Genetics Final study set

Outline how DNA replication occurs at a replication fork

(see Figure 11.7, Table 11.1). Functions of key proteins involved with bacterial DNA replication •DNA Helicase breaks the hydrogen bonds between the DNA strands. •Topoisomerase II alleviates positive supercoiling. •Single-stranded binding proteins keep the parental strands apart •Primase synthesizes an RNA primer •DNA polymerase III synthesizes a daughter strand of DNA •DNA polymerase I excises the RNA primers and fills in with DNA (not shown) •DNA ligase covalently links the Okazaki fragments together

Explain the results of Mendel's single-factor crosses and how they allowed him to conclude that genes are unit factors that may dominant or recessive, and they segregate during gamete formation

(see Figure 2.5, 2.6) For all seven characteristics studied •The F1 generation showed only one of the two parental traits •The F2 generation showed an ~ 3:1 ratio of the two parental traits These results refuted a blending mechanism of heredity Indeed, the data suggested a particulate theory of inheritance Mendel postulated the following: •A pea plant contains two discrete hereditary factors for a given character, one from each parent •The two factors may be identical or different When the two factors of a single character are different and present in the same plant •One variant is dominant and its effect can be seen •The other variant is recessive and is not seen •Particulate theory of inheritance : the genetic determinants that govern traits are inherited as discrete units that remain unchanged as they are passed from parent to offspring During gamete formation, the paired factors for a given character segregate randomly so that half of the gametes receive one factor and half of the gametes receive the other •This is Mendel's Law of Segregation

Explain how meiosis is related to the law of segregation and the law of independent assortment

(see Figures 3.15, 3.16). Chromosomes contain the genetic material Chromosomes are replicated and passed from parent to offspring The nuclei of most eukaryotic cells contain chromosomes that are found in homologous pairs, they are diploid •During meiosis, each homolog segregates into one of the two daughter nuclei During the formation of gametes, different types of (nonhomologous) chromosomes segregate independently Each parent contributes one set of chromosomes to its offspring •The sets are functionally equivalent •Each carries a full complement of genes The chromosome theory of inheritance allows us to see the relationship between Mendel's laws and chromosome transmission •Mendel's law of segregation can be explained by the homologous pairing and segregation of chromosomes during meiosis •Mendel's law of independent assortment can be explained by the relative behavior of different (nonhomologous) chromosomes during meiosis

Outline the general organization of bacterial chromosome sequences

10.1 Key features: •Most, but not all, bacterial species contain circular chromosomal DNA. •Most bacterial species contain a single type of chromosome, but it may be present in multiple copies. •A typical chromosome is a few million base pairs in length. •Several thousand different genes are interspersed throughout the chromosome. The short regions between adjacent genes are called intergenic regions. •One origin of replication is required to initiate DNA replication. •Repetitive sequences may be interspersed throughout the chromosome.

Describe the basic structural features that underlie DNA replication, and what the products are

11.11 DNA replication relies on the complementarity of DNA strands •A pairs with T and G pairs with C •The AT/GC rule or Chargaff's rule The process is: •The two complementary DNA strands come apart •Each serves as a template strand for the synthesis of new complementary DNA strands •The two newly-made DNA strands = daughter strands •The two original DNA strands = parental strands

Describe the ability of ncRNAs to bind to other molecules

17.1 Some genes do not encode polypeptides, but are transcribed into non-coding RNAs (ncRNAs) •Estimates of the number of ncRNAs in humans range from several thousand to tens of thousands. •In most cell types, ncRNAs are more abundant than mRNAs. ncRNAs can bind to DNA or RNA due to complementary base pairing RNA molecules can also form stem-loop structures, which may bind to pockets on the surface of proteins

Explain how nucleotides are connected in a DNA strand

9.7 Nucleotides are covalently linked together by phosphodiester bonds •A phosphate connects the 5' carbon of one nucleotide to the 3' carbon of another Therefore the strand has directionality •5' to 3' •In a strand, all sugar molecules are oriented in the same direction The phosphates and sugar molecules form the backbone of the nucleic acid strand •The bases project from the backbone

Incomplete dominance

Figure 4.5, 4.6 inheritance: this pattern occurs when the heterozygote has a phenotype that is an intermediate between either corresponding homozygote. (For example, a cross between red flowers and white flowers results in pink flowers) molecular explanation: 50% of the protein, produced by a single copy of the functional allele, is not sufficient to produce the same trait as the homozygote making 100%

Be able to set up a Punnett square for single-factor and two-factor crosses

ch 2 you know how to do this, can't really make a flashcard out of it though :/

List the four criteria that the genetic material must fulfill

ch 9 To fulfill its role, the genetic material must meet several criteria 1.Information: It must contain the information necessary to make an entire organism 2.Transmission: It must be passed from parent to offspring 3.Replication: It must be copied •In order to be passed from parent to offspring 4.Variation: It must be capable of changes •To account for the known phenotypic variation in each species

Outline the relationships between the coded information in a gene and the synthesis of a polypeptide

fig 13.3 Polypeptide synthesis has a directionality that parallels the 5' to 3' orientation of mRNA During each cycle of elongation, a peptide bond is formed between the carboxyl group of the last amino acid in the polypeptide chain and the amino group in the amino acid being added The first amino acid has an exposed amino group •Said to be N-terminal or amino terminal end The last amino acid has an exposed carboxyl group •Said to be C-terminal or carboxyl terminal end

List the general functions of ncRNAs

pg 413 Scaffold •A ncRNA binds a group of proteins Guide •A ncRNA binds to a protein and guides it to a specific site in the cell Alteration of Protein Function or Stability •A ncRNA binds to a protein and alters that protein's structure •the ability of the protein to act as a catalyst •the ability of the protein to bind to another molecule •the stability of the protein Ribozyme •RNA molecules with catalytic function Blocker •An ncRNA physically prevents or blocks a cellular process from happening Decoy •A ncRNA recognizes another ncRNA and sequesters it

Define selective breeding and selection limit

see Figures 28.9-28.11). •Selective breeding is the modification of phenotypes in plant and animal species of economic importance -It is also called artificial selection -The primary difference between artificial and natural selection is how the parents are chosen •Natural selection is due to natural variation in reproductive success •In artificial selection, the breeder chooses individuals with traits that are desirable from a human perspective •After many generations, the population will eventually become monomorphic for all or most of the desirable alleles in question •At this point, additional selective breeding will have no effect. This is known as a selection limit

Define mutagen and distinguish between chemical and physical mutagens

table 19.6 Agents that alter the structure of DNA and thereby cause mutations are called mutagens The public is concerned about mutagens for two main reasons: 1.Mutagens are often involved in the development of human cancers 2.Mutagens can cause gene mutations that may have harmful effects in future generations An enormous array of agents can act as mutagens Mutagenic agents are usually classified as chemical or physical mutagens Chemical mutagens come into three main types: 1.Base modifiers •Some covalently modify base structure •Others disrupt pairing by alkylating bases 2.Intercalating agents •Directly interfere with replication process 3.Base analogues •Incorporate into DNA and disrupt structure •Some tautomerize at high rate Physical mutagens include radiation: •X-rays, gamma rays, UV light

Describe the action of chromatin remodeling enzymes

15.9 ATP-dependent chromatin remodeling refers to dynamic changes in chromatin structure These changes range from a few nucleosomes to large scale changes Carried out by diverse multiprotein machines that reposition and restructure nucleosomes (a) Change in nucleosome position (b) Histone eviction (c) Replacement with histone variants

Be able to use the Hardy-Weinberg equation to calculate allele frequencies and genotype frequencies

27.4, 27.5 •p + q = 1 p^2 + 2pq + q^2 = 1 allele frequency (p, q) = # of copies of alleles in a pop. / total # of alleles in the pop. genotype freq (p^2, 2pq, q^2) = # with a particular genotype / total # of individuals in pop.

Also be able to predict the outcome of a cross if you already know the map distance between two genes.

ch 6 map distance = (# of recombinant offspring / total offspring) x 100 •One map unit is equivalent to 1% recombination frequency

Describe the basic structural features of the DNA double helix

(see Figure 9.12, 9.13, 9.14). Two strands are twisted together around a common axis •There are 10 base pairs and 3.4 nm per complete turn of the helix The two strands are antiparallel •One runs in the 5' to 3' direction and the other 3' to 5' The helix is right-handed •As it spirals away from you, the helix turns in a clockwise direction The double-bonded structure is stabilized by: 1.Hydrogen bonding between complementary bases •A bonded to T by two hydrogen bonds •C bonded to G by three hydrogen bonds 2.Base stacking •Within the DNA, the bases are oriented so that the flattened regions are facing each other (see Figure 9.13a) There are two asymmetrical grooves on the outside of the helix 1.Major groove 2.Minor groove Certain proteins can bind within these grooves •They can thus interact with a particular sequence of bases

Describe the two key characteristics of stem cells and why they may be useful medically, but don't memorize Table 22.4

(see Figures 22.8, 22.9, 22.10). Stem cells supply the cells that construct our bodies from a fertilized egg •In the adult, stem cells also replenish damaged cells Stem cells have two common characteristics •They have the capacity to divide •They have the capacity to differentiate into one or more specialized cell types •totipotent cells, like fertilized eggs, can give rise to all cell types •pluripotent cells can differentiate into almost every cell, but can't give rise to an entire, intact individual •multipotent cells can differentiate into several cell types •unipotent cells can only differentiate into one cell type When a stem cell divides, one may remain undifferentiated, while the other can differentiate into a specialized cell type •Thus the population of stem cells remains constant In mammals, stem cells are commonly categorized based on their developmental stage and their ability to differentiate Embryonic stem cells (ES cells) are found in the inner cell mass of the blastocyst Embryonic germ cells (EG cells) are the germ-line cells found in the gonads during the fetal stage Stem Cells Have the Potential to Treat a Variety of Diseases In 2006: Shinya Yamanaka and colleagues showed that adult mouse fibroblasts could become pluripotent •By injection of four different genes that encode transcription factors •Called induced pluripotent stem cells (iPS cells)

Describe the concept of a genetic map, and be able to use testcross data to make a linkage map

(see Figures 6.7, 6.8, 6.9). Genetic mapping is also known as gene mapping or chromosome mapping Its purpose is to determine the linear order of linked genes along the same chromosome. •each gene has its own unique locus Uses of genetic maps: 1.They allow us to understand the overall complexity and genetic organization of a particular species 2.They can help molecular geneticists to clone genes 3.They improve our understanding of the evolutionary relationships among different species 4.They can be used to diagnose, and perhaps, someday to treat inherited human diseases 5.They can help in predicting the likelihood that a couple will produce children with certain inherited diseases 6.They provide helpful information for improving agriculturally important strains through selective breeding programs Genetic maps allow us to estimate the relative distances between linked genes, based on the likelihood that a crossover will occur between them Experimentally, the percentage of recombinant offspring is correlated with the distance between the two genes •If the genes are far apart à many recombinant offspring •If the genes are close à very few recombinant offspring map distance = (# of recombinant offspring / total offspring) x 100 The units of distance are called map units (mu) •They are also referred to as centiMorgans (cM) •One map unit is equivalent to 1% recombination frequency

Outline the central dogma of genetics

12.1 DNA replication: makes DNA copies that are transmitted from cell and from parent to offspring. Chromosomal DNA: stores information in units called genes. Transcription: produces an RNA copy of gene. Messenger RNA: a temporary copy of a gene that contains information to make a polypeptide. Translation: produces a polypeptide using the information in mRNA. Polypeptide: becomes part of a functional protein that contributes to an organism's traits.

Outline how eukaryotic genes may be regulated at many points along the gene expression pathway

15.1 Eukaryotic gene regulation is necessary to ensure 1.They can respond to changes in nutrient availability 2.They can respond to environmental stresses 3.Expression of genes in an accurate pattern during the various developmental stages of the life cycle •Some genes are only expressed during embryonic stages, whereas others are only expressed in the adult 4.Differences among distinct cell types •Nerve and muscle cells look so different because of gene regulation rather than differences in DNA content regulation can occur during transcription, RNA processing, translation, and post-translational modifications

Describe the components of nucleotides but you don't need to memorize their structures. Distinguish between pyrimidines and purines, and between the components found in DNA nucleotides versus RNA nucleotides

9.4, 9.5 DNA and RNA are large macromolecules with several levels of complexity •Nucleotides form the repeating unit of nucleic acids •Nucleotides are linked to form a linear strand of RNA or DNA •Two strands can interact to form a double helix •The 3-D structure of DNA results from folding and bending of the double helix. Interaction of DNA with proteins produces chromosomes within living cells The nucleotide is the repeating structural unit of DNA and RNA It has three components •A phosphate group •A pentose sugar •Ribose in RNA (OH on 2' C) •Deoxyribose in DNA (H on 2' C) •A nitrogenous base purines are double rings (A and G), pyrimidines are single rings (C and T (in DNA) and U (in RNA)

Describe the concept of combinatorial control

ch 15 Common factors contributing to combinatorial control are: •One or more activator proteins may stimulate transcription •One or more repressor proteins may inhibit transcription •Activators and repressors may be modulated by: •binding of small effector molecules •protein-protein interactions •covalent modifications •Regulatory proteins may alter nucleosomes near the promoter (part 2 of Chapter 15 lecture) •DNA methylation may inhibit transcription (part 3 of Chapter 15 lecture) •prevent binding of an activator protein •recruiting proteins that compact the chromatin •Various combinations of these factors can contribute to the regulation of a single gene

Explain what a transgenic organism is

ch 22 •Genetically modified organisms (GMOs) have received genetic material via recombinant DNA technology •An organism that has integrated recombinant DNA from another species into its genome is called transgenic

Compare and contrast mutations that produce oncogenes (gain-of-function) versus those that affect tumor-suppressor genes (loss-of-function).

ch 25 Oncogenes promote abnormal cell growth Proto-oncogenes are normal cellular genes that can be mutated into an oncogene Expression becomes abnormally active •This is a gain-of-function mutation This can occur in three common ways: 1.The amount of the encoded protein is greatly increased. 2.A change occurs in the structure of the encoded protein that causes it to be overly active. 3.The encoded protein is expressed in a cell type where it is not normally expressed. •By studying proto-oncogenes, researchers have found that this occurs in four main ways: •1. Missense mutations •2. Gene amplifications •3. Chromosomal translocations •4. Viral integration Tumor- suppressor mutations: 1.A mutation in the tumor-suppressor gene itself •The promoter could be inactivated •An early stop codon could be introduced in the coding sequence 2.DNA methylation or other epigenetic changes •The methylation of CpG islands near the promoters of tumor-suppressor genes, inhibits transcription 3.Aneuploidy •Chromosome loss may contribute to the progression of cancer if the lost chromosome carries one or more tumor-suppressor genes

Explain the results of Mendel's two-factor crosses and how they allowed him to propose the law of independent assortment

(see Figures 2.7, 2.8, 2.9). The F2 generation contains seeds with novel combinations (i.e.: not found in the parentals) •Round and Green •Wrinkled and Yellow These are called nonparentals Their occurrence contradicts the linkage model If the genes, on the other hand, assort independently •Then the predicted phenotypic ratio in the F2 generation would be 9:3:3:1 Mendel's data was very close to segregation expectations. Thus, he proposed the law of Independent Assortment: •During gamete formation, the segregation of any pair of hereditary determinants is independent of the segregation of other pairs

Describe the general organization of the different types of transposable elements

20.9 DNA sequences within transposable elements are organized in several different ways •Direct repeats (DRs) are identical sequences found on both sides of all TEs. •Inverted repeats (IRs) are at the ends of some transposable elements. (all simple transposons) •Long terminal repeats (LTRs) are regions containing a large number of tandem repeats. (some but not all retrotransposons)

Explain how gene number and the environment affect the overlap between genotypes and phenotypes for polygenic traits

28.4 Polygenic Inheritance and Environmental Factors May Produce a Continuum of Phenotypes •Many polygenic traits are difficult or impossible to categorize into several discrete genotypic categories •This is especially true when -1. The number of genes controlling the trait increases -2. The influence of the environment increases Genotypes and Phenotypes may Overlap for Quantitative Traits

Describe the four types of changes in chromosome structure

8.2 There are two primary ways in which the structure of chromosomes can be altered •The total amount of genetic material in the chromosome can change •Deficiencies/Deletions •Duplications •The total amount of genetic material remains the same, but is rearranged •Inversions •Translocations Deficiency (or deletion) : The loss of a chromosomal segment Duplication : The repetition of a chromosomal segment compared to the normal parent chromosome Inversion : A change in the direction of part of the genetic material along a single chromosome Translocation : A segment of one chromosome becomes attached to a different chromosome •Simple translocations : A piece of a chromosome is attached to another chromosome •Reciprocal translocations : Two different types of chromosomes exchange pieces, producing two abnormal chromosomes with translocations

Define homology

ch 24 Homologous Genes Are Derived from the Same Ancestral Gene •When two homologous genes are found in different species, these genes are termed orthologs •When two homologous genes are found in a single organism, these genes are termed paralogs -A gene family consists of two or more copies of homologous genes within the genome of a single organism

Explain the concept of gene regulation, and where it may occur in the gene expression pathway

14.1 The term gene regulation means that the level of gene expression can vary under different conditions Genes that are unregulated are termed constitutive •They have essentially constant levels of expression •Frequently, constitutive genes encode proteins that are continuously necessary for the survival of the organism The benefit of regulating genes is that encoded proteins will be produced only when required Gene regulation is important for cellular processes such as 1.Metabolism 2.Response to environmental stress 3.Cell division Regulation can occur at any of the points on the pathway to gene expression transcription, translation, post-translation

Explain how epigenetic changes can occur during development via genomic imprinting

16.4 Genomic imprinting is a form of gene regulation in which an offspring expresses the copy of a gene from one parent but not both (chapter 5) •Example: in mammals, only the Igf2 gene inherited from the father is expressed •The Igf2 gene is de novo methylated during sperm formation but not during egg formation (refer to Figure 16.4) •The methylation occurs at two sites: the imprinting control region (ICR) and a differentially methylated region (DMR) •Methylation inhibits the binding of a protein called the CTC-binding factor, which allows the Igf2 gene to be stimulated by a nearly enhancer. •In contrast, CTC-binding factor binds to the unmethylated gene and inhibits transcription by forming a loop

Be able to calculate heritability using correlation coefficients

28.8 •There are several ways to estimate narrow sense heritability. A common strategy is the following: hN2 = robs /rexp robs is the observed phenotypic correlation between related individuals rexp is the expected correlation based on the known genetic relationship •For siblings, rexp = 0.5, unless identical twins, rexp = 1.0 •For parent-offspring relationship, rexp = 0.5 •For uncle-niece relationship, rexp = 0.25

Apply the product rule and binomial expansion equation in genetic problems. (product rule)

ch 2 product rule: The probability that two or more independent events will occur is equal to the product of their respective probabilities •Independent events are those in which the occurrence of one does not affect the probability of another Step 1: Calculate the individual probabilities •This can be obtained via a Punnett square Step 2: Multiply the individual probabilities 1/4 × 1/4 × 1/4 = 1/64 1/64 can be converted to 0.016, or 1.6% •Therefore 1.6% of the time, the first three offspring of a heterozygous couple, will all have congenital analgesia

Explain the concept of a chromosome, and how genes are found within chromosomes.

ch 3 Chromosomes are structures within living cells that contain the genetic material •They contain the genes Biochemically, chromosomes are composed of •DNA, which is the genetic material •Proteins, which provide an organized structure •In eukaryotes the DNA-protein complex is called chromatin

Explain what the BLAST program does

table 24.5 •BLAST starts with a genetic sequence and then locates homologous sequences in a large database -Homology among protein sequences is easier to identify than is DNA sequence homology -The amino acid sequence of human phenylalanine hydroxylase was used as a "query sequence" by the BLAST program -Within minutes, BLAST can search the entire database and determine which sequences are the closest matches •Small E-Value indicates that similarity is unlikely to be due to random events they may be homologous •E-values depend on length of query, number of gaps in alignment and database size Homologous Genetic Sequences Can Identify Conserved Sites that Are Functionally Important

Outline the organization of the lac operon, and why it makes a bacterial cell more efficient in the utilization of lactose

14.3 1.The lac operon DNA elements •Promoter: Binds RNA polymerase •Operator: Binds the lac repressor protein •CAP site: Binds the Catabolite Activator Protein (CAP) Protein-encoding genes •lacZ : Encodes b-galactosidase •Enzymatically cleaves lactose and lactose analogues •Also converts lactose to allolactose (an isomer) •lacY : Encodes lactose permease •Membrane protein required for transport of lactose and analogues •lacA : Encodes galactoside transacetylase •Covalently modifies lactose and analogues The lacI gene •Not considered part of the lac operon •Has its own promoter, the i promoter •Constitutively expressed at fairly low levels •Encodes the lac repressor •The lac repressor protein functions as a tetramer •Only a small amount of protein is needed to repress the lac operon

Compare and contrast gene modification and gene addition

22.2 The genomes of animals can be altered by Gene Modification or Gene Addition. The introduction of a cloned gene into a cell can lead to one of two outcomes •Gene modification alters the sequence of a gene •Gene addition •In mice, researchers can produce a gene knockin, in which a gene of interest has been inserted into a particular site in the mouse genome Genes can be added from different species •Gene knockouts are used to study how the loss of normal gene function affects the organism •Frequently have effects on the phenotype of a mouse indicating the function of the gene within a particular organ or stage of development •A gene knockout with no obvious phenotypic effect may be due to gene redundancy

Explain the concept of linkage and how it is affected by crossing over

fig 6.2 The term synteny means two or more genes are located on the same chromosome and are physically linked Genetic Linkage is the phenomenon that genes close together on a chromosome tend to be transmitted as a unit, which influences inheritance patterns Chromosomes are called linkage groups •They contain a group of genes that are linked together The number of linkage groups is the number of types of chromosomes of the species •For example, in humans •22 autosomal linkage groups •An X chromosome linkage group •A Y chromosome linkage group Genes that are far apart on the same chromosome may independently assort from each other due to crossing over •A two-factor cross studies linkage between two genes •A three-factor cross studies linkage between three genes In diploid eukaryotic species, linkage can be altered during meiosis as a result of crossing over Crossing over •Occurs during prophase I of meiosis •replicated sister chromatid homologues associate as bivalents •Non-sister chromatids of homologous chromosomes exchange DNA segments Crossing Over May Produce Recombinant Genotypes Two haploid cells may contain a combination of alleles NOT found in the original chromosomes These are nonparental or recombinant cells Recombinant offspring are produced by the exchange of DNA between two homologous chromosomes during meiosis in one or both parents, leading to a novel combination of genetic material. •IMPORTANT POINT: In this chapter, the definition of recombinant offspring is that they are offspring that have inherited a chromosome this is the product of a crossover.

Apply the product rule and binomial expansion equation in genetic problems (binomial expansion)

Represents all of the possibilities for a given set of unordered events P = (n! / x! (n-x)!) * p^x * q^1-x Where P = probability that the unordered outcome will occur n = total number of events x = number of events in one category p = individual probability of x q = individual probability of the other category Note : p + q=1 •The symbol ! denotes a factorial •n! is the product of all integers from n down to 1 Step 1: Calculate the individual probabilities •This can be obtained via a Punnett square Step 2: Determine the number of events •n = total number of children = 5 •x = number of blue-eyed children = 2 Step 3: Substitute the values for p, q, x, and n in the binomial expansion equation

Explain the underlying causes of dominant patterns of inheritance; they are usually due to haploinsufficiency, gain-of-function, or a dominant negative mutation.

ch 25 Three common molecular explanations for dominant disorders •Haploinsufficiency •The heterozygote has 50% of the normal protein •This is not sufficient for a normal phenotype •Gain-of-function mutations •Mutation changes protein so it gains a new function •Dominant negative mutations •The altered gene product acts antagonistically to the normal product

Outline the general organization of eukaryotic chromosome sequences and their functions

10.7 Eukaryotic species contain one or more sets of chromosomes •Each set is composed of several different linear chromosomes •Humans have 2 sets of 23 chromosomes Each chromosome contains a single, linear molecule of DNA •Typically tens to hundreds of millions of base pairs •Typically a few hundred to several thousand genes Three types of DNA sequences are required for chromosomal replication and segregation •Origins of replication •Chromosomal sites necessary to initiate DNA replication •Centromeres •Regions that play a role in segregation of chromosomes •Telomeres •Specialized regions at the ends of chromosomes •Important in replication and for stability

Explain how epigenetic changes can occur via environmental factors

16.8, 16.9 Many environmental agents have been shown to cause epigenetic changes. These include dietary effects as well as toxins in the environment. Examples include •Dietary effects on the Agouti gene in mice Toxins that contribute to cancer. When pregnant mice were fed a diet that contained chemicals that tend to increase DNA methylation, the offspring tended to be have darker fur. This result is consistent with the idea that DNA methylation inhibits the Agouti gene Female honeybees can be: •Queen bees: larger, live for years, and produce up to 2000 eggs each day •Worker bees: small, sterile, typically live only for weeks •Nurse bees: produce a secretion called royal jelly •Bees that eat royal jelly into adulthood become queens •Larvae injected with a DNA methyltransferase inhibitor became queen bees

Codominance

4.9 inheritance: this pattern occurs when the heterozygote expresses both alleles simultaneously without forming an intermediate phenotype. (AB blood for example) molecular explanation: the codominant alleles encode proteins that function slighty differently from each other, and the function of each protein in the heterozygote affects the phenotype uniquely

List the seven observations that suggest a disease is caused (at least in part) by a genetic component.

ch 25 1. When an individual exhibits a disease, the disorder is more likely to occur in blood relatives than in the general population 2. Identical twins share the disease more often than fraternal twins •Identical twins are also called monozygotic (MZ) twins •They are formed from the same sperm and egg 3. The disease does not spread to individuals sharing similar environmental situations 4. Different populations tend to have different frequencies of the disease 5. The disease tends to develop at a characteristic age §Many genetic disorders exhibit a specific age of onset 6. The human disorder resembles a genetic disorder that has a genetic basis in an animal (Figure 25.1) 7. A correlation is observed between a disease and a mutant human gene or a chromosomal alteration

Define epigenetic inheritance, dosage compensation, and genomic imprinting.

ch 5 Epigenetic inheritance refers to a pattern in which a modification occurs to a nuclear gene or chromosome that alters gene expression •However, the expression is not permanently changed over the course of many generations •That is because the DNA sequence does not change Epigenetic changes are caused by DNA and chromosomal modifications •These can occur during oogenesis, spermatogenesis or early embryonic development Dosage compensation: The purpose of dosage compensation is to offset differences in the number of active sex chromosomes - in mammals, it occurs by the inactivation of a single x chromosome in females genomic imprinting: Genomic imprinting is a phenomenon in which a segment of DNA is marked and the effect is maintained throughout the life of the organism inheriting the marked DNA Imprinted genes follow a non-Mendelian pattern of inheritance Depending on how the genes are "marked", the offspring expresses either the maternally-inherited or the paternally-inherited allele •Not both •This is termed monoallelic expression At the cellular level, imprinting is an epigenetic process that can be divided into three stages: 1.Establishment of the imprint during gametogenesis 2.Maintenance of the imprint during embryogenesis and in the adult somatic cells 3.Erasure and reestablishment of the imprint in the germ cells It may involve •A single gene •A part of a chromosome •An entire chromosome •Even all the chromosomes from one parent •It can be used for X inactivation in some species Genomic imprinting must involve a marking process At the molecular level, the imprinting of several genes is known to involve an imprinting control region (ICR) located near the imprinted gene •The ICR is methylated either in the oocyte or sperm •Not both •The ICR contains binding sites for one or more transcription factors that regulate the imprinted gene •For most genes, methylation causes inhibition of transcription

Be able to predict the outcome of crosses for genes that exhibit a maternal effect

figure 5.1 Maternal effect refers to an inheritance pattern for certain nuclear genes in which the genotype of the mother directly determines the phenotype of her offspring •Surprisingly, the genotypes of the father and offspring themselves do not affect the phenotype of the offspring This phenomenon is due to the accumulation of gene products that the mother provides to her developing eggs This Non-Mendelian Inheritance Pattern can be Explained by the Process of Oogenesis Maturing animal oocytes are surrounded by maternal cells that provide them with nutrients •These nurse cells are diploid, whereas the oocyte becomes haploid The nurse cells express mRNA and/or protein from genes of the mother Maternal effect genes encode RNA and proteins that play important roles in the early steps of embryogenesis •For example-Cell division, Cleavage pattern, Body Axis orientation Accumulation of maternal effect gene products before fertilization allows these steps to proceed very quickly after fertilization

Be able to analyze a pedigree and determine if an inheritance pattern for a disease is or is not consistent with autosomal recessive, autosomal dominant, X-linked recessive, or X-linked dominant inheritance

see Figures 25.2, 25.3, 25.4). ***males are squares, females are circles Four common features of autosomal recessive inheritance are: 1.Frequently, an affected offspring will have two unaffected parents 2.When two unaffected heterozygotes have children, the percentage of affected children is (on average) 25% 3.Two affected individuals will have 100% affected children 4.The trait occurs with the same frequency in both sexes Five common features of autosomal dominant inheritance are as follows: 1.An affected offspring usually has one or both affected parents •Can be altered by reduced penetrance 2.An affected individual with only one affected parent is expected to produce (on average) 50% affected offspring 3.Two affected, heterozygous individuals will have (on average) 25% unaffected offspring 4.The trait occurs with the same frequency in both sexes 5.For most dominant disease-casing alleles, the homozygote is more severely affected with the disorder Three common features of X-linked recessive inheritance are as follows: 1.Males are much more likely to exhibit the trait 2.The mothers of affected males often have brothers or fathers who are affected with the same trait 3.The daughters of affected males will produce (on average) 50% affected sons X-linked dominant inheritance is rare. Characteristics of such disorders are: •Males are often more severely affected •Females may be less affected due to wild-type copy on second X •Females are more likely to exhibit the trait when it is lethal to males •Affected mothers have a 50% chance of passing the trait to daughters

Explain the five factors that govern microevolution

table 27.1 Microevolution describes changes in a population's gene pool from generation to generation •Driven by: •Mutation •Random genetic drift •Migration •Natural Selection •Nonrandom mating genetic drift: Genetic drift refers to random changes in allele frequencies due to random fluctuations In other words, allele frequencies may drift from generation to generation as a matter of chance Over the long run, genetic drift favors either the loss or the fixation of an allele •The fixed allele is monomorphic and cannot fluctuate •The rate depends on the population size In nature, a population can be reduced dramatically in size by a natural disaster for example (Bottle neck effect) A small group of individuals separates from a larger population and establishes a colony in a new location (founder effect) •This has two important consequences 1.The founding population is expected to have less genetic variation than the original population 2.The founding population will have allelic frequencies that may differ markedly from those of the original population, as a matter of chance Migration between two different established populations can alter allele frequencies Gene flow is the transfer of alleles from donor population to recipient population, changing its gene pool. In nature, it is common for individuals to migrate between populations in both directions This bidirectional migration has two important consequences 1.It tends to reduce allele frequency differences between populations 2.It can enhance genetic diversity within a population New mutations in one population can be introduced to neighboring group Mating and phenotypes •Assortative mating occurs when individuals do not mate randomly •Positive assortative mating occurs when individuals are more likely to mate due to similar phenotypic characteristics •Negative assortative mating occurs when individuals with dissimilar phenotypes mate preferentially

Describe how eukaryotic chromosomes come in sets, and that most species are diploid, which means they have homologous pairs of chromosomes

(see Figure 3.3). Most eukaryotic species are diploid •Have two sets of chromosomes Members of a pair of chromosomes are called homologs •The two homologs form a homologous pair The two chromosomes in a homologous pair •Are nearly identical in size •Have the same banding pattern and centromere location •Have the same genes •But not necessarily the same alleles The DNA sequences on homologous chromosomes are also very similar •There is usually less than 1% difference between homologs Nevertheless, these slight differences in DNA sequence provide the allelic differences in genes The sex chromosomes (X and Y) are not homologous •They differ in size and genetic composition •However, they do have short regions of homology •The physical location of a gene on a chromosome is called its locus.

Compare and contrast the four common types of natural selection

(see Figures 27.6, 27.7, 27.8, 27.9, 27.10, 27.11, 27.12, 27.13). Natural selection acts on phenotypes (which are derived from an individual's genotype) With regard to many traits, there are four ways that natural selection may operate 1.Directional selection 2.Balancing selection 3.Disruptive (or diversifying) selection 4.Stabilizing selection Directional Selection affects the Hardy-Weinberg equilibrium and allele frequencies by favoring the extreme phenotype A polymorphism may reach an equilibrium where opposing selective forces balance each other •The population is not evolving toward allele fixation or elimination •Such a situation is known as balancing selection It can occur because of different reasons 1. The heterozygote has a higher fitness than either homozygote, called heterozygote advantage 2. A species occupies a region that contains heterogeneous environments Negative frequency-dependent selection is another mechanism of balancing selection •Common individuals lose fitness because of being common •Rare individuals gain fitness because of being rare •Selection always favors less numerous genotype •Always drives towards balance Disruptive selection favors the survival of two or more different genotypes with different phenotypes •Also known as diversifying selection •Caused by fitness values for a given genotype that vary in different environments •Typically acts on traits that are determined by multiple genes In stabilizing selection, the extreme phenotypes are selected against •Intermediate phenotypes have the highest fitness values •Tends to decrease genetic diversity •Eliminates alleles that cause variation in phenotypes •Laying eggs is an example •Too many eggs drains resources to care for young •Too few eggs does not contribute many individuals to the next generation

Be able to predict the outcome of crosses for imprinted genes and know how genomic imprinting is controlled by methylation

(see Figures 5.8, 5.9, 5.10, 5.11). The Igf2 gene encodes a growth hormone called insulin-like growth factor 2 •A functional Igf2 gene is necessary for a normal size Imprinting results in the expression of the paternal but not the maternal allele •The paternal allele is transcribed into RNA •The maternal allele is not transcribed Igf2- is a loss-of-function allele that does not express a functional Igf2 protein •This may cause a mouse to be dwarf depending on whether it inherits the mutant allele from its father or from its mother At the cellular level, imprinting is an epigenetic process that can be divided into three stages: 1.Establishment of the imprint during gametogenesis 2.Maintenance of the imprint during embryogenesis and in the adult somatic cells 3.Erasure and reestablishment of the imprint in the germ cells Genomic imprinting occurs in several species including insects, mammals and flowering plants It may involve •A single gene •A part of a chromosome •An entire chromosome •Even all the chromosomes from one parent •It can be used for X inactivation in some species Genomic imprinting must involve a marking process At the molecular level, the imprinting of several genes is known to involve an imprinting control region (ICR) located near the imprinted gene •The ICR is methylated either in the oocyte or sperm •Not both •The ICR contains binding sites for one or more transcription factors that regulate the imprinted gene •For most genes, methylation causes inhibition of transcription Both parents inherit one methylated and one unmethylated gene, which is maintained in somatic cells. Methylation is removed in gamete forming cells de novo methylation in sperm, maintenance methylation in somatic cells

Don't memorize the genetic code, but explain its purpose and that there are exceptions

(see Tables 13.1, 13.2). •In genetics, the nucleotide language of mRNA is translated into the amino acid language of proteins Translation relies on the genetic code The genetic information is coded within mRNA in groups of three nucleotides known as codons Special codons: •AUG (which specifies methionine) = start codon •This defines the reading frame for all following codons •AUG specifies additional methionines within the coding sequence •UAA, UAG and UGA = termination, or stop, codons The code is degenerate •More than one codon can specify the same amino acid •For example: GGU, GGC, GGA and GGG all code for glycine •In most instances, the third base is the variable base The code is nearly universal •Only a few rare exceptions have been noted Ex: Selenocysteine and pyrrolysine are sometimes called the 21st and 22nd amino acids •Found in specialty enzymes •Encoded by UGA and UAG codons, respectively •Attached by tRNAs that carry them to the ribosome •Codon and downstream sequences in mRNA are needed to incorporate these amino acids

Describe the levels of compaction leading to a metaphase

10.17 The compaction of linear DNA in eukaryotic chromosomes involves interactions between DNA and several different proteins The DNA-protein complex is called chromatin •Proteins bound to DNA are subject to change during the life of the cell •These changes affect the degree of chromatin compaction The repeating structural unit within eukaryotic chromatin is the nucleosome It is composed of a double-stranded segment of DNA wrapped around an octamer of histone proteins •A histone octamer is composed of two copies each of four different histone proteins •146 bp of DNA make 1.65 negative superhelical turns around the octamer Nucleosomes associate with each other to form a more compact structure termed the 30 nm fiber Histone H1 plays a role in this compaction The 30 nm fiber shortens the total length of DNA another seven-fold Its structure has proven difficult to determine •The DNA conformation may be substantially altered when extracted from living cells •Two models have been proposed •Solenoid model (stacked) •Three-dimensional zigzag model The two events we have discussed so far have shortened the DNA about 50-fold A third level of compaction involves interaction between the 30 nm fiber and the nuclear matrix The nuclear matrix is composed of two parts •Nuclear lamina •Fibers that line the inner nuclear membrane •Internal matrix proteins •Connected to nuclear lamina and fills interior of nucleus •Structural and functional role remains controversial The attachment of radial loops to the nuclear matrix is important in two ways •It plays a role in compaction •It serves to organize the chromosomes within the nucleus •Each chromosome in the nucleus is located in a discrete and nonoverlapping chromosome territory The third mechanism of DNA compaction involves the formation of radial loop domains •25,000 - 200,000 bp Matrix-attachment regions (MARs), or scaffold-attachment regions (SARs), anchored to nuclear matrix In metaphase chromosomes the radial loops are highly compacted and stay anchored to a scaffold •The scaffold is formed from the nuclear matrix Histones are needed for the compaction of radial loops Two multiprotein complexes help to form and organize metaphase chromosomes •Condensin •Plays a critical role in chromosome condensation •Cohesin •Plays a critical role in sister chromatid alignment Both contain a category of proteins called SMC proteins •Acronym = Structural Maintenance of Chromosomes •SMC proteins use energy from ATP to catalyze changes in chromosome structure

Explain the two mechanisms for the compaction of bacterial chromosomes

10.3 and 10.4 loop domains: To fit within the bacterial cell, the chromosomal DNA must be compacted about a 1000-fold •This involves the formation of loop domains •Microdomains emanate from the core and typically contain about 10,000 bp of DNA. •In some species, like E. coli, about 80 to 100 microdomains are organized into macrodomains, but this is not evident in electron micrographs. DNA supercoiling: Because the two strands within DNA already coil around each other, the formation of additional coils due to twisting forces is referred to as DNA supercoiling. Both underwinding and overwinding of the DNA double helix can cause supercoiling •Example from Figure 10.4: DNA given a turn that unwinds the helix •can cause fewer turns OR can cause a negative supercoil to form •DNA given a turn that overwinds the helix •either more turns OR the formation of a positive supercoil •These forms are topoisomers

Compare and contrast the synthesis of the leading and lagging strands

11.10 and 11.11 The two new daughter strands are synthesized in different ways and opposite orientations Leading strand •One RNA primer is made at the origin •DNA pol III attaches nucleotides in a 5' to 3' direction as it slides toward the opening of the replication fork Lagging strand •Synthesis is also in the 5' to 3' direction •However it occurs away from the replication fork •Many RNA primers are required •DNA pol III uses the RNA primers to synthesize small DNA fragments (1000 to 2000 nucleotides in bacteria, 100-200 in eukaryotes) •These are termed Okazaki fragments after their discoverers •DNA pol I removes the RNA primers and fills the resulting gap with DNA •It uses a 5' to 3' exonuclease activity to digest the RNA and 5' to 3' polymerase activity to replace it with DNA •After the gap is filled a covalent bond is still missing •DNA ligase catalyzes the formation of a phosphodiester bond •Thereby connecting the DNA fragments

With regard to splicing, explain how pre-mRNA splicing occurs in eukaryotes

12.19 and 12.20 •Three different splicing mechanisms have been identified •Group I intron splicing •Group II intron splicing •Spliceosome All three cases involve •Removal of the intron RNA •Linkage of the exon RNA by an ester bond •We will not cover Group I and Group II splicing in this course, but we will cover the spliceosome. In eukaryotes, the transcription of structural genes produces a long transcript known as a pre-mRNA •This RNA is altered by splicing and other modifications, before it leaves the nucleus Splicing requires the aid of a multicomponent structure known as the spliceosome •It is composed of several subunits known as snRNPs (pronounced "snurps") •Each snRNP contains small nuclear RNA and a set of proteins The subunits of a spliceosome carry out several functions 1.Bind to an intron sequence and precisely recognize the intron-exon boundaries 2.Hold the pre-mRNA in the correct configuration 3.Catalyze the chemical reactions that remove introns and covalently link exons

Describe the general organization of a protein-encoding gene

12.2 DNA: Regulatory sequences: site for the binding of regulatory proteins; the role of regulatory proteins is to influence the rate of transcription. Regulatory sequences can be found in a variety of locations. Promoter: site for RNA polymerase binding; signals the beginning of transcription. Terminator: signals the end of transcription. mRNA: Ribosome-binding site: site for ribosome binding; translation begins near this site in the mRNA. In eukaryotes, the ribosome scans the mRNA for a start codon. Start codon: specifies the first amino acid in a polypeptide sequence, usually a formylmethionine (in bacteria) or a methionine (in eukaryotes). Codons: 3 nucleotide sequences within the mRNA that specify particular amino acids. The sequence of codons within mRNA determines the sequence of amino acids within a polypeptide. Stop codon: specifies the end of polypeptide synthesis. Bacterial mRNA may be polycistronic, which means it encodes two or more polypeptides.

Describe the three stages of transcription

12.3 Transcription occurs in three stages •Initiation •Elongation •Termination Initiation: The promoter functions as a recognition site for transcription factors (not shown). The transcription factors enable RNA polymerase to bind to the promoter. Following binding, the DNA is denatured into a bubble known as the open complex. The binding of the RNA polymerase to the promoter forms the closed complex Then, the open complex is formed when the TATAAT box in the -10 region is unwound •A-T bonds are more easily separated A short RNA strand is made within the open complex •The sigma factor is released at this point •This marks the end of initiation The core enzyme now slides down the DNA to synthesize an RNA strand Elongation/synthesis of the RNA transcript: RNA polymerase slides along the DNA in an open complex to synthesize RNA. Key Points: RNA polymerase slides along the DNA, creating an open complex as it moves. The DNA strand known as the template strand is used to make a complementary copy of RNA, resulting in an RNA-DNA hybrid. RNA polymerase moves along the template strand in a 3' to 5' direction, and RNA is synthesized in a 5' to 3' direction using nucleoside triphosphates as precursors. Pyrophosphate is released (not shown). The complementary rule is the same as the AT/GC rule except that U is substituted for T in the RNA. Termination: A terminator is reached that causes RNA polymerase and the RNA transcript to dissociate from the DNA. Termination is the end of RNA synthesis •It occurs when the short RNA-DNA hybrid of the open complex is forced to separate •This releases the newly made RNA as well as the RNA polymerase E. coli has two different mechanisms for termination (both involve stem loop formation of RNA and pausing of RNA pol) 1.rho-dependent termination •Requires a protein known as r (rho) 2.rho-independent termination •Does not require r - ρ-independent termination is facilitated by two sequences in the RNA 1.A uracil-rich sequence located at the 3' end of the RNA 2.A stem-loop structure upstream of the uracil-rich sequence

Describe the three stages of translation

13.14 Translation can be viewed as occurring in three stages •Initiation •Elongation •Termination Initiation: The binding of mRNA to the 30S subunit is facilitated by a ribosomal-binding site or Shine-Dalgarno sequence •This is complementary to a sequence in the 16S rRNA •Interact via hydrogen bonds The mRNA, initiator tRNA, and ribosomal subunits associate to form an initiation complex •This process requires three Initiation Factors The initiator tRNA recognizes the start codon in mRNA •In bacteria, this tRNA is designated tRNAfmet •It carries a methionine that has been covalently modified to N-formylmethionine •The start codon is AUG, but in some cases GUG or UUG •In all three cases, the first amino acid is N-formylmethionine In eukaryotes, the assembly of the initiation complex is similar to that in bacteria •However, additional factors are required •Note that eukaryotic Initiation Factors are denoted eIF •Refer to Table 13.6 The initiator tRNA is designated tRNAmet •It carries a methionine rather than a formylmethionine An initiation factor protein complex (eIF4) binds to the 5' cap in mRNA These are joined by a complex consisting of the 40S subunit, tRNAmet, and other initiation factors The entire assembly moves along the mRNA scanning for the right start codon Once it finds this AUG, the 40S subunit binds to it The 60S subunit joins This forms the 80S initiation complex elongation: During this stage, amino acids are added to the polypeptide chain, one at a time The addition of each amino acid occurs via a series of steps outlined in Figure 13.17 This process, though complex, can occur at a remarkable rate •In bacteria à 15-20 amino acids per second •In eukaryotes à 2-6 amino acids per second After peptide bond formation, tRNAs at the P and A sites move into the E and P sites The 23S rRNA (a component of the large subunit) is the actual peptidyl transferase •Thus, the ribosome is a ribozyme! termination: The final stage occurs when a stop codon is reached in the mRNA •In most species there are three stop or nonsense codons •UAG •UAA •UGA •These codons are not recognized by tRNAs, but by proteins called release factors •Indeed, the 3-D structure of release factors mimics that of tRNAs Bacteria have three release factors •RF1 recognizes UAA and UAG •RF2 recognizes UAA and UGA •RF3 does not recognize any of the three codons - required for the termination process Eukaryotes only have two release factors •eRF1 recognizes all three stop codons •eRF3 required for termination process

Explain how the lac repressor protein works

14.4 14.5 In the absence of the inducer allolactose, the repressor protein is tightly bound to the operator site, thereby inhibiting the ability of RNA polymerase to transcribe the operon. When allolactose is available, it binds to the repressor. This alters the conformation of the repressor protein, which prevents it from binding to the operator site. Therefore, RNA polymerase can transcribe the operon. 1.When lactose becomes available, a small amount of it is taken up and converted to allolactose by β-galactosidase. The allolactose binds to the repressor, causing it to fall off the operator site. 2.lac operon proteins are synthesized. This promotes the efficient uptake and metabolism of lactose. 3.The lactose is depleted. Allolactose levels decrease. Allolactose is released from the repressor, allowing it to bind to the operator site. 4.Most proteins involved with lactose utilization are degraded. The lac repressor must bind to two of the three operators to cause repression Can bind to O1 and O2 or O1 and O3 Binding of the lac repressor to two operator sites requires that the DNA form a loop •A loop in the DNA brings the operator sites closer together

Describe how histones are displaced when a gene is transcriptionally active

15.13 Many genes are flanked by nucleosome-free regions (NFR) and well-positioned nucleosomes Binding of activators: Activation protein binds to enhancer sequences. The enhancer may be close to the transcriptional start site (as shown here) or far away. Chromatin remodeling and histone modification: An activator protein recruits a chromatin remodeling complex (such as SWI/SNF) and a histone-modification enzyme (such as histone acetyltransferase). Nucleosome may be moved, and histone may be evicted or replaced with variants. Some histones are subjected to covalent modification, such as acetylation. Formation of the preinitiation complex: General transcription factors and RNA polymerase II are able to bind to the core promoter and form a preinitiation complex. Elongation: During elongation, histones ahead of the open complex are covalently modified by acetylation and evicted. Behind the open complex, histones are deacetylated and become tightly bound to the DNA.

Explain how activators and repressors bind to enhancers and silencers to modulate gene expression

15.2 •A regulatory protein that increases the rate of transcription is termed an activator •The sequence it binds is called an enhancer •A regulatory protein that decreases the rate of transcription is termed a repressor •The sequence it binds is called a silencer The activator/coactivator complex recruits TFIID to the core promoter and/or activates its function. Transcription will be enhanced (a) Transcriptional activation via TFIID The repressor protein inhabits the binding of TFIID to the core promoter or inhabits its function. Transcription is silenced. (b) Transcriptional repression via TFIID The activator protein interacts with mediator. This results in the phosphorylation of the carboxyl-terminal domain of RNA polymerase. Some general transcription factors are released, and RNA polymerase proceeds to the elongation phase of transcription. (a) Transcriptional activation via mediator The repressor protein interacts with mediator in a way that prevents the phosphorylation of RNA polymerase. Therefore, it cannot proceed to the elongation phase of transcription. (b) Transcriptional repression via mediator

Describe the mechanism of RNA interference (RNAi)

17.4 Fire and Mello used the term RNA interference (RNAi) to describe the phenomenon in which double-stranded RNA causes the silencing of mRNA microRNAs •Transcribed from endogenous eukaryotic genes •Regulate gene expression •Commonly, a single type of miRNA inhibits the translation of several different mRNAs through partial complementarity •As many as 60% of human genes may be regulated by microRNAs Small interfering RNAs •ncRNAs that usually originate from exogenous sources (not normally made by cells) •such as viruses First pri-miRNA is made •Forms a hairpin recognized by Drosha and DGCR8 Cleaved to 70-nucleotide pre-miRNA and exported Cut by dicer to 20-25 bp (both miRNA and snRNA) RISC is composed of •A double-stranded RNA molecule that is 20-25 bp long produced from pre-miRNAs and pre-siRNAs by dicer •One strand will be degraded •Proteins Outcomes of RISC complex binding to mRNA: •Inhibit translation without degrading the mRNA •Common for miRNAs, often partially complementary to their mRNAs •RISC-mRNA complex may remain in a cellular structure called a processing body (P-body) •Degradation of the mRNA through cleavage by Argonaute •Common with siRNAs, typically a perfect match to their target mRNAs

Be able to describe the Holliday model for homologous recombination

20.2 1.Homologous recombination •Occurs between DNA segments that are homologous •Segments break and rejoin to form new combinations The Holliday model can account for the general properties of recombinant chromosomes in meiosis 1. Both chromosomes are nicked at identical locations 2. The DNA strandsto the left of the nicks invade the homologous chromatids and covalently link to the strands to the right of the nicks (forms a Holliday junction) 3. The Holliday junction migrates from left to right (called branch migration). It creates two heteroduplex regions. 4a. When the strands that were originally nicked are broken, the strands reconnect to form nonrecombinant chromatids with short heteroduplex regions 4b. When the strands that were not originally nicked are broken, the strands are connected to make recombinant chromosomes with short heteroduplex regions. More detailed studies of genetic recombination have led to a refinement of the Holliday model •In particular, more recent models have modified the initiation phase of recombination •Two nicks in the same location on two strands is unlikely •Rather, it is more likely for one DNA helix to incur a single nick or a break in both strands •Either of these is enough to initiate recombination

Be able to describe the two mechanisms that account for gene conversion

20.4, 20.5 Genetic recombination can cause two different alleles to become identical alleles •This process, whereby one of the alleles is converted to the other has been termed gene conversion Gene conversion can occur in one of two ways 1.DNA mismatch repair 2.DNA gap repair synthesis mismatch repair: branch migration travels over a region with a minor sequence difference. This creates a heteroduplex (the bases don't pair correctly). The mismatch repair yields 4 possible combinations (2 with gene conversion, 2 with no gene conversion) Gene conversion by gap repair synthesis according to the double-strand break model: double strand break occurs. The region adjacent to the double strand break is digested away, eliminating one allele. Strand invasion causes D-loop formation. Gap repair synthesis uses the strands from the not broken DNA to fill the vacant region in the broken DNA. The intertwined strands are resolved. Both chromosomes now carry the same allele

Explain the two general ways that transposable elements can move

20.8 Two general types of transposition pathways have been identified •Simple or conservative transposition (cut and paste) •Retrotransposition (copy and paste) Simple: The enzyme transposase catalyzes the removal of a TE and the its reinsertion at another location Transposase recognizes the inverted repeats at the ends of a TE and brings them close together (two different subunits). Transposase cleaves outside the inverted repeats which excises the TE from the chromosome. Transposase cleaves the DNA at staggered sites and insert the TE. DNA gap repair synthesis results in the creation of direct repeats outside the TE Retrotransposons: Retrotransposons use an RNA intermediate in their transposition mechanism The movement requires two key enzymes: •Reverse transcriptase •Integrase Transcription of TE to RNA, reverse transcription of mRNA to DNA (using the DNA as a primer since the TE RNA can base pair to it), integrase to cut the chromosomal DNA to insert the TE DNA into the chromosome

Describe how to clone genes into vectors

21.1, 21.2 The vectors commonly used in gene cloning were originally derived from two natural sources 1.Plasmids •Many naturally occurring plasmids have selectable markers •Typically, genes conferring antibiotic resistance to the host cell 2.Viruses Recognized sequences are typically palindromic •The sequence is identical when read in the opposite direction in the complementary strand DNA of interest and plasmid are cut with restriction enzymes to make recombinant DNA. Ends are 'sticky' in that they are short, single-stranded regions of DNA that can base-pair with another piece of DNA with complementary sequence •The vector carries two important genes •ampR: Confers antibiotic resistance to the host cell •Identifies cells that have taken up the vector •lacZ: Encodes β-galactosidase •Provides a means by which bacteria that have picked up the cloned gene can be identified Cells that are able to take up DNA are called competent cells Transformation refers to uptake of plasmid vectors by a bacterial cell Transfection refers to introduction of viral vectors into a host cell All bacterial colonies growing on the plate must have picked up the vector and its ampR gene •Now need to differentiate between the colonies that have a recircularized vector from those with a hybrid vector •This is where the lacZ gene comes into play In the hybrid vector, the chromosomal DNA inserts into the lacZ gene, thereby disrupting it The growth media contains two relevant compounds: •IPTG (isopropyl-b-D-thiogalactopyranoside) •A lactose analogue that can induce lacZ gene expression •X-Gal (5-bromo-4-chloro-3-indolyl-b-D-galactopyranoside) •A colorless compound that is cleaved by β-galactosidase into a blue dye •The color of bacterial colonies will therefore depend on whether or not the β-galactosidase enzyme is functional •If it is, the colonies will be blue •If not, the colonies will be white In this experiment •Bacterial colonies with recircularized vectors form blue colonies •While those with hybrid vectors form white colonies The net result of gene cloning is to produce an enormous amount of copies of a gene •During transformation, a single bacterial cell usually takes up a single copy of a vector •Amplification of a cloned gene occurs in two ways: •The vector gets replicated by the host cell many times •This will generate a lot of copies per cell (25-50 for plasmids)

Describe the technique of dideoxy DNA sequencing

21.10, 21.11 Based on DNA replication but with a clever twist DNA polymerase connects adjacent deoxynucleotides by covalently linking the 5'-P of one to the 3'-OH another Nucleotides without a 3'-OH can be synthesized Called dideoxyribonucleotides or ddNTPs 2', 3' - Dideoxyadenosine triphosphate (ddA) Sanger reasoned that if a dideoxynucleotide is added to a growing DNA strand, the strand can no longer grow •Called chain termination Prior to DNA sequencing, the DNA to be sequenced must be obtained in large amounts •This is accomplished using cloning or PCR techniques In many sequencing experiments, the target DNA is cloned into the vector at a site adjacent to a primer annealing site The target DNA (recombinant vector DNA in Figure 21.11) is heat denatured into single strands A single tube contains all four dideoxyribonucleotides •Each type (ddA, ddT, ddG, and ddC) has a different-colored fluorescent label attached •After polymerization, the sample is loaded into a single lane of a gel •The color of fluoresence indicates the last base in each strand •Read the sequence of the sequencing ladder The procedure is automated using a laser and fluorescence detector The fragments are separated by gel electrophoresis •The DNA fragments are electrophoresed off the end of the gel As each band comes off the bottom of the gel, the fluorescent dye is excited by the laser •The fluorescence emission is recorded by the fluorescence detector •The detector reads the level of fluorescence at four wavelengths

Define cDNA and explain how it is made

21.3 The enzyme reverse transcriptase is used to clone DNA from RNA •Uses RNA as a template to make a complementary strand of DNA •Used by retroviruses to copy their RNA genome to DNA •DNA made from RNA is complementary DNA (cDNA) Add polyT primer to bind to polA tail of mRNA Add reverse transcriptase and dNTPs to create a complementary DNA strand Add RNaseH to cut up RNA and generate RNA primers Add DNA polymerase and DNA ligase to create second DNA strand -> double stranded cDNA From a research perspective, an important advantage of cDNA is that it lacks introns This has two ramifications •It allows researchers to focus their attention on the coding sequence of a gene •It allows the expression of the encoded protein •Especially, in cells that would not splice out the introns properly (e.g., a bacterial cell)

Be able to describe the technique of PCR

21.5, 21.6 A way to copy a specific sequence of DNA is called polymerase chain reaction (PCR) •Developed by Kary Mullis in 1985 PCR can copy DNA without the aid of vectors and host cells •Must know enough about the gene of interest to have the sequence of 2 short primers 1. Denaturation: separate DNA strands with high temperature 2. Primer annealing: lower temperature which allows primers to bind to template DNA 3. Primer extension: incubate at a slightly higher temperature, which allows DNA synthesis to occur starting material: 1.Template DNA •Contains the region that needs to be amplified 2.Oligonucleotide primers •Complementary to sequences at the ends of the DNA fragment to be amplified •These are synthetic and about 15-20 nucleotides long 3.Deoxynucleoside triphosphates (dNTPs) Provide the precursors for DNA synthesis 4.Taq polymerase •DNA polymerase isolated from the bacterium Thermus aquaticus •This thermostable enzyme is necessary because PCR involves heating steps that inactivate most other DNA polymerases PCR is carried out in a thermocycler All the reagents are placed in one tube The experimenter sets the temperature range and number of cycles •The sequential process of denaturing-annealing-synthesis is then repeated for many cycles A typical PCR run is likely to involve 20 to 30 cycles of replication •This takes a few hours to complete After 20 cycles, a target DNA sequence will increase 220-fold (~ 1 million-fold) After 30 cycles, a target DNA sequence will increase 230-fold (~ 1 billion-fold) •This assumes 100% efficiency

Describe what a DNA microarray is, and explain how it is used to study gene expression patterns

24.1 •Researchers have developed an exciting technology called DNA microarrays (also called gene chips) -This new technology makes it possible to monitor thousands of genes simultaneously •A DNA microarray is a small silica, glass or plastic slide that is dotted with many sequences of DNA -Each of these sequences corresponds to a known gene •These fragments are made synthetically •These sequences of DNA will act as probes to identify genes that are transcribed •The DNA fragments can be either -Amplified by PCR and then spotted on the microarray -Synthesized directly on the microarray itself -A single slide contains tens of thousands of different spots in an area the size of a postage stamp •The relative location of each spot is known •Once a DNA microarray has been made, it is used as a hybridization tool Convert mRNA to fluorescently labelled single stranded cDNA using reverse transcriptasde, poly T primers, and fluorescently labelled dNTPs. Hybridize cDNA to microarray, visualize which mRNA is in the sample by looking at which spots are fluorescent

Describe how most cancers involve a progression of multiple mutations

25.16 Most forms of cancer involve multiple genetic changes leading to malignancy Many cancers begin with a benign mutation that, with time and more mutations leads to malignancy •Furthermore, a malignancy can continue to accumulate genetic changes that make it even more difficult to treat Some genes affect growth directly, others may enable metastasis which allows expansion to new locations, giving the cells a growth advantage Estimated that 300 different genes may play a role in development of human cancer •over 1% of our genes Chromosomal abnormalities are often associated with cancer •Missing chromosomes may have carried a tumor suppressor gene •Duplicated chromosomes may overexpress proto-oncogenes •Translocations fuse or disrupt genes

x-linked inheritance

4.11 inheritance: this pattern involves the inheritance of genes that are located on the x-chromosome. (males only have 1 X, females have 2 - in mammals and fruit flies anyway) molecular explanation: If a pair of x-linked alleles shows a simple dominant/recessive relationship, 50% of the protein, produced by a single copy of the dominant allele in a heterozygous female, is sufficient to produce the dominant trait (in the female). Males only have 1 copy of x-linked genes, and therefore express the copr they carry

sex- influenced inheritance

4.13 inheritance: this pattern refers to the effect of sex on the phenotype of the individual. Some alleles are recessive in one sex and dominant in the other molecular explanation: sex hormones may regulate the expression of some genes. This can influence the phenotypic effect of some alleles genes are autosomal*

Compare and contrast variation in euploidy versus aneuploidy

8.15 Chromosome numbers can vary in two main ways •Euploidy •Variation in the number of complete sets of chromosome •Aneuploidy •Variation in the number of particular chromosomes within a set •Euploid variations occur occasionally in animals and frequently in plants •Aneuploid variations, on the other hand, are regarded as abnormal conditions Aneuploidy commonly causes an abnormal phenotype •It leads to an imbalance in the amount of gene products •Three copies of a gene will lead to 150% production •A single chromosome can have hundreds or even thousands of genes Imbalance of gene products in Trisomic and Monosomic individuals •In most cases, these effects are detrimental •They produce individuals that are less likely to survive than a euploid individual Most species of animals are diploid In many cases, changes in euploidy are not tolerated •Polyploidy in animals is generally a lethal condition In many animals, certain body tissues display normal variations in the number of sets of chromosomes Diploid animals sometimes produce tissues that are polyploid •This phenomenon is termed endopolyploidy •Liver cells, for example, can be triploid, tetraploid or even octaploid (8n) •May enhance ability of cell to produce specific gene products In contrast to animals, plants commonly exhibit polyploidy Polyploids with an odd number of chromosome sets are usually sterile •These plants produce highly aneuploid gametes

Be able to carry out a chi square test to evaluate the validity of a hypothesis.

ch 2 A statistical method used to determine goodness of fit •Goodness of fit refers to how close the observed data are to those predicted from a hypothesis Note: The chi square test does not prove that a hypothesis is correct •It evaluates whether or not the data and the hypothesis have a good fit x^2 = ∑ [(O - E)^2 / E] where •O = observed data in each category •E = expected data in each category based on the experimenter's hypothesis •∑ = Sum of the calculations for each category •Step 1: Propose a hypothesis that allows us to calculate the expected values based on Mendel's laws •The two traits are independently assorting •Straight wings and gray body are dominant traits Step 2: Calculate the Expected Values of the Four Phenotypes, based on the Hypothesis Step 3: Apply the Chi Square Formula Step 4: Interpret the Chi Square Value The calculated chi square value can be used to obtain probabilities, or P values, from a chi square table •These probabilities allow us to determine the likelihood that the observed deviations are due to random chance alone Low chi square values indicate a high probability that the observed deviations could be due to random chance alone. High chi square values indicate a low probability that the observed deviations are due to random chance alone. PERHAPS OUR HYPOTHESIS IS WRONG! If the chi square value results in a probability that is less than 0.05 (ie: less than 5%), the hypothesis is rejected Before we can use the chi square table, we have to determine the degrees of freedom (df) •The df is a measure of the number of categories that are independent of each other •df = n - 1 •where n = total number of categories •In our experiment, there are four phenotypes/categories •Therefore, df = 4 - 1 = 3

List the three common ways that the function of tumor-suppressor genes is lost.

ch 25 1.A mutation in the tumor-suppressor gene itself •The promoter could be inactivated •An early stop codon could be introduced in the coding sequence 2.DNA methylation or other epigenetic changes •The methylation of CpG islands near the promoters of tumor-suppressor genes, inhibits transcription 3.Aneuploidy •Chromosome loss may contribute to the progression of cancer if the lost chromosome carries one or more tumor-suppressor genes

Explain the underlying cause of most recessive patterns of inheritance

ch 25 Molecular basis: Often due to a loss-of-function mutation Disorders that involve defective enzymes typically have an autosomal recessive mode of inheritance •The heterozygote carrier has 50% of the normal enzyme •This is sufficient for a normal phenotype

List the four common types of genetic changes that produce oncogenes.

ch 25 •1. Missense mutations •2. Gene amplifications: Increase in copy number •Expected to increase the amount of protein •3. Chromosomal translocations: Translocation creates an oncogene that encodes an abnormal fusion protein •4. Viral integration: •Certain viruses integrate into host DNA as part of their life cycle •This can cause activation of a cellular proto-oncogene •Direct transcription from viral promoter •Activation of cellular promoter by viral enhancer

Define population and gene pool

ch 27 A population is a group of individuals of the same species that occupy the same region and (for sexually reproducing species) can interbreed with each other A large population is usually composed of smaller groups called local populations •Members of a local population are far likelier to breed with each other than with members of the general population •Local populations are often separated from each other by moderate geographic barriers Conceptually, all of the alleles of every gene in a population make up the gene pool •Only individuals that reproduce contribute to the gene pool of the next generation

Define quantitative genetics

ch 28 •Quantitative genetics is the study of traits that can be described numerically -Such traits are usually controlled by more than one gene and are significantly influenced by environmental factors •These are called complex traits •A quantitative trait is any trait that varies measurably in a given species •Quantitative traits such as height are continuous traits-they do not fall into discrete categories

Be able to apply a chi square test to evaluate if two genes are linked.

ch 6 This statistical method is frequently used to determine if the outcome of a dihybrid cross is consistent with linkage or independent assortment Step 1: Propose a hypothesis •The genes for eye color and body color are assorting independently •Even though the observed data appear inconsistent with this hypothesis, it allows us to calculate expected values •Since we do not know the probability of crossing over, we cannot calculate an expected value for a hypothesis based on linkage •Indeed, we actually anticipate that the chi square analysis will allow us to reject this hypothesis in favor of a linkage hypothesis Step 2: Based on the hypothesis, calculate the expected values of each of the four phenotypes •An independent assortment hypothesis predicts that each phenotype has an equal probability of occurring Step 3: Apply the Chi Square Formula Step 4: Interpret the Calculated Chi Square Value This is done with a chi square table (Chapter 2) There are four experimental categories (n = 4) •However, the way to calculate degrees of freedom (df) is not agreed upon by everyone •Some calculate it as n -1 = 3 •Others say there are really two categories, recombinant and nonrecombinant; in this case n =2 so n -2 = 1 (I prefer the second approach) If p > 0.05, •Therefore, we reject the hypothesis that the two genes assort independently •In other words, we accept the hypothesis that the genes are linked

Explain how mutations in the coding sequence of a gene may affect polypeptide structure and function

table 19.1 Gene mutations are molecular changes in the DNA sequence of a gene A point mutation is a change in a single base pair •It can involve a base substitution •A transition is a change of a pyrimidine (C, T) to another pyrimidine or a purine (A, G) to another purine •A transversion is a change of a pyrimidine to a purine or vice versa •Transitions are more common than transversions Mutations in the coding sequence of a structural gene can have various effects on the polypeptide •Silent mutations are those base substitutions that do not alter the amino acid sequence of the polypeptide •Due to the degeneracy of the genetic code •Missense mutations are those base substitutions in which an amino acid change does occur •Example: Sickle-cell anemia (Refer to Figure 19.1) •If the substituted amino acid has no detectable effect on protein function, the mutation is said to be neutral. This can occur if the new amino acid has similar chemistry to the amino acid it replaced •Nonsense mutations are those base substitutions that change a normal codon to a stop codon •Frameshift mutations involve the addition or deletion of a number of nucleotides that is not divisible by three •This shifts the reading frame so that translation of the mRNA results in a completely different amino acid sequence downstream of the mutation

Be able to describe the general features of various DNA repair systems. We will focus on nucleotide excision repair

table 19.7, fig 19.18 •Photolyase can repair thymine dimers •It splits the dimers restoring the DNA to its original condition •Uses energy of visible light for photoreactivation Base excision repair (BER) involves a category of enzymes known as DNA N-glycosylases •These enzymes can recognize an abnormal base and cleave the bond between it and the sugar in the DNA Depending on the species, this repair system can eliminate abnormal bases such as •Uracil; Thymine dimers •3-methyladenine; 7-methylguanine An important general process for DNA repair is nucleotide excision repair (NER) This type of system can repair many types of DNA damage, including •Thymine dimers and chemically modified bases •missing bases, some types of crosslinks NER is found in all eukaryotes and prokaryotes •However, its molecular mechanism is better understood in prokaryotes In E. coli, the NER system requires four key proteins •These are designated UvrA, UvrB, UvrC and UvrD •Named as such because they are involved in Ultraviolet light repair of pyrimidine dimers •They are also important in repairing chemically damaged DNA •UvrA, B, C, and D recognize and remove a short segment of damaged DNA •DNA polymerase and ligase finish the repair job **UvrB flanked by two UvrA in a complex scan DNA for thymine dimer. After damage is detected, UvrA is released and UvrC binds. UvrC makes a cut on both sides of the dimer, UvrD which is a helicase removes the damages DNA, UvrB and C are released A base mismatch is another type of abnormality in DNA The structure of the DNA double helix obeys the AT/GC rule of base pairing •However, during DNA replication an incorrect base may be added to the growing strand by mistake DNA polymerases have a 3' to 5'proofreading ability that can detect base mismatches and fix them If proofreading fails, the mismatch repair system comes to the rescue. Mismatch repair systems are found in all species An important aspect of these systems is that they are specific to the newly made strand Molecular mechanism of mismatch repair in E. coli •Three proteins, MutL, MutH and MutS detect the mismatch and direct its removal from the newly made strand •The proteins are named Mut because their absence leads to a much higher mutation rate than normal