mutations change allelic frequencies, but eventually the rate of forward and reverse mutations will reach equilibrium

93
Q

Implications of migration in the Hardy Weinberg equilibrium:

A

Migration will affect a population’s gene pool if the number of migrating individuals is significant and if the allele frequencies are quite different.

94
Q

Implications of not having a large population size:

A

Genetic drift: large changes in allele frequencies in a short period of time.

Small populations are much more likely to experience genetic drift than large populations.

95
Q

Certain translocations are associated with certain types of cancer.

A
  • Common in CML (Chronic Myelogenous Leukemia) , this translocation creates a fusion protein that is overactive.
  • Reciprocal translocation between Chromosome 9 and 22 (Philadelphia Chromosome) causes CML (Chronic Myelogenous Leukemia)
96
Q

Explain clearly why tortoiseshell cats are almost always female, and why they have a patchy distribution of orange and black fur. Be sure to give the genotype of such a cat and explain the alleles involved.

A
  • Although many genes contribute to coat color and pattern in domestic cats, a single X-linked locus determines the presence of orange color. There are two possible alleles at this locus: X+, which produces non-orange (usually black) fur, and, Xo, which produces orange fur. Males are hemizygous and thus may be black (X+Y) or orange (XoY) but not black and orange.
  • Females may have the tortoiseshell (X+Xo) pattern arising from a patchy mixture of black and orange fur. Each orange patch is a clone of cells derived from an original cell in which the black allele is inactivated (Barr body) and each black patch is a clone of cells derived from an original cell in which the orange allele is inactivated (Barr body).
97
Q

Hairlessness in American rat terriers is recessive to the presence of hair. Suppose that you have a rat terrier with hair. How can you determine whether this dog is homozygous or heterozygous for the hairy trait?

A

Use h for the recessive hairless allele and H for the dominant allele for the presence of hair. Because H is dominant over h, a rat terrier with hair could be either homozygous (HH) or heterozygous (Hh). To determine which genotype is present in the rat terrier with hair, cross this dog with a hairless rat terrier (hh). If the terrier with hair is homozygous (HH), then no hairless offspring will be produced by the test cross. However, if the terrier is heterozygous (Hh), then ½ of the offspring will be hairless.

98
Q

Describe the sequence or (rare) events that are required in order for a recombinant bacterial chromosome to result from conjugation between an Hfr cell and F- cell

A

Hfr x F- = Hfr + F- (no change)

  1. In order to get an Hfr I need to conjugate F+ with an F- first. Then the F factor (produces the sex pili) is integrated into the bacterial chromosome in the F- cells. This takes place with cross over between the F factor and the chromosome.
  2. When this plasmid with the sex pili is part of the bacterial chromosome we get the Hfr.
  3. Then this bacteria is expressing the gene for the pili which can then use it to donate DNA to an F- bacteria by using the rolling over or unidirectional copying. But the chromosome is now too long and the cells won’t stay attach through the pili in order to transmit the whole chromosome.
  4. So the transmission is halted and only a section of the chromosome is transferred to the F-. Which can contain only part of the pili sex genes and another mutant gene.
  5. The transfer strand replicates → then 1) restriction enzymes can degrade the new strand or 2) crossing over can take place between the donated Hfr chromosome and the original chromosome of the F- cell.
  6. This crossing over leads to the recombination of alleles and F- cell contains the mutant allele while the rest of the linear chromosome is degraded.
99
Q

Explain in a short paragraph what is happening genetically in the following pedigree, which shows ABO blood phenotypes of a certain family. Use the correct terminology, and give possible genotypes when appropriate, to demonstrate your ideas.

A
  • The Bombay phenotype suppresses the expression of alleles at the ABO locus. In most people, a dominant allele (H) encodes an enzyme that makes H, a molecule necessary for the production of antigens.
  • People with the Bombay phenotype are homozygous for a recessive mutation (h) that encodes a defective enzyme.
  • The defective enzyme is incapable of making H and, because H is not produced, NO ABO antigens are synthesized.
  • People with genotype hh, who would normally have A, B, or AB blood types, do not produce antigens and therefore express an O phenotype. In this example, the alleles at the ABO locus are hypostatic to the recessive h allele. The gene that does the masking is called an epistatic gene.
100
Q

Why do extra copies of individual genes (or whole chromosomes containing many genes) in a cell sometimes cause drastic phenotypic effects? Use a hypothetical scenario to illustrate your point.

A
  • For some genes, having an extra copy may not cause any detectable problem. For others though, it can upset the carefully regulated interactions inside the cell, and create a problem.
  • This drastic phenotypic effects are due to that gene products from the extra chromosome or individual genes alter the delicate regulated interactions among gene products inside the cell and among cells.
  • Pateau Syndrome which has an extra copy of chromosome 13 ** (47, XX or XY, +13)** will cause more genetic material from this chromosome to produce more proteins responsible for appropriate heart muscle development and neural tube or the roof of the palate. This in turn will create ventricular septal defect, or neural tube defects, cleft palate and other defects. This imbalance in trisomy 13 leads to too many defects for a baby to live past 3 days. Very rare cases make it to 6 months.
101
Q

Explain the process of protein secretion, incorporating Blobel’s signal hypothesis

A
  • Blobel postulated (then demonstrated) that proteins secreted out of the cell contain an intrinsic signal that governs them to and across membranes and allows for their localization in the correct location of the cell.
  • The idea that secretory proteins are synthesized with an amino terminal extension that is recognized by a cytosolic factor and that, together, they are responsible for targeting to the endoplasmic reticulum.
102
Q

Inherited forms of ALS have a dominant mutation in an allele of the gene encoding SOD1. Describe the nature of a common mutation at the DNA level, the effect the mutation has on the protein, and explain how this relates to the symptoms of ALS.

A

Amyotrophic lateral sclerosis (ALS) is a devastating neurodegenerative disease that causes muscle weakness, disability and eventually death. ALS is inherited in most cases in an autosomal dominant pattern. Many patients would consider assisted suicide. The disease severity, treatment ineffectiveness and increasing dependence on caregivers give rise to thoughts to terminate life among patients, especially those at an advance stage. In ALS1, mutation in the Cu-Zn Superoxide Dismutase 1 gene (SOD1) on chromosome 21q22.1 correlates with the ALS phenotype. SOD1 is an enzyme that catalyzes the destruction of O2- free radicals, thus conferring protection to cells from the harmful effects of superoxide free radicals. The mutations in the SOD1 gene are thought to be pathogenic through a gain in toxic property (e.g. by misfolding and aggregation) rather than a loss of function of the protein. Disease duration or rate of disease progression does correlate with some mutations. Particularly the A4V mutation (Deng, 1998) causes about 50% of ALS1 in North America families being consistently associated with a rapid course from symptoms onset until death. This is a missense mutation where a single base pair substitution results in the translation of a different amino acid at that position. The amino acid alanine is replaced by valine in the genetic code, introducing an incorrect amino acid into the protein sequence. Another mutation is I113T, which also occurs at the SOD1 dimer interface. These mutations exhibited reduced superoxide scavenging ability due to misfolding of the homo-dimer at the interface. Currently there are two hypothesis that describe the mechanism by which mutations in SOD1 causes familial ALS. The oligomerization hypothesis, in which normally soluble, functional proteins misfold to form some toxic multimeric species. These aggregated proteins can be also bind to other essential proteins inside the neurons making them unavailable to perform their function. The second hypothesis is the oxidative damage where mutant CuZnSOD proteins are mutated at the metal-binding regions. This mutant CuZnSOD catalyzes reactions with hydrogen peroxide that damages important substrates for cell survival. In summary oligomerization in CuZnSOD proteins seem to be a very important explanation for the toxicity of these proteins in the neurons in patients with ALS. The upper motor neurons and the lower motor neurons degenerate or eventually die, and stop sending messages to muscles. Unable to function, the muscles gradually weaken and atrophy.

103
Q

Avery, McLeod, and McCarty
(1944)

A
  • Discovered that “R” bacteria could become deadly simply when combined with inert lethal “S” form in a test tube—the mice were unnecessary to the equation.
  • They, like Griffith, attempted to transform the R strain into the S strain by incubating living R and heat-killed S. They pre-treated the heat-killed bacteria (S strain) by making 3 samples one with RNase (destroys RNA), the second with Protease (destroys proteins) and the third one with DNase (destroys DNA).
  • Then they mixed the 3 samples with R strain and got the following
  • 1st R strain + RNas = Type R and type S
  • 2nd R strain + Protease = Type R and type S
  • 3rd R strain + DNase = Type R only
  • They concluded that DNA was the transforming susbtance –> Genetic information resides in DNA
104
Q

Hershey and Chase (1952)

A
  • In their experiments, they used two radioactive markers to label the proteins and the DNA of the phages. The proteins were labeled with 35S (a radioactive form of sulfur) and the DNA was labeled with 32P, a radioactive form of phosphorus. This allowed the researchers to easily differentiate between a sample that contained protein (thus would have 35S present) and a sample that only contained DNA (thus would have 32P present).
  • During a phage infection, it was hypothesized that some part of the phage was injected into the bacterium and it was this injected material that conveyed the genetic material necessary to produce new phages. Hershey and Chase determined that the phage injected only the DNA into the bacterium and concluded that DNA must be the genetic material in phages.
  • This was done by a series of two experiments in which different sets of non-radioactive bacteria were incubated with phages that had either their protein or their DNA labeled (as described above).
  • They allowed the phages to infect the bacteria for a short time, they agitated the incubations to dislodge any loose parts of the phages.
  • The bacteria cells were then pelleted (sedimentar) in a centrifuge and the location of the radioactivity. They found that the radioactive DNA was always found with the bacteria cells and that the radioactive protein was always in the supernatant. This suggested that the DNA was injected into the bacteria but the protein coat was not. Thus all of the information needed to produce new viruses was contained in the DNA and not the protein.
105
Q

Amyotrophic lateral sclerosis (ALS)

A
  1. Amyotrophic lateral sclerosis (ALS) is a devastating neurodegenerative disease that causes muscle weakness, disability and eventually death. ALS is inherited in most cases in an autosomal dominant pattern. Many patients would consider assisted suicide.
  2. The disease severity, treatment ineffectiveness and increasing dependence on caregivers give rise to thoughts to terminate life among patients, especially those at an advance stage.
  3. In ALS1, mutation in the Cu-Zn Superoxide Dismutase 1 gene (SOD1) on chromosome 21q22.1 correlates with the ALS phenotype. SOD1 is an enzyme that catalyzes the destruction of O2- free radicals, thus conferring protection to cells from the harmful effects of superoxide free radicals. The mutations in the SOD1 gene are thought to be pathogenic through a gain in toxic property (e.g. by misfolding and aggregation) rather than a loss of function of the protein.
  4. Disease duration or rate of disease progression does correlate with some mutations. Particularly the A4V** mutation (Deng, 1998) causes about 50% of ALS1 in North America families being consistently associated with a rapid course from symptoms onset until death. This is a missense mutation where a single base pair substitution results in the translation of a different amino acid at that position. **The amino acid alanine is replaced by valine in the genetic code, introducing an incorrect amino acid into the protein sequence.
  5. Another mutation is I113T (substitution of the amino-acid Isoleucine 113to the amino-acid threonine)**, which also occurs at the SOD1 dimer interface. These mutations exhibited reduced superoxide scavenging ability due to_misfolding of the homo-dimer at the interface._**
  6. Currently there are two hypothesis that describe the mechanism by which mutations in SOD1 causes familial ALS.
  7. The oligomerization hypothesis, in which normally soluble, functional proteins misfold to form some toxic multimeric species. These aggregated proteins can be also bind to other essential proteins inside the neurons making them unavailable to perform their function.
  8. The second hypothesis is the oxidative damage where mutant CuZnSOD proteins are mutated at the metal-binding regions. This mutant CuZnSOD catalyzes reactions with hydrogen peroxide that damages important substrates for cell survival.
  9. In summary oligomerization in CuZnSOD proteins seem to be a very important explanation for the toxicity of these proteins in the neurons in patients with ALS. The upper motor neurons and the lower motor neurons degenerate or eventually die, and stop sending messages to muscles. Unable to function, the muscles gradually weaken and atrophy.
106
Q

What is the primary structure of a protein?

A

The linear sequence of amino acids (stabilized by peptide bonds between adjacent amino acids)

107
Q

What is the secondary structure of protein?

A

The alpha helix and beta-pleated sheet (stabilized by hydrogen bonds between backbone atoms of non-adjacent amino acids; no R groups are involved in bonding at the level)

108
Q

What is the tertiary structure of a protein?

A

The 3D shape (bonding is between R groups of nonadjacent amino acids; involves disulfide bridges between cysteine residues, ionic or salt bridges, hydrogen bonds, hydrophobic interactions)

109
Q

What is the quaternary structure of a protein?

A

Involves more than one polypeptide chain serving as subunits in a larger complex (all bond types may be involved:. Not all proteins have a quaternary structure.

110
Q

Name two types of tertiary structures.

A

Globular or fibrous

111
Q

Anagenesis

A

Is evolution within a lineage with the passage of time.

112
Q

Cladogenesis

A

Is the splitting of one lineage into two

113
Q

Evolution is genetic change over time. While many things can cause genetic change in a population (mutation, migration, genetic drift), only ____________ ___________ is consistent at producing adaptive changes.

A

natural selection

114
Q

Types of reproductive isolating mechanisms

A

1) Prezygotic

  • Ecological: differences in habitat; individuals do not meet
  • Temporal: reproduction takes place at different times
  • Mechanical: anatomical differences prevent copulation
  • Behavioral: differences is mating behavior prevent mating
  • Gametic: gametes incompatible or not attracted to each other.

2) Postzygotic

  • Hybrid inviability: hybrid zygote does not survive to reproduction
  • Hybrid sterility: hybrid is sterile
  • Hybrid breakdown: F1 hybrids are viable and fertil, but F2 are inviable or steril.
115
Q

Allopatric Speciation

A
  • speciation by geographic isolation,
  • Isolation might occur because of great distance or a physical barrier, such as a desert or river
  • In order for a speciation even to be considered “allopatric,” gene flow between the soon-to-be species must be greatly reduced—but it doesn’t have to be reduced completely to zero.
116
Q

Sympatric Speciation

A
  • Arises when reproductive isolation existe in the absece of any geographic barrier.
117
Q

Phylogenetic trees

A

depict evolutionary relationships based on current data

118
Q

Phylogeny

A

The evolutionary relationship among a group of organisms

119
Q

Synonymous substitutions” are the same as

A

“silent mutations”

120
Q

depict evolutionary relationships based on current data

A

Phylogenetic trees

121
Q

The evolutionary relationship among a group of organisms

A

Phylogeny

122
Q

2 Proto- Oncogenes

3 Tumor-suppressor genes

(studied in class)

A
  • Proto- Oncogene
  1. ras: (1/3 of all tumors have a ras mutation). Located in cell membrane. Function: GTP binding and GTPase.
  2. myc: Located in nucleus. Function: transcription factor
  • Tumor-suppressor genes
  1. p53: (1/2 of all tumors have a p53 mutation) transcription factor, regulates apoptosis
  2. RB (retinoblastoma): transcription factor
  3. WT-1 Transcription factor (e.g. children’s kidney cancer)

Genetics (15 decks)

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