Chapter 6: Identifying Gene Function

Question 1

Module 5

Homologous genes

Answer 1

• share a common evolutionary ancestor, revealed by sequence similarities between the genes

Question 2

Module 5

Orthologous genes

Answer 2

• homologs present in different organisms
• common ancestor predates the split between the species
• usually have the same or very similar functions

Question 3

Module 5

Paralogous genes

Answer 3

• present in the same organism
• often as members of a recognized multigene family
• common ancestor may or may not predate the species in which the genes are now found

Question 4

Module 5

A homology search can be conducted with a _____ sequence, but usually a tentative gene sequence is converted into an _____ _____ sequence before the search is carried out

Answer 4

• DNA
• amino acid

Question 5

Module 5

Why in a homology search, is the sequence converted into an amino sequence?

Answer 5

• there are 20 different amino acids in proteins
• only four nucleotides in DNA
• genes that are unrelated usually appear to be more different from one another when their amino acid sequences are compared

Question 6

Module 5

how is a homology search performed

Answer 6

• by making alignments between the query sequence and sequences from the databases
• For each alignment, a score is calculated
• two ways of generating the score
• simplest programs
  
  • count the number of positions at which the same amino acid is present in both sequences
  • when converted into a percentage, gives the degree of identity between two sequences
• uses chemical relatedness between nonidentical amino acids to assign a score
  
  • higher score for identical or closely related amino acids
  • lower score for less closely related amino acids
• To achieve the highest possible score, the algorithm introduces gaps at various positions in one or both sequences
  
  • parallels processes thought to occur during the evolution of genes

Question 7

Module 5

standard BLAST program is efficient at identifying homologous genes that have more than _____% sequence similarity but is less effective at recognizing evolutionary relationships if the similarity is _____

Answer 7

• 40
• lower

Question 8

Module 5

PSI-BLAST (position-specific iterated BLAST)

Answer 8

identifies more distantly related sequences by combining the homologous sequences from a standard BLAST search into a profile, the features of which are used to identify additional homologous sequences that were not detected in the initial search.

Question 9

Problems with BLAST

Answer 9

the presence in the databases of genes whose stated functions are incorrect.

Question 10

may be possible to deduce at least some part of the function of the gene by searching the amino acid sequence for _____ that encode _____ _____ of known function

Answer 10

• motifs
• protein domains

Question 11

Zinc fingers are _____-_____ structures, so identification in an unknown gene of an amino acid sequence that can encode a zinc finger indicates that the gene codes for

Answer 11

• DNA-binding
• a DNA-binding protein

Question 12

searching for motifs called _____ _____, which direct proteins to organelles such as the nucleus or mitochondria or might specify that the protein is secreted from the cell

Answer 12

sorting sequences

Question 13

The presence of a shared domain indicates that two proteins can perform a similar _____ _____, but that does not necessarily mean that the proteins have

Answer 13

• biochemical activity
• similar overall functions

Question 14

Identification of a domain sequence in an unknown gene therefore identifies a specific ____ _____ to be identified, but on its own this does not enable the actual _____ of the gene to be assigned

Answer 14

• biochemical activity
• function

Question 15

Module 5

If the starting point is the gene, rather than the phenotype, then one strategy is

Answer 15

• to mutate the gene and identify the phenotypic change that results
• basis of most of the techniques used to assign functions to unknown genes

Question 16

Module 5

Gene Inactivation

The easiest way to inactivate a specific gene is to disrupt it with an unrelated segment of DNA. This can be achieved by _____ _____. The vector carries two segments of DNA that match the _____ of the gene to be inactivated. These end segments _____ with the chromosomal copy of the target gene. As a result, the target gene becomes disrupted.

Answer 16

• homologous recombination
• ends
• recombine

Question 17

Module 5

Gene Inactivation w/Homologous Recombination

Deletion Cassette

Answer 17

• deletion cassette consists of
  
  • promoter sequence
  • followed by an antibiotic-resistance gene
  • new segments of DNA are attached to either end
    
    • they have identical parts of the yeast gene to be inactivated
  • flanked by two restriction sites
• start and end segments of the target gene are inserted into the restriction sites, and the vector is introduced into yeast cells
• Recombination between the gene segments in the vector and the chromosomal copy of the target gene results in disruption of the latter
• Cells in which the disruption has occurred are identifiable because they now express the antibiotic-resistance gene and will grow agar medium containing geneticin

Question 18

Module 5

Gene Inactivation w/Homologous Recombination

barcode deletion

Answer 18

• A high-throughput version of the gene inactivation method
• uses a modified version of the basic deletion cassette
• also includes two 20-nucleotide barcode sequences
• different for each deletion, that act as tags for that particular mutant
• Each barcode is flanked by the same pair of sequences and so can be amplified by a single PCR
• groups of mutated yeast strains, each with a different inactivated gene, can be mixed together and their phenotypes can be screened in a single experiment
• the relative abundance of each barcode indicates the abundance of each mutant after growth in glucoserich medium
• Barcodes that are absent or present only at low abundance indicate mutants whose inactivated genes were needed for growth under these conditions.

Question 19

Module 5

Gene Inactivation w/Homologous Recombination

model organism, knockout

Answer 19

• engineered embryonic stem cell is injected into a mouse embryo
• mouse embryo develops and gives rise to a chimera (mouse whose cells are a mixture of mutant ones), derived from the engineered ES cells embryonic site and nonmutant ones, derived from all the other cells in the embryo
• This is still not quite what we want, so the chimeric mice are allowed to mate with one another
• Some of the offspring result from fusion of two mutant gametes producing knockout mice
• works well for many gene inactivations, but some are lethal and so cannot be studied in a homozygous knockout mouse
• Instead, a heterozygous mouse is obtained in the hope that the phenotypic effect of the gene inactivation will be apparent even though the mouse still has one correct copy of the gene being studied

Question 20

Module 5

Gene Inactivation w/out Homologous Recombination

RNA interference, or RNAi

Answer 20

• natural processes by which short RNA molecules influence gene expression in living cells
• provides a means of silencing the expression of a target gene
• doesn’t disrupt gene itself but destroys its mRNA
• short double-stranded RNA molecules, whose sequences match that of the mRNA being targeted are introduced into the cell
• double-stranded RNAs are broken down into shorter molecules, which induce degradation of the mRNA

Question 21

Module 5

Gene Inactivation w/out Homologous Recombination

programmable nuclease

Answer 21

• a nuclease that can be directed to a specific site in a genome
• can be programmed to make a double-stranded cut in a selected gene
• stimulates nonhomologous end-joining
• results in a short insertion or deletion which will inactivate the gene
• creates a true knockout
• example: Cas9 endonuclease

Question 22

Module 5

Gene Inactivation w/out Homologous Recombination

RNA interference, or RNAi issues

Answer 22

1. does not always result in complete silencing of the target gene
   
   • Often the silencing is incomplete and is referred to as knockdown rather than knockout
2. so short that off­target effects are possible
   
   • interfering RNAs bind to mRNAs other than the targets, resulting in silencing of more than one gene
3. In mammals it often results in activation of signaling proteins called interferons, which stimulate an antiviral defense resulting in phenotypic changes that can mask the specific change occurring due to silencing of the target gene.

Question 23

Module 5

Gene overexpression

Answer 23

• used to assess function
• organism is engineered w/test gene much more active than normal (gain of function) to determine what changes, if any, this has on the phenotype
• a multicopy vector is used, that multiplies inside the host organism to 40–200 copies per cell and also contain a highly active promoter sequence
• must be treated w/caution because gene product may synthesized in excessive amounts, possibly in tissues in which the gene is normally inactive making it difficult to distinguish a phenotype change that is due to the specific function of an overexpressed gene

Question 24

Module 5

The critical aspect of a gene inactivation or overexpression experiment is the need to identify a phenotypic change. This can be much more difficult than it sounds. Why?

Answer 24

• the range of phenotypes that must be examined is immense
• effect of gene inactivation can be very subtle and may not be recognized
• many gene inactivations appear to give no discernible phenotypic change
• effects of these genes only become apparent, if at all, when the cells are grown under a range of different conditions or when groups of genes that contribute to the same phenotype are co-inactivated
• In the human genome, there appears to be a subset of several hundred genes that are nonessential, both copies of which can be inactivated, due to natural mutation, without any discernible effect on the health of the individual.
• These observations suggest that a complete functional annotation of the genomes of many species will not be achievable by approaches that are based solely on gene inactivation or overexpression.

Question 25

# Module 6 Additional insights into gene function can be obtained by identifying in which tissues, and at what times, a gene is expressed and by direct examination of the

Answer 25

protein coded by the gene.

Question 26

# Module 5 Reporter Genes

Answer 26

* used to determine the pattern of gene expression within an organism * cells that express the reporter gene may become blue, fluoresce, or give off some other visible signal * must be subject to the same regulatory signals as the test gene, so ORF of the test gene is replaced w/the ORF of the reporter gene

Question 27

# Module 5 Reporter Genes * It is important to know two things: which cells a gene is expressed and * Reporter genes cannot help here because the DNA sequence upstream of the gene—the sequence to which the reporter gene is attached—is not involved in targeting the protein product to its correct intracellular location. Instead, the _____ \_\_\_\_\_\_ ______ of the protein itself contains the targeting information. * the only way to determine where the protein is located is to

Answer 27

* what position within the cell where the protein coded by the gene is found is important. ie. mitochondria, in the nucleus, or on the cell surface * amino acid sequence * search for it directly

Question 28

# Module 5 immunocytochemistry

Answer 28

* used to locate the position within a cell where the protein can be found * uses a labeled antibody that is specific for the protein of interest * antibody binds only to this protein * label allows protein to be visualized via fluorescent labeling and confocal microscopy or electron microscopy

Question 29

# Module 5 site-directed or in vitro mutagenesis

Answer 29

* deletes or alters the relevant part of the gene sequence leaving the bulk unmodified so that the protein is still synthesized and retains the major part of its activity * makes subtle changes to gene * used to find things like: it might be suspected that part of a gene codes for an amino acid sequence that directs its protein product to a particular compartment in the cell or is responsible for the ability of the protein to respond to a chemical

Question 30

# Module 5 site-directed or in vitro mutagenesis Oligonucleotide-directed mutagenesis

Answer 30

* an oligonucleotide containing a single base-pair mismatch, corresponding to the desired mutation, is annealed to a single-stranded version of the relevant gene * it primes a strand-synthesis reaction that continues all the way around the circular template molecule * DNA replication produces numerous copies of this recombinant DNA molecule * Half of these are copies of the original strand of DNA, and half are copies of the strand that contains the mutated sequence * mutant ones are identified by hybridization probing with the original oligonucleotide

Question 31

# Module 5 site-directed or in vitro mutagenesis PCR

Answer 31

* only one mutation can be created per experiment * one normal primer: forms a fully base-paired hybrid with the template DNA * one mutagenic primer: contains a single basepair mismatch * mutation is initially present in two PCR products, each corresponding to one half of the starting DNA molecule * The two PCR products are then mixed together and a final PCR cycle is carried out to construct the full-length, mutated DNA molecule.

Question 32

# Module 5 site-directed or in vitro mutagenesis * The problem w/a site-directed mutagenesis experiment is * The answer is to use a more complex, two-step gene replacement. explain.

Answer 32

* that we must be sure that any change in the activity of the gene being studied is the result of the specific mutation that was introduced into the gene, rather than the indirect result of changing its environment in the genome by inserting a marker gene next to it * two-step gene replacement: * target gene is replaced with the marker gene * cells where recombination takes place are identified by selecting for the marker gene phenotype * the marker gene in these cells is replaced by the mutated gene * we now look for for cells that have lost the marker gene phenotype

Question 33

# Module 5 reverse genetics

Answer 33

* reverse of the conventional approach to genetics * starts with a gene and attempts to discover its function * functional annotation of a genome

Question 34

# Module 5 forward genetics

Answer 34

* conventional approach * starts with a phenotype and attempts to discover the gene or genes responsible for that phenotype * a means of identifying human genes responsible for inherited diseases

Question 35

# Module 6 What is omics?

Answer 35

* Technologies that measure some characteristic of a large family of cellular molecules, such as genes, proteins, or small metabolites * Have been named by appending the suffix “-omics,” as in “genomics.” * refers to the collective technologies used to explore the roles, relationships, and actions of the various types of molecules that make up the cells of an organism. * Began with genomics * Omics = the biology of everything

Question 36

# Module 6 Serial Analysis of Gene Expression (SAGE) Process

Answer 36

* oligo(dT) strands are attached to cellulose beads * mRNA is immobilized in a chromatography column by annealing the poly(A) tails present at the 3' ends to the beads * The mRNA is converted into doublestranded cDNA via restriction enzyme * treated with a restriction enzyme AluI that recognizes a 4 bp target site (5'–AGCT–3'.) and so cuts frequently in each cDNA * 4⁴ cut is every 256 bp * leaves a blunt cut * terminal restriction fragment of each cDNA remains attached to the cellulose beads * all the other fragments to be eluted and discarded * short ds oligonucleotide containing a recognition sequence for BsmFI, is attached to the free end of each cDNA * BsmFI is restriction enzyme that rather than cutting within its recognition sequence, it cuts 10–14 nucleotides downstream * BsmFI therefore removes a fragment with an average length of 12 bp from the end of each cDNA * fragments are collected, ligated head-to-tail to produce a concatamer, and sequenced * the concatamer that is formed is made up partly of sequences derived from the BsmFI oligonucleotides * to avoid this the oligonucleotide can be designed so that the end that ligates to the cDNA contains the recognition sequence for a third restriction enzyme * Treatment with this enzyme cleaves the oligonucleotide from the cDNA fragment

Question 37

# Module 6 Serial Analysis of Gene Expression (SAGE)

Answer 37

* First efforts to measure abundance mRNA of all genes * Based on idea that each gene has unique identifying barcode: the UTRs at the end * Since UTR's are non-coding regions, then they are unlikely to become conserved among members of gene families * Predates Next Gen Sequencing and Sequencing by Synthesis * Amt of sequence that can be obtained was a limitation * Mass amount of nucleic acid could not be done * meant to be used with the Dideoxy chain termination method * necessary to extract the barcodes at the 3' prime UTR for sequencing, means less sequencing could be performed if you extracted the UTR * Rather than studying complete cDNAs, SAGE yields short sequences, as little as 12 bp in length * 12 base sequence found in every 412 = 16,277,216 bases * Average mRNA is 1500 bp * 16,277,26 / 1500 = 10851 mRNAs can be uniquely from the UTR region * number is higher than the number of transcripts expected in all except for the most complex transcriptomes, so the 12 bp should be able to identify the genes coding for all the mRNAs that are present

Question 38

# Module 6 issues with microarray or chip to study one or more transcriptomes Further complications arise if the objective is to compare two or more transcriptomes. name an alternative method.

Answer 38

* design the experiment so that the two transcriptomes can be directly compared, in a single analysis using a single array * done by labeling the cDNA preparations with different fluorescent probes * then scanning the array at the appropriate wavelengths to determine the relative intensities of the two fluorescent signals at each position * determines the differences between the mRNA contents of the two transcriptomes

Question 39

# Module 6 issues with microarray or chip to study one or more transcriptomes Further complications arise if the objective is to compare two or more transcriptomes. why?

Answer 39

* For comparisons to be valid, differences between the hybridization intensities for the same gene with two different microarrays or chips must represent genuine differences in mRNA amount * data analysis must include normalization procedures that enable results from different array experiments to be accurately compared * data analysis include negative controls so that the background can be determined in each experiment, as well as positive controls that should always give identical signals * actin gene is often used as a positive control as its expression level tends to be fairly constant in a particular tissue

Question 40

# Module 6 issues with microarray or chip to study one or more transcriptomes with all but the simplest transcriptomes, hybridization analysis will have insufficient specificity to distinguish between every mRNA that is present. why?

Answer 40

* because two different mRNAs might have similar sequences * when two or more paralogous genes are active in the same tissue * when two or more different mRNAs are derived from the same because of of alternative splicing * exons from a pre-mRNA are assembled in different combinations to give a series of related but different mRNAs * The array must be designed very carefully if all of these variants are to be detected and accurately quantified

Question 41

# Module 6 Using a microarray or chip to study one or more transcriptomes When a transcriptome is analyzed, there are two key objectives. Name them and how are they addressed w/a microarray and a chip

Answer 41

* identify the genes whose mRNAs are present * demands that every relevant gene be represented by at least one probe in the array * W/a microarray it's achieved by using PCR products or cDNAs that are derived from the genes of interest * W/a DNA chip it's achieved by synthesizing at each position a mixture of oligonucleotides which match different positions in the relevant gene * determine the relative amounts of these various mRNAs * met because the microarray or chip contains up to 10⁹ copies of the probe molecules * higher than the anticipated copy number for any mRNA * no position ever becomes saturated (i.e., so much hybridization that every probe molecule is base-paired to a target molecule)

Question 42

# Module 6 microarray or chip analysis Compared with SAGE, this approach has the advantage that

Answer 42

* a rapid evaluation of the differences between two or more transcriptomes can be made by hybridizing different cDNA preparations to identical arrays and comparing the hybridization patterns * can probe the arrya w/cDNA that has been prepared from the mRNA bound to ribosomes rather than from total mRNA * bound mRNAs correspond to the part of the transcriptome that is actively directing protein synthesis * gives a slightly different picture of genome activity

Question 43

# Module 6 microarray or chip analysis hierarchical clustering

Answer 43

* identifies genes that display similar expression profiles that are likely to be ones with related functions * involves comparing the expression levels of every pair of genes in every transcriptome that has been analyzed, and assigning a value that indicates the degree of relatedness between those expression levels. * data can be expressed as a dendrogram, in which genes with related expression profiles are clustered together * The dendrogram gives a clear visual indication of the functional relationships between genes.

Question 44

# Module 6 microarray or chip analysis * DNA chips and microarrays can also be used to study \_\_\_\_\_. * The chip carries an array of _____ \_\_\_\_\_ synthesized in situ on the surface of a wafer of glass or silicon, and microarrays comprise _____ \_\_\_\_\_ —usually PCR products or cDNAs—that have been spotted onto the surface of a glass slide or nylon membrane. * mRNAs is converted into a mixture of \_\_\_\_\_, labeled with a _____ \_\_\_\_\_, and applied to the microarray or chip, and the positions at which hybridization occurs are detected

Answer 44

* transcriptomes * immobilized oligonucleotides * DNA molecules * cDNAs * fluorescent marker)

Question 45

# Module 6 steady state is reached when its rate of synthesis

Answer 45

when its rate of synthesis equals its rate of degradation

Question 46

# Module 6 To increase the amount of an RNA in a transcriptome, the rate of

Answer 46

synthesis must increase or the rate of degradation must decrease.

Question 47

# Module 6 The composition of a transcriptome is determined by the balance between

Answer 47

synthesis and degradation of the individual RNAs that it contains

Question 48

# Module 6 To reduce the amount of an RNA,

Answer 48

the rate of synthesis must decrease or the rate of degradation must increase

Question 49

# Module 6 Understanding how RNAs are synthesized and turned over is therefore central to understanding how the _____ of a transcriptome responds to external stimuli, such as the presence of _____ or an alteration in the \_\_\_\_\_, and also provides us with an appreciation of how gene expression patterns change during

Answer 49

* composition * hormones * environment * differentiation, development, and disease

Question 50

# Module 6 phage display

Answer 50

* methods for studying protein–protein interactions * cloning vector used for phage display is a bacteriophage genome with a unique restriction site located within a gene for a coat protein * technique was originally carried out with the gene III coat protein of the filamentous phage called M13, but has now been extended to other phages including λ * DNA sequence coding for the test protein is ligated into the restriction site so that a fused reading frame is produced—one in which the series of codons continues unbroken from the test gene into the coat protein gene. * After transformation of Escherichia coli, this recombinant molecule directs synthesis of a hybrid protein made up of the test protein fused to the coat protein * Phage particles produced by these transformed bacteria display the test protein in their coats.

Question 51

# Module 6 phage display library

Answer 51

* The test protein is immobilized within a well of a microtiter tray * phage display library added * After washing, the phages that are retained in the well are those displaying a protein that interacts with the test protein

Question 52

# Module 6 activators

Answer 52

* responsible for controlling the expression of genes in eukaryotes * activator must bind to a DNA sequence upstream of a gene and stimulate the RNA polymerase enzyme that copies the gene into RNA * DNA-binding and polymerase activation—are specified by different parts of the activator * some activators will work even after cleavage into two segments, one segment containing the DNA-binding domain and one containing the activation domain. In the cell, the two segments interact to form the functional activator

Question 53

# Module 6 two-hybrid system

Answer 53

* methods for studying protein–protein interactions * uses Saccharomyces cerevisiae strain that lacks an activator for a reporter gene → gene is switched off * Hibryd I * artificial gene coding for the activator DNA-binding domain ONLY is ligated to another gene coding the protein of interest * gene of interest can come from any organism * Hibryd I is introduced into yeast * recombinant yeast strain cannot express the reporter gene because it's missing the activation domain for Hibryd I * it cannot influence the RNA polymerase, it can only bind to DNA * Hibryd II * artificial gene encoding the activation domain ONLY fused to a DNA fragment that specifies a protein able to interact with the protein from Hibrid I * gene is ligated with a mixture of DNA fragments so that many different constructs are made * Hibryd II is introduced into recombinant yeast strain * cells are plated out * A colony in which the reporter gene is expressed contains fusion proteins whose human segments interact, thereby bringing the DNA-binding and activation domains into proximity and stimulating the RNA polymerase

Question 54

# Module 6 phage display and two-hybrid system drawback

Answer 54

* reveals only basic level of protein–protein interaction * cannot handle multiprotein complexes which are composed of core and ancillary proteins * Identifying core and ancillary proteins is a critical step toward understanding how these complexes carry out their functions * large proteins disrupt the phage replication cycle * necessary to display a short peptide, representing part of a cellular protein * short peptide may be unable to interact with all members of the complex because it lacks some of the protein–protein attachment sites present in the intact form

Question 55

# Module 6 affinity chromatography

Answer 55

* works with intact proteins * test protein is attached to a chromatography resin and placed in a column * cell extract is passed through the column in a low-salt buffer, which allows formation of the hydrogen bonds that hold proteins together in a complex * proteins that interact with the bound test protein are retained in the column & others are washed away * interacting proteins are then eluted with a high-salt buffer * disadvantage: need to purify protein, which is time-consuming and difficult * identities of the purified proteins are determined by mass spectrometry

Question 56

# Module 6 tandem-affinity purification (TAP)

Answer 56

* gene for the test protein is modified so that the protein, when synthesized, has a C–terminal extension that binds to a second protein called calmodulin * cell extract prepared under gentle conditions so multiprotein complexes do not break down * extract is passed through an affinity chromatography column packed with a resin containing calmodulin molecules * results in immobilization both of the test protein and others with which it is associated * identities of the purified proteins are determined by mass spectrometry

Question 57

# Module 6 disadvantage with affinity chromatography

Answer 57

* a single member of a multiprotein complex is used as the “bait” for isolation of other proteins from that complex * if a member of a complex does not interact directly with the bait, then it may not be isolated * it identify groups of proteins that are present in a complex, but do not necessarily provide the total protein complement of the complex

Question 58

# Module 6 coimmunoprecipitation

Answer 58

* cell extract is prepared under gentle conditions so that complexes remain intact * An antibody specific for the test protein is added * this results in precipitation of this protein and all other members of the complex

Question 59

multi-dimensional protein identification technique (MudPIT)

Answer 59

* combines various chromatography techniques (e.g., reversed-phase liquid chromatography with either cation-exchange or size exclusion chromatography) in order to isolate intact complexes * components of a complex can then be identified by mass spectrometry

Question 60

# Module 6 Proteins do not need to form _____ \_\_\_\_\_ with one another in order to have a functional interaction. Some biochemical pathways never form physical interactions with one another, and if studies were to be based solely on detection of physical associations between proteins then many _____ \_\_\_\_\_ would be overlooked

Answer 60

* physical associations * functional interactions

Question 61

# Module 6 Several methods can be used to identify proteins that have functional interactions, most of which do not involve direct study of the proteins themselves Comparative genomics

Answer 61

* used in various ways to identify groups of proteins that have functional relationships * observation that pairs of proteins that are separate molecules in some organisms are fused into a single polypeptide chain in others * or based on examination of bacterial operons. * operon consists of two or more genes that are transcribed together and which usually have a functional relationship * identities of genes in bacterial operons infer functional interactions between proteins coded by homologous genes in a eukaryotic genome

Question 62

# Module 6 Several methods can be used to identify proteins that have functional interactions, most of which do not involve direct study of the proteins themselves Transcriptome studies

Answer 62

* can identify functional interactions between proteins * mRNAs for functionally related proteins often display similar expression profiles under different conditions.

Question 63

# Module 6 Several methods can be used to identify proteins that have functional interactions, most of which do not involve direct study of the proteins themselves Gene inactivation studies

Answer 63

* can be informative * If a change in phenotype is observed only when two or more genes are inactivated together, then it can be inferred that those genes function together in generation of the phenotype

Question 64

interactome networks

Answer 64

* Protein interaction maps * display all of the interactions that occur between the components of a proteome

Question 65

interactome networks hubs

Answer 65

* each network is built up around a small number of proteins that have many interactions forming hubs in the network, along with a much larger number of proteins with few individual connections * thought to minimize the effect on the proteome of the disruptive effects of mutations which might inactivate individual proteins * Only if a mutation affects one of the proteins at a highly interconnected node will the network as a whole be damaged

Question 66

interactome networks hubs are divided into 2 groups

Answer 66

* party hubs * hub proteins that interact with all their partners simultaneously * their removal has little effect on the overall structure of the network * party hubs work within individual biological processes, and do not contribute greatly to the overall organization of the proteome * date hubs * interacts with different partners at different times * their removal breaks the network into a series of small subnetworks * the key players that provide an organization to the proteome by linking biological processes to one another

Question 67

The proteome is part of the final link that connects the genome with the _____ of the cell.

Answer 67

* biochemistry

Question 68

metabolome

Answer 68

* the complete collection of metabolites present in a cell or tissue under a particular set of conditions * biochemical blueprint * characterized by infrared spectroscopy, mass spectrometry, and nuclear magnetic resonance spectroscopy which, individually and in combination, can identify and quantify small molecules that make up the metabolites in a cell

Question 69

metabolomics or biochemical profiling

Answer 69

* gives a precise description of the biochemistry underlying different physiological states, including disease states, that can be adopted by a cell or tissue * By converting the biochemistry of a cell into an itemized set of metabolites, metabolomics provides a dataset that can be directly linked to the equivalent from proteomics and other studies of genome expression * most advanced with organisms such as bacteria and yeast whose biochemistries are relatively simple

Question 70

metabolic flux

Answer 70

* rate of flow of metabolites through the network of pathways that make up the cellular biochemistry * Changes in the metabolome can then be defined in terms of perturbations in the flux of metabolites through one or more parts of the network, providing a very description of the biochemical basis to changes in the physiological state

Question 71

metabolic engineering

Answer 71

changes are made to the genome by mutation or recombinant DNA techniques in order to influence the cellular biochemistry in a predetermined way

Question 72

# Module 6 Systems biology

Answer 72

* an approach in biomedical research to understanding the larger picture—be it at the level of the organism, tissue, or cell—by putting its pieces together. * It's in stark contrast to decades of reductionist biology, which involves taking the pieces apart.