Final Flashcards

Question

Meiosis stages

Answer 1

Meiosis I 1. Prophase I – most complex; produces a lot of variation a. Many of the same processes as mitosis i. The chromatin coils & becomes visible as chromosomes (sister chromatids) ii. The nuclear membrane and the nucleolus disappear iii. The spindle apparatus forms – grows from centrioles b. Synapsis occurs – new i. Homologous chromosome pair up & form tetrads - Each chromosome in homologous pair will attach to only one microtubule ii. Tetrad – 4 sister chromatids are present; 4 individual, unique chromosomes will eventually result - Will have 23 tetrads – mitosis has 46 “boxes” c. Crossing over occurs – must be between homologous pairs i. Chiasmata – attachment point between homologous pairs; place of cross over - The lateral chromatids do not participate – too far apart ii. Responsible for genetic variation – would otherwise have 2 sets of identical haploid cells - Variation is produced because the homologues can carry alternate versions of a gene - Ex: one homologue encodes brown hair and the other encodes blonde hair - Crossing over occurs and the homologue encoding brown hair may now encode blonde hair and vice versa - The homologues contain genes other than hair color so crossing over has created a different combination of gene types on the same chromatid 2. Metaphase I a. Chromosome tetrads are aligned at the metaphase plate down the center of the cell i. The sister chromatids remain attached by their centromeres b. Spindle microtubules are attached to the kinetichore in preparation for cell division i. Chromosomes are only attached to one spindle - Spindle microtubules from one pole of the cell are attached to one homologue - Spindle microtubules from the other pole are attached to the other homologue c. Homologous pairs are held together in metaphase at chiasmata sites 3. Anaphase I a. Homologous pairs separate at chiasmata from one another – NOT chromatids i. They are completely random in their separation – mother and father version will be randomly divided ii. The sister chromatids remain attached by the centromere b. 2 haploid cells result from 1 diploid cell – only have one chromosome set due to the separation of homologues 4. Telophase I & Cytokinesis a. Telophase may be incomplete – cell is going to divide again right away i. Normally involves the reforming of the nuclear membrane and nucleolus - May reform partially ii. Spindle apparatus does not disassemble b. During telophase I – haploid chromosome set have arrived at opposite poles of the cell i. Each pole has 23 distinct chromosomes – present as sister chromatids c. Cytokinesis occurs simultaneously – the cytoplasm and its contents are divided into two and the haploid daughter cells are formed Interphase does not occur between Meiosis I & II a. There is enough chromosomal content created in interphase preceding Meiosis I for 4 daughter cells o They can also produce more after division o There is no chromosomal duplication b. The chromosomes may briefly uncoil and become active – more often goes directly into prophase II Meiosis II – very similar to mitosis except cell that is entering is haploid 1. Prophase II – looks the same as mitosis a. Chromosomes will recoil if they have resumed activity since Meiosis I b. The spindle apparatus will attach to the centromere of each duplicated chromosome i. 2 kinetochore proteins & microtubules are attached per centromere – unlike meiosis I c. Chromatids are NOT identical because crossover occurred in prophase I 2. Metaphase II a. 23 sister chromatids are aligned down the metaphase plate i. Mitosis – has 46 sister chromatids ii. Metaphase I – has 23 tetrads 3. Anaphase II a. Centromeres of the sister chromatids separate b. Sister chromatids of each pair (individual chromosomes) move toward opposite poles of the cell 4. Telophase II & Cytokinesis: a. Telophase – the nuclear membrane and nucleolus reform, chromosomes uncoil b. Cytokinesis simultaneously occurs – separates cytoplasmic contents i. Cleavage furrow c. Results in 2 non identical haploid cells – due to crossing over

Answer 2

Differences 1. Mitosis – asexual reproduction a. Functions in growth, repair and asexual reproduction b. Has one cell division • Daughter cells are diploid and genetically identical to the parent cell – no variability unless errors occur 2. Meiosis – sexual reproduction a. Has 2 cell divisions • 4 daughter cells are haploid and genetically distinct from the 2n parent cell b. You have a random combo of mother and fathers • Even if they are equal from mother and father – still not identical due to crossing over c. All events unique to meiosis occur in meiosis I • Meiosis II is essentially mitosis except haploid – you need an indicator Similarities o Mitosis and meiosis are both preceded by chromosomal duplication during S phase of interphase

Answer 3

genetic variation of sexual reproduction arises in a number of ways 1. Independent Chromosomal Orientation a. Arrangement of chromosomes in metaphase I is random i. This arrangement effects the composition of gametes that result ii. The chromosomes from the mother will often carry different versions of genes than the chromosomes from the father b. During tetrad formation there is a 50/50 chance that the chromosome of maternal origin will end up on one side of the cell over the other c. Two possible chromosome arrangements – there are four possible gametes that can arise i. Total number of chromosome combinations = 2^n (n = haploid number) - Human sperm + human egg = 64 trillion possibilities 2. Random Fertilization a. Homologues carry the exact same genes at the exact same loci i. Ex: C (brown) and c (white) are alternate alleles for mouse coat color and E (black) and e (pink) are alternate alleles for mouse eye color b. Differences in gene alleles differentiate gametes from one another – any sperm gamete can be used in fertilization i. Egg – will be whichever is that month ii. Sperm – will have variability between each other & any of them can fertilize the egg - Creates more potential for genetical variability 3. Crossing Over a. The exchange of corresponding segments between homologous tetrads in prophase I i. Chiasma – where they contact and cross over - 2 nonsister chromatids attach - plural = chiasmata) ii. Produces recombinant – new chromosomal combinations b. At chiasma – each gene on one homologue is precisely aligned with the same gene on the other homologue i. There are many genes carried on a single chromosome ii. In humans approx. 1-3 crossing over events occur per chromosome per meiotic division

Answer 4

Molecular control systems exist to make sure errors do not occur – must have high levels accuracy o Timing and the rate of cell division are critical to normal cell development o Unregulated growth – cancerous Frequency/rate of cell division varies considerably 1. Varies depending on type of cells a. Nerve, some muscle cells & cardiac cells don’t divide at all – anything that damages will compromise the organisms ability to survive • Building muscle – adding more contractile protein within cell; you haven’t actually added more fibres, only fibrils b. Skin cells are constantly dividing Differences in division are introduced by molecular regulatory mechanisms o Signaling molecules in the cytoplasm are important – typically proteins that trigger and control key events o Checkpoints in G1, S, and G2 – internal and external

Answer 5

Checkpoints – occurs at G1, S, and G2 1. G1 is the most important – most effective because it’s at the beginning of the cycle & prevents wasting energy in subsequent stages a. If the cell passes this checkpoint – it will move on and be committed to divide b. If the cell does not pass this checkpoint – it will enter into G0 and not go on to divide i. Some cells remain in G0 for life – cells will no longer divide (ex. neurons) 2. Some cells are pulled out of G0 based on external cues and enter into normal division a. Ex: growth factors that are released due to injury – tissue must be replenished Cyclin and cyclin dependent kinases (Cdks) – 2 classes of regulatory molecules that work together 1. Cyclin dependent protein kinases (Cdk) a. Kinases are phosphorylators – activate and deactivate proteins via phosphorylation of molecules - Important for go signals at G1 and G2 phases - Addition of neg charged group changes the protein structure – will now have repulsive forces b. Cdk is always present in the cell at constant concentration i. Inactive – requires the attachment of cyclin to become activated (cyclin dependent) ii. Cyclin is not always present - Degraded after division – must be regenerated for division to occur again - Activity will increase or decrease based on the presence of the cyclin partner 2. Cyclin increases in G2 a. MPF – Cdk attached to cyclin; activated form of kinase enzyme - Triggers the movement of the cell past G2 in mitosis - MPF phosphorylates and activates proteins and other kinases b. MPF initiates mitosis - Proteins that cause the fragmentation of the nucleus arrive and are activated – results in dissolution of membrane - Assists with chromosome condensation and spindle formation c. MPF switches itself off in anaphase - Activates a process that causes destruction of its own cyclin – inactivates - Kinase is still present within cell at high concentration as an inactive Cdk Internal and external signals – used as signals at checkpoints 1. External – detected by receptors on membrane a. Ex. cancers that originate from foods 2. Internal – during anaphase sister chromatids must be perfectly aligned down the metaphase plate in order for the cycle to move forward a. Regulatory complex is activated when they’re properly aligned i. Activates a pathway that ends in separase activation (enzyme) - Separase cleaves cohesins between sister chromatids - Protects the chromosomal content of daughter cells b. If it’s not aligned – cannot go into anaphase Physical and chemical factors – external factors 1. Physical – presence of absence of certain things a. Lack of an essential nutrient will stop cell division – even if all other conditions are favourable 2. Chemical – growth factors are proteins released by cells to stimulate other cells to divide a. >50 known types b. Ex: platelet derived growth factor (PDGF) - Made by platelets – not cells; broken up megakaryocyte cell fragments - Needed for fibroblast division – cells in CT - PDGF receptor is located on the plasma membrane of responsive cells - Binding of PDGF to the receptor causes the cell to pass the G1 checkpoint and divide - Occurs in the body to allow wound healing - PDGF is not always present – will not stimulate cells without the receptor - If you cut yourself – PDGF presence initiates mitosis Density dependent inhibition – physical external factor a. Crowded cells will stop dividing – contact on all sides of cell stops division process - Division will continue until a single layer of cells forms on the inside of a container - Removal of some of the cells cause those located around the border to begin to divide - Repair until density inhibition is restored and cells are present on all sides again b. Cell surface protein binds to the counterpart on the adjacent cell and inhibits the growth of both cells Anchorage dependence – external physical factor 1. Cell division will not occur unless cells are attached to a solid support a. Ex: extracellular matrix or the bottom of a container - Controlled by plasma membrane proteins & linked cytoskeleton - Cytoskeleton is attached to cell membrane proteins and subsequently extracellular matrix 2. If not attached – inhibitory signal to prohibit division

Answer 6

Cancer cells do not exhibit density dependence inhibition or anchorage dependence 1. Results in excessive division – invade healthy tissues a. Lack normal growth factors – may have their own growth factors • When cancer cells stop dividing – random points; atypical checkpoints 2. Immortal – with unlimited nutrients they will divide indefinitely a. Normal cells – divide 20-50 times before aging and dying Develop when a normal cell undergoes a transformation to cancer cell 1. We have cancerous cells all the time – your immune system kills it 2. If the immune system doesn’t recognize the abnormal cell it will proliferate becoming a Tumor – mass of abnormal cells a. Benign – stays in one spot b. Malignant – undergoes changes and moves to other parts of the body • Metastatic cancer • You have no control – the longer its left, the harder it is to get rid of Cancer cells have an unusual chromosome number – they divide so often; errors are more likely to occur 1. Loose their ability to attach to surrounding cells and the extracellular matrix a. Spread to nearby tissues & may enter blood/lymph and spread to other locations • This is why lymph nodes are often removed in cancer patients – if cancer cell is present in lymph it indicates spreading May produce signaling molecules that allow blood vessels to grow inward a. Angiogenesis – blood vessel development b. If there is an influx of blood – there has been an increase of blood vessels & often indicates presence of cancer cells Metabolic ability and normal function are lost May be treated with: 1. High energy radiation – more specific to cancer cells because they loose their ability to repair DNA damage - Once dna is damaged – can’t be fixed 2. Chemotherapy – drugs are administered into the blood that target the cell cycle in actively dividing cells a. Target anything that are actively dividing – health and cancer cells • Causes alot of side effects b. Poor selective toxicity

Answer 7

Garden peas have 1. Many different versions available – more statistics can be produced a. Ex: purple, white variants 2. Have very short generation times & produce a large number of offspring a. Not the case for humans – 9 month generation; small number of offspring; ethical violations Characters vs traits a. Characters – varied heritable features; can be pass down generation - Ex: flower color, hair color, eye color b. Traits – the variant of a character - Ex: purple flowers vs white flowers, brown eyes vs blue eyes, brown hair vs blonder hair - There is an infinite amount of possibilities Pea plants are self-fertilizing – mates with itself with only its own gene pool; no new genes and traits coming in 1. Each flower has male and female parts a. Egg bearing carpel (only 1) b. Pollen producing stamen 2. Mendel removed the immature stamen – allowed him to control gametes used in fertilization a. He dusted the carpel with pollen taken from a different stamen – flower sperm was transferred 3. True breeding individuals – mate with itself; only produces more of itself a. Always homozygous – can be dominant or recessive • Many rounds of self pollination – will produce the parent variety only; all future generations will only produce the same colour as parent • No variability in F2 generation if they are self pollinating b. Self pollination – male and female gametes are from the same parent • No genetic variation Mendel studied characters that occurred in two traits - Didn’t consider things that had more traits – kept simple - Ex: purple and white flowers Crossed two different true breeding plants – hybridized a. 2 true breeding individuals form the parent generation (P gen) b. F1 generation – hybrids from P gen c. F1 self pollinated to produce F2 generation

Answer 8

Blending hypothesis – phenotypes of offspring will be the intermediate of the parent 1. If the blending hypothesis were to hold true – a cross of a purple flower with a white flower would produce a daughter population with light purple colored flowers 2. Blending was not observed in Mendel experiment a. Purple flower and white flower are P gen – both are homozygous - Purple is dominant, white is recessive b. F1 generation – original purple colour - All are heterozygous (Punnett square) – purple and white coexist - Self pollinated c. F2 generation – purple flowers are still present, but white flower reappears - 2 heterozygous Punnett square – 3 purple & 1 white - 705 were purple : 224 were white - Blending has not occurred because white remains – it was only masked by dominant gene The same experiment with 6 different characters & 2 traits each showed the same outcome o Always 3:1 in F2 generation from dominant vs recessive

Answer 9

1. Alternate alleles of genes account for variation in inherited characteristics a. Ex: gene for flower color exists as white or purple alleles b. Nucleotide sequence of DNA at the same locus of homologous pairs may be nonidentical due to differing alleles i. Homologous pair – genes are at same locus ii. Gives rise to a different allele of the same character – many alleles are possible depending on gene - Similar enough to be for the same character but may code for different traits 2. An organism will have two copies of each character a. One from each parent – one chromosome in a homologous pair b. Homozygous – same allele i. Can be dominant (PP) or recessive (pp) ii. True breeding are all homozygous - PP + PP = PP - pp + pp = pp c. Heterozygous – different allele - Not true breeding - Gametes fusing can be Pp, PP, pp 3. Dominant allele will determine phenotype in heterozygous organism a. The recessive allele will have no visible affect – you don’t see it; it’s hidden in genome i. Can appear in future generation if organism receives recessive allele from both parents b. Ex. i. The purple allele of Mendel’s pea plants is the dominant allele ii. The white allele of Mendel’s pea plants is the recessive allele – it ‘disappeared’ in the F1 generation 4. Law of segregation a. Two alleles for a heritable character separate during gamete formation i. Each gamete will have allele either from mother or father – no longer has both alleles after homologues separate ii. Alleles separate into different cells during meiosis – separate gametes - Homozygous – 4 daughter cells will have same allele (Not considering crossing over) - Heterozygous – 2 will have dominant allele and 2 will have recessive b. Gametes come together randomly during fertilization – any combination of dominant and recessive alleles present is possible i. Determined by which gametes (egg and sperm) are present ii. Any combination is possible -> determined by which gametes are used - Depending on sperm and egg

Answer 10

Punnett Squares – allow prediction of the possible allele combinations & expression in phenotype a. Dominant vs recessive o Dominant alleles are shown with capital letters o Recessive alleles are shown with lower case letters b. Phenotype – observable traits o Will always be dominant allele c. Genotype – genetic makeup o Determines phenotype Test Crosses – used to determine genotype for a particular phenotype 1. Dominant trait is expressed in heterozygous or homozygous dominant a. Can cross unknown with homozygous recessive organism i. If unknown is heterozygous – recessive allele is expressed • Half of offspring will express recessive and half will express dominant ii. If unknown is homozygous dominant – recessive is not expressed • Would make 4 heterozygous offspring b. Unethical with humans – can experiment with other plant organisms

Answer 11

Law of Independent Assortment – each pair of alleles will sort independently of each other allele during gamete formation 1. Monohybrid crosses – analyze one character at a time a. Ex: flower color - Trait would be purple or white 2. Dihybrid crosses – analyze two characters simultaneously a. Ex: seed color and seed shape - Yellow seeds (Y-dominant) - Green seeds (y-recessive) - Round seeds (R dominant) - Wrinkled seeds (r-recessive) b. A cross of two plants: i. P gen (true breeding) • Homozygous dominant for color and shape: YYRR -> True breeding • Homozygous recessive for color and shape: yyrr -> True breeding ii. F1 plants will be all heterozygous if 2 above parents are mated • Yellow and round YyRr Dependent vs independent assortment of alleles 1. Dependent assortment – passed along together a. Gametes of homozygous P gen would be YR or yr b. Gametes of heterozygous F1 would be YR or yr • Passed on as units • if only YR and yr are observed in gametes (remember these are 2 different traits) then it is dependent assortment c. Phenotypic ratio of dependent assortment = 3:1 /4 • Phenotypic ratio will match genotypic ratio 2. Independent assortment – passed along as individual units; obeys the law of independent assortment a. Gametes of homozygous parents would be YR or yr regardless • Does not need to be heterozygous to sort independently but we won’t observe any variation in homozygous anyway b. Gametes of heterozygous F1 could be YR, Yr, yR, yr i. Punnett square will have 4 x 4 = 16 possible gamete combinations during cross of 2 F1 organisms ii. 4 possible phenotypes in F2 generation (Mendel examined) o 9 yellow, round plants -> 9/16 o 3 yellow, wrinkled plants -> 3/16 o 3 green, round plants -> 3/16 o 1 green, wrinkled plant -> 1/16 iii. Phenotypic ratio of independent assortment = 9:3:3:1 /16 c. Each allele segregates as though it is participating in a monohybrid cross

Answer 12

Complete dominance – dominant allele has the same phenotypic effect whether it is present in one or two copies 1. Mendel’s Law only explains complete dominance a. BB or Bb will both have brown hair – because B is completely dominant • Realistically Bb could have other phenotypes if incomplete dominance was operational Incomplete dominance – F1 hybrids from homozygous dominant and homozygous recessive will produce heterozygous with intermediate phenotype (all will be heterozygous) 1. Ex: cross a red snap dragon with a white snap dragon and all of the F1 generation is pink a. The F1 generation has less red pigment than the homozygous dominant parent generation 2. Still goes against blending hypothesis – 2 heterozygous F1 mating will produce F2 generation that will show reappearance of red and white (1 red, 1 white, 2 pink)

Answer 13

Most genes in the population have multiple alleles 1. Ex: the ABO blood groups – A, B O are superscripts a. Three alleles exist for the blood type gene: IA, IB, i • i indicates O type • Each parent provides one blood type allele There are 4 different blood types from the possible allele combinations 1. Type A blood: IAIA or IAi 2. Type B blood: IBIB or IBi 3. Type AB blood: IAIB a. IA and IB alleles are co-dominant because heterozygotes express both on RBC – neither is masked • Codominance is not incomplete dominance 4. Type O blood: ii RBC have antigens on surface (carbohydrates) 1. Type A blood – will have A antigens a. Can only receive O or A 2. Type B – have B antigens a. Can receive B or O 3. Type AB – have A and B antigens on RBC a. Can receive A, B, AB, or O – universal acceptor 4. Type O – has no antigens; smooth a. Universal donor – you have give them to any b. Can only receive O

Answer 14

Phenotypic expression of a gene at one locus affects phenotypic expression of a gene at a second locus Ex: Labs 1. Gene #1 determines coat color a. Brown is recessive (b) b. Black is dominant (B) 2. Gene #2 determines whether the pigment will be deposited a. E is dominant • E deposits black or brown pigment • Can be EE or Ee and have pigment deposition b. e is recessive • Homozygous recessive (ee) has no pigment deposition regardless of black or brown genotype 3. 16 outcomes a. 9 black : 3 brown : 4 golden b. Different than ratio without epistasis

Answer 15

One gene that affect many phenotypes; common in the body Ex: sickle-cell disease in humans 1. Results when someone is homozygous recessive for allele a. Heterozygous – generally normal; will have some sickle-shaped cells • Alleles are codominant – both are expressed b. Homozygous dominant will not express any 2. Leads to the production of abnormal hemoglobin – RBC are sickle shaped a. Uncharacteristic are destroyed by the body i. Anemia – inadequate amounts of RBC and hemoglobin - Inadequate o2 transport for cellular respiration and energy - Low energy is a characteristic symptom b. Deformed shape of the cell causes it to accumulate in vessels creating blockages – obstruct flow to certain areas resulting in fever and pain 3. Kills 100,000 people worldwide per year a. Most common inherited disorder in African populations compared to other parts of the world – more similar genetics • Diseases are often more common in a smaller gene pool – smaller, isolated communities Heterozygotes are resistant to malaria – caused by parasite

Answer 16

One character influenced by many genes; spectrum of possibility for certain traits; reverse of pleiotropy Poly = many in Greek; genic = gene Not everything is black and white with respect to phenotype Ex: Human skin colour or height 1. Assume that human skin color is controlled by three genes which are inherited separately (independent assortment) a. A, B, and C are the dark skin alleles -> incompletely dominant to the a, b, and c light skin alleles i. Incomplete dominance - AA, Aa, aa – will all be different phenotypes ii. Very large Punnett square – 3 genes - 3 alleles from each parent b. A person who is AABBCC would be very dark, a person who is aabbcc would be very light and a person who is AaBbCc would be somewhere in the middle

Answer 17

Many genetic factors are affected by the environment a. There are many more varieties of skin color than were predicted by the Punnett square previously - Sun exposure also has an effect on skin color (thus the environment) Traits often result from a combination of heredity and environment o Ex. sun exposure and skin colour; experience on intelligence; nutrition on height Twin studies – Identical twin that were separated display that genetics and the environment have an effect on a person’s individual features

Answer 18

chromosomes are vesicles of genetic heredity Mendelian genes have special positions (loci) on their chromosome – exhibit independent assortment and undergo segregation 1. Law of segregation – the two alleles at the same locus for each gene separate during gamete formation; meiotic behavior of homologous chromosomes 2. Law of independent assortment – alleles of genes on nonhomologous chromosomes assort independently during gamete formation; meiotic behavior of non-homologous chromosomes - Account for independent assortment of alleles for 2 or more genes on different chromosomes Morgan provided evidence to associate a particular gene with a specific chromosome – which recipe on which chromosome 1. Studied Drosophila melanogaster – fungi-eating insect that produces hundreds of offspring in brief period of time a. Quick gestational period – 2 weeks b. Four chromosomal pairs – humans have 23 • 3 autosomal • 1 sex pair (female: XX, Male XY) 2. He carried out many matings, analyzing offspring for the appearance of natural variants a. He discovered male flies that had white eyes b. Usually flies had red eyes – wild type 3. Wild type – most commonly occurring phenotype is the wild type; the most prevalent all the time a. Alteration of the wild type produces mutants – change/variation of the wild type that will give different phenotypes

Answer 19

Allele behavior correlates with behavior of the chromosomal pair – sometimes the sex chromosome Mate a white eyed male (recessive) with a red eyed female (dominant) 1. F1 generation all have red eyes – dominant inheritance pattern • F1 offspring produces the typical 3:1 ratio seen in 2 heterozygous mating 2. The F2 offspring with white eyes were all males – could also be red • All F2 females had red eyes The eye color of the flies must have been linked to sex – gene is carried on X chromosome 1. Females carrying a recessive allele may have it masked by a dominant allele a. XX – the white eyed allele will be masked by allele on other X chromosome i. Will only show up in phenotype if both inherited X chromosomes carry recessive allele – less common - Father must have white eyes and mother must be a carrier (or homozygous recessive) b. This is why females didn’t have white eyes 2. Males carrying a recessive allele do not have a second X chromosome to mask the recessive allele a. XY – males can only have 1 version; what is present on X cannot be masked by another X chromosome - Y can’t mask because they have different genes b. Males with white eyes will transit it to all his daughters and no sons i. Mating with homozygous female will result in all heterozygous daughters & no white eyed sons o Daughters will all be carriers ii. Mating 2 heterozygotes results in 3:1 ratio Sex chromosomes carry genes unrelated to sex a. Sex linked gene – any gene located on a sex chromosome (ex. fly eye colour) - Will have different inheritance pattern for males - Females are more similar to autosomal Punnett squares Common phenomenon for disease and non disease states 1. Sex-linked disorders mostly affect the male population – because they only get one X; will more likely be in their phenotype a. Males are hemizygous – neither homozygous or heterozygous; the X and the Y chromosome are non-identical but both are sex chromosomes 2. Women will more likely be phenotypically normal a. Carriers – woman who carries the recessive gene but does not express it 3. Disease states are recessive – Sex-linked traits include: - Hemophilia – inability to clot blood - Red-green colorblindness - Duchenne muscular dystrophy

Answer 20

Only 1 X chromosome in females is active – 2 X chromosomes carry different genetic info 1. Barr body – supercoiled and inactivated X chromosome (coiled too tightly to make mRNA) a. Barr body is determined early on in embryonic development and every daughter cell resulting from mitosis will have the same Barr body as parent cell b. Barr bodies are determined at random during embryonic development – not all cell will choose the same one 2. A heterozygous female for an X linked gene – will have one allele turned on in some cells and the other allele in different cells a. Ex. female cats have spots because both alleles are being expressed in phenotype Occurs to prevent different amounts of protein production in males and females – because females have double the amount; turned off to make the protein production equal o The Y chromosome in males does not contain many genes – is not comparable to second chromosome Autosomal chromosomes are not inactivated & does not exhibit same differences – will only present the phenotype for dominant

Answer 21

1. Sex determined by male a. Humans – females are XX & males are XY i. It is not 100% clear why the Y chromosome determines male offspring - The Y chromosome has 78 genes which encode 25 different proteins (not a 1:1 ratio) a. Female has no recipe for these genes ii. The SRY (sex determining region) is a gene located on the Y chromosome – determines sex of fertilized egg - Triggers the development of the testes/testicles – produce proteins necessary for - When absent – the ovaries develop by default (XX) because no SRY is present iii. X is much larger than Y – do not have same genes, loci, or centromere position b. Grasshoppers and some other insects have an X-O system (O = zero) i. Females are XX and males are XO (where O symbolizes the lack of a second sex chromosome) - There is only 50% of sex chromosomes present in males – no Y chromosome ii. Meiotic division in males produces 4 gametes – only two will contain a sex chromosome - Egg will always give X – will be male if fertilized with 0 sex chromosome 2. Some animals have sex determined by the female – some fish, butterflies and birds a. Males have the genotype ZZ – homozygous pair b. Females have the genotype ZW – heterozygous i. W will cause development of female ii. Female gametes (eggs) will either give W or Z 3. Some organisms do not have sex chromosomes – develop based on chromosome number (ex. bees & ants) a. Can artificially stimulate the development of haploid organisms that are normally diploid (only experimentally – cannot be done under normal circumstances) i. Ex. rabbits b. Males will develop from unfertilized eggs -> haploid c. Females develop from fertilized eggs -> diploid

Answer 22

Sex linked genes – exhibit dependent assortment; genes that are located near to one another on the same chromosome are inherited together – dependent assortment 1. Do not follow Mendel’s Laws of Independent Assortment a. Meiosis of unlinked genes will produce an equal amount of all four possible genotypes • b + vg +, b + vg, b vg +, b vg b. unlinked have 9:3:3:1 2. During meiosis there is a high proportion of gametes with the b+ vg+ and b vg genotypes produced because they are sorted as a unit a. Do not have 9:3:3:1 ratio Ex. Studies of flower color and pollen shape 1. Sex linked inheritance a. Flower color – P: purple & p: red b. Pollen shape – L: long & l: round 2. Would expect a 3:1 ratio according to Mendel’s Law when crossing of two heterozygotes for just one trait (either PpXPp or LlxLl) a. Would produce 3 purple long & 1 red round in F2 offspring (9:3:3:1 ratio) • PL, Pl, pL, pl • 9:3:3:1 ratio – similar to independently assorting dihybrid crosses b. This is not what is observed when crossing 2 heterozygotes • 75% of plants were purple and long – more than expected 9/16 • 14% of plants were red and round – more than expected 1/16 • 11% have heterozygous phenotypes (purple round and red long) – were much fewer than expected from 3/16 + 3/16 ratio 3. Observed differ from expected because independent assortment is not present a. P gen is true breeding – all gametes will be homozygous recessive or dominant (PL or pl) b. In flies – high proportion of gray normal and black vestigial traits seen in the experiment were a consequence of fertilization between gametes of the b+ vg + and b vg genotype Crossing over during prophase I gives rise to variation that results in 4 different alleles – if crossing over did not occur, 2 games would have same combination of alleles 1. Recombinant genes – produced via the separation of linked genes; follows crossing over of two gametes will be of the parental phenotype a. b+ vg+ and b vg are initially linked – b+ vg and b vg+ are the recombinant genotypes of the other two gametes b. Fertilization with the recombinant gametes – gives rise to the small proportion of heterozygous offspring (less than the 3/16+3/16 ratio expected) 2. Crossing over unlinks those genes during meiosis Ex. Drosophila (fruit flies) were used by Thomas Morgan in genetic studies 1. Easily and inexpensively grown & produces lots of generations in a matter of weeks or months – can study quickly 2. Mated a wild-type fruit fly (gray body and long wings; BL or b+ vg+) with a black fly with undeveloped wings (bl or b vg) a. b+: gray body (dominant) b: black body (recessive) b. vg+: long wings (dominant) vg: undeveloped wings (recessive) 3. Genes are linked – majority of F2 generation were of the parental phenotype (b vg or b+ vg+) a. 17% were recombinant phenotypes – either b vg+ or b+ vg • The recombination frequency is equal to the percentage of recombinants produced from crossing over

Answer 23

2 hypotheses by one of Morgan’s students 1. Crossing over is equally likely at all points on a chromosome 2. The further apart two genes are located on a chromosome the greater the likelihood that crossing over will occur between them (not equally likely) a. Rationalized that the increased distance allowed a greater number of possibilities for crossing over to occur i. Created a linkage map – using frequencies at which the crossing over took place • Distance between genes is in map units -> 1 map unit is equal to a one percent recombination frequency o Ex. distance between gene 1 and 2 is 7 map units = 7% recombinant frequency b. Tested recombinant data from fruit flies – used to map out the location of various genes i. Looked at 3 genes on the same chromosome • Linked genes for body color (black: b) and wing type (undeveloped wings: vg) • Recessive allele for eye color (cn: bright red) that was located on the same chromosome as the body color and wing type genes ii. Crossing over frequencies determined from experiment: • b & vg: 17% (17 map units between the 2 genes = greater chance that crossing over will occur) • b & cn: 9% • vg & cn: 9.5% iii. Testing supported hypothesis further apart genes resulted in greater crossing over c. Also hypothesized that since the recombination frequency between b & vg was approx. two times distance of b & cn and vg &cn -> b & vg must be located farthest apart from one another with cn located somewhere between b and vg - According to the primary hypothesis that recombination frequency should be higher between genes that are farthest apart from one another than between those that are near to one another - Helped to show where they were on the chromosome – be able to position genes on chromosomes It was later shown that crossing over is not equally likely at all points on a chromosome

Answer 24

Non-disjunction – 2 types 1. Meiosis I – members of homologous pairs do not separate; both homologous pairs go to same side and other side gets none a. None of the cells will be normal in these gametes 2. Meiosis II – sister chromatids do not separate a. 2/4 gametes will be normal b. 2/4 will have wrong amounts • One will have n + 1 (2 copies of the same chromosome) • One will have n – 1 (no copies of that chromosome) 3. Usually leaves the other chromosome pairs distributed normally Aneuploidy 1. A gamete with an abnormal chromosome number is used in fertilization a. 2 types i. Monosomic – gamete is missing a chromosome; zygote is 2n-1 o Diploid – 1 (only one copy of chromosome where error occurred) ii. Trisomic – the gamete has an extra chromosome; zygote is 2n + 1 o Diploid + 1 (3 copies of chromosome where error occurred) b. Mitotic divisions of embryonic cells are all abnormal chromosome numbers i. Ex: Down Syndrome o Trisomy 21: three copies of chromosome 21 in all body cells (only the 21st is abnormal) 2. Can also occur in mitosis – impact depends on when it occurs a. earlier on in embryonic development – will be transmitted to many cells with large effect • initial zygote is still 2n (normal) • subsequent division of cell will all be abnormal Polyploidy – more than 2 complete chromosome sets are present 1. Common in plants – would greatly affect humans 2. Triploidy: 3n a. Can occur when one gamete in fertilization has undergone a nondisjunction of all chromosome pairs – none of the homologous pairs separated during meiosis I • Result of meiosis I – creates 2n (diploid) and 0n b. When used in fertilization – creates 3n 3. Tetraploidy: 4n a. Can occur when the diploid zygote completes S phase in preparation for mitotic division and doesn’t divide • Fertilized egg undergoes interphase but does not divide – 4n

Answer 25

can occur due to meiotic divisions or damaging agents that break chromosome structures 4 types: 1. Deletion – chromosomal fragments are lost -> genes are missing a. Ripping out pages of recipe book – you loose recipes 2. Duplication – the duplicated fragment reattaches to the sister chromatid a. May also reattach on a non-sister chromatid – differing alleles may create non-duplicates • New material added to chromosome 3. Inversion – the broken fragment reattaches to the same chromosome backwards a. Changes the sequence of genes on chromosome 4. Translocation – the broken fragment reattaches to a non-homologous chromosome a. Differs from duplication onto non-sister chromatid in that there is no duplication of genetic material before it moves Deletions/duplications – can occur due to unequal size exchange during crossing over o Suppose to be taking off equal amounts and exchanging the same amount of genetic material Structural alteration of chromosomes can cause severe disorders – especially if abnormal chromosomes are used in fertilization

Answer 26

Occurs when there are three copies of chromosome number 21 -> 47 total chromosomes a. Called trisomy 21 -> affects 1:700 children b. Abnormal chromosome numbers often leads to spontaneous abortion during development o Trisomy 21 doesn’t upset development as much -> fetus continues to develop ``` ‘syndrome’ – characteristic set of symptoms that arise as a result o Round face o Skin fold at inner corner of the eye o Flattened nose bridge o Small and irregular teeth – dentition o Heart defects – can lead to open heart surgeries o Short stature o Reduced life span – can be in 30s ``` 50% of a woman with Down Syndrome’s eggs will have 3 copies of chromosome 21-> a 50% chance that her offspring will be affected

Answer 27

either 2n+1 or -1 Tends to create less of an issue than aneuploidy of autosomal chromosome a. The Y chromosomes has very few genes – will not alter development as much b. Additional X chromosomes are inactivated – extra X will be inactivated as Barr body - Still creates problems even through additional would be inactivated and lacking one would have been inactivated anyway Effects a. XXY: Klinefelter’s Syndrome - Small testicles, sterile (can’t reproduce), breast enlargement, decreased intelligence b. XYY: - Normal sex development, taller - Height isn’t always associated with Y (women can be tall) c. XXX: Trisomy X - Learning disabilities, taller, otherwise normal - Having too many X tends to affect cognitive development d. X: Turner’s Syndrome - Sterile, immature sex organs, normal intelligence - Too few X doesn’t impact intellect as much as too many - Estrogen therapy allows for development of secondary sex characteristics

Answer 28

Originally thought that either DNA or protein may have been the genetic material o Infection of bacterial cells with virus was crucial to the determination of DNA as the genetic material Griffith 1928 – used two different strains of Streptococcus pneumoniae (bacterial organism) to try and vaccinate against pneumonia 1. 70% of people have in respiratory system naturally – often must have comorbidity in order to actually affect organism 2. 2 strains a. Strain #1: pathogenic, disease causing bacteria b. Strain #2: non-pathogenic bacteria 3. Griffith heat killed the pathogenic strain – mixed remaining cellular material with nonpathogenic a. Some cells became pathogenic – was passed onto daughter cells following cell division • Heritable – It integrated into chromosomal content b. Transformation – change to genotype/phenotype of the cell due to the uptake of naked (free floating; not inside cell) DNA into the cell • Term was coined by Griffith Avery – wanted to know what materials were causing transformation 1. Materials that are present within the cell with info in their composition - DNA - RNA - Protein 2. Extracted RNA, DNA, and protein from heat killed bacteria - Separated DNA, RNA, and proteins and added to cultures of nonpathogenic cells - Found that only active DNA was able to transform heat killed bacterial cells into virulent cells (able to cause disease) All living things house genetic material as DNA 1. Bacteria contain a single circular chromosome a. E. coli contains a single circular double stranded chromosome • 4.6 million base pairs within • Can replicate in as little as 20 minutes 2. Human nuclei have 46 chromosomes a. 6 billion nucleotide pairs – between 46 chromosomes • Approx. 1 in every 10 billion nucleotide additions an error is made – highly accurate process b. Takes ~3 hours in total to replicate (occurs in S phase) • Over 12 proteins and enzymes used – collectively called the replisome

Answer 29

viruses that infect bacterial cells Viruses are much smaller than bacteria – can have DNA or RNA as genetic material (never both) a. Viruses are not living – living cells always have DNA as genetic material b. Nucleic acid is enclosed within a protein coat – capsomeres c. Obligate intracellular parasites -> rely on host cell machinery for replication - Has an obligation to be inside cell - Parasitic – virus will do well; cell will not Hershey and Chase – showed that DNA was the genetic material of the T2 bacteriophage (DNA virus) a. T2 phage specifically Infects E. coli – can program the host cell to produce high quantities of T2 phage - Wanted to determine if it was DNA or proteins coat material with this ability b. Added radioactive tag to the protein in one T2 batch and DNA in another – could see where the protein and DNA goes - Protein – added the label to sulfur - DNA – added the label to phosphate c. Two E. coli cultures were used – one was infected by DNA, the other protein - Found that only the phage DNA could enter E. coli – protein could not enter - Determined that DNA was responsible for transforming and moving genetic material into bacterial cell

Answer 30

1. Equal numbers of complimentary nucleotide between double strands a. Adenine residues = thymine residues - 2 hydrogen bonds – there are 2 partial pos/neg areas on structure where hyd bonds can form b. Guanine residues = cytosine residues - 3 hydrogen bonds – there are 3 partial pos/neg area on each structure where hyd bonds can form c. Pyrimidine (C & T) paired with a purine (A & G) – always 3 rings across - Purines – 2 rings - Pyrimidines – 1 ring 2. Composition of base pairs varies between species a. Ex: bacteria vs eukaryotes b. Nucleotide sequence of the ladder is random – will be more similar within same species and between more similar species - Determines the genetic make up of the individual 3. Structure of DNA a. Double stranded helical structure – this is the lowest energy and most stable confirmation i. Each single strand has a sugar (ribose) phosphate backbone • Negatively charged phosphate groups are found on the exterior – minimizes repulsion ii. Two strands are anti-parallel – 5’ and 3’ end on each side • Carbons have primes b. Each full turn of the helix is 3.4nm in length - Bases are 0.34nm apart - 10 base pairs per full turn of the helix

Answer 31

Two strands of DNA must separate for DNA replication to occur – must occur with a high level of accuracy a. High levels of accuracy – mistakes will result in proteins being formed differently and may not function Theories of DNA replication 1. Semi conservative – each DNA strand becomes a template for the synthesis of a new DNA strand; new DNA will contain one parent strand and one daughter strand a. Pool of free nucleotides in the cell is used to synthesize the new DNA strand – results in two double stranded DNA strands from the original parent strand • Enzymes are used to link the newly added nucleotides together – catalyze phosphodiesterase covalent bond between nucleotides once nucleotide has been added b. This is what actually occurs 2. The dispersive model – all four DNA strands consist of a mixture of new and old DNA 3. The conservative model – one double stranded DNA molecule is entirely newly synthesized DNA & the other is entirely parental DNA

Answer 32

Bacterial replication is the best studied (eukaryotic replication is same concept) 1. Begins at the origins of replication (ori) – unique short nucleotide sequences read by enzymes to indicate where to begin a. Initiator proteins – attach to the ori and open up the double stranded DNA molecule; molecule become single stranded around ori i. Forms a replication bubble – replication proceeds in both directions until the entire double stranded molecule is completed - Decreases time required to complete replication b. Eukaryotic chromosomes may have 100s of origins of replication i. Multiple bubbles form and fuse accelerating the copying process c. Replication forks – Y shaped ‘zipper’ ahead of each bubble (position of zipper); parental strand is changing double to single stranded 2. Several proteins participate in the unwinding – replisome a. Helicase – has 2 functions i. Untwists DNA at the replication fork ii. Separates the two strands b. Single stranded binding protein (SSBP) – binds to separated strands and prevents them from reattaching i. Physically prevents hyd bonds from forming c. Topoisomerase – breaks, swivels and rejoins the DNA strand upstream of where it is being twisted i. Forces chromosome to stay linear in double stranded conformation d. DNA polymerase – catalyzes the addition of the new DNA strand by reading strand and adding nucleotides one by one to 3’ end i. There are many DNA polymerases - DNA polymerase I & DNA polymerase III are most important - Different isoforms – different nucleotide sequences in their genes ii. Adds: - 500 nucleotides/second in bacterial cells – errors can still occur but not as high risk because it’s less complex - 50 nucleotides/second in human cells – accuracy is highly important because errors can compromise entire organism 3. RNA primer – needed to begin the process of DNA replication a. Synthesized by primase – occurs at ori i. Can’t add dna unless primase is in position ii. Adds RNA nucleotides one at a time according to the DNA template b. New DNA strand begins forming at the 3’ end of the RNA primer i. RNA is only temporary and must be deleted later – RNA is not present in DNA ii. Chain is elongated off of RNA template -> gives a place for DNA to start -> is deleted later and refilled 4. Nucleoside triphosphates – the starting nucleotide material (do not confuse with ATP) a. 2 inorganic phosphate (Pi) molecules are removed – nucleoside monophosphate that will be added to the growing DNA strand i. An exergonic reaction – due to high energy phosphate bonds being broken - Energy is used to build DNA – catalyzed by DNA polymerase 5. Antiparallel elongation – DNA are arranged oppositely to each other a. DNA polymerase – can only add nucleotides to the 3’ end of the daughter DNA strand or the RNA primer i. Enzymes are highly specific and do not recognize 5’ b. Leading strand – 3’ to 5’ end i. The daughter DNA strand is synthesized continuously c. Lagging strand – 5’ to 3’ end (watch this part again) i. Synthesized discontinuously – enzyme is constantly falling off and on ii. Produces Okazaki fragments – a series of small fragments of DNA and RNA hybrids - 1000-2000 nucleotides long in E. coli - 100-200 nucleotides long in eukaryotes iii. DNA ligase – fills in the gaps between the Okazaki fragments according to the DNA template - Joins the sugar phosphate backbone together

Answer 33

1: 10^5 nucleotide errors made – DNA polymerase proofreads nucleotides once they’ve been added to remove errors; can remove errors and continue to synthesize DNA a. Mismatch repair enzymes – repair mismatched base pairs in DNA that the polymerase has missed b. Nucleases – cut out damaged DNA and replace the segment according to the undamaged template - DNA ligase and DNA polymerase will fill in the gaps c. 1:1010 nucleotide errors will remain in the final DNA molecule Skin cells have enzymes to repair damage to DNA due to UV light o DNA can be damaged by carcinogenic chemicals

Answer 34

Lagging strand will have an RNA primer that is removed from one end that cannot be filled in via normal mechanisms because DNA cannot be added to 5’ end o Repeated rounds of DNA replication will provide shorter and shorter DNA strands o Not an issue with the circular chromosomes of prokaryotes – they attach in a circle Telomere sequences – linear DNA sequences have 100-1000 short nucleotide sequences at their ends; replications that do not contain genes o Serve to postpone the erosion of genes located near the end of the DNA molecule - Ex: TTAGGG Telomerase – catalyzes the lengthening of telomeres in germ cells & restores the original length and prevents shortening o This process is not necessary and therefore inactive in somatic cells

Answer 35

Bacteria – single, circular, double stranded chromosomes 1. Only a small amount of protein – mostly nucleic acid a. Requires protein to allow DNA to coil/supercoil so that it occupies only the nucleotide region (no membrane) • DNA in E. coli is 500x the length of the cell • No membrane around nucleoid b. Supercoiling – extremely compact Eukaryotes – linear chromosomes 1. Large amount of protein a, Histone proteins – allows them to condense & stay organized 2. Chromatin – DNA complexed together with protein; macromolecule a. Euchromatin – fully unwound structure; allows for full chromosomes access b. During interphase – chromatin is present as a diffuse mass when replication occurs in nucleus • Heterochromatin – intermediate level of packaging; transcriptional machinery cannot access the structure c. Coils into chromosomes in preparation for mitosis – short, fat metaphase chromosomes 3. 1/46 eukaryotic chromosomes contain 1.5 x 10^8 base pairs & stretched out it is ~4cm in length – all must fit into nucleus

Answer 36

Gene – contains info for amino acid sequence to create protein 1. One gene = one polypeptide a. Protein = multiply polypeptide units • Ex. hemoglobin – 4 polypeptide sequences b. Multiple genes come together to form protein quaternary structure 2. Give rise to both mRNA and functional RNA (tRNA & rRNA) – all are utilized in protein synthesis; only mRNA is read 3. Single gene can be used to produce closely related polypeptides via alternative splicing (substitutes) RNA polymerase – separates two DNA strands & covalently joins nucleotides according to DNA sequence 1. DNA must be linearized a. Only one strand has gene info and is read b. Does not come apart completely like in DNA replication 2. Can only join nucleotides in a 5’ -> 3’ direction – starts at 3’ end of DNA a. DNA is read 3’ to 5’ end b. Does not require a primer 3. RNA sequence is complimentary to the DNA template a. Differences – has uracil instead of thymine & has ribose sugar instead of deoxyribose sugar Begins at promoters – sites on the DNA template where RNA polymerase has the opportunity to attach; properly align to initiate transcription o Always upstream of the gene – not part of the gene; extends several dozen nucleotides upstream from the transcriptional start point (not short) o Section before the gene starts – not part of the gene Transcription unit – stretch of DNA transcribed; gene sequence; contains critical info 1. Proks – have only one type of RNA polymerase; all have same binding strength and detail 2. Euks – have more than one RNA polymerase a. RNA polymerase II assembles mRNA b. RNA polymerase I and III assemble RNA that will not be made into protein – functional RNA (tRNA, rRNA) Terminator – sequence of DNA that trigger the end of transcription in prokaryotes o Downstream o More complex in eukaryotes

Answer 37

1. Initiation a. Promoter – contains the transcription starting point; RNA polymerase binds and begins transcription i. RNA polymerase must bind at a precise location and orientation ii. Determines which of the two strands will serve as the template – only 1 of 2 strands will be what is actually transcribed b. In prokaryotes – RNA polymerase binds to the promoter sequence directly c. In eukaryotes – transcription factors mediate the binding of RNA polymerase and the beginning of transcription i. Transcription initiation complex – many transcription factors & the RNA polymerase - Transcription factors must be in position on promoter for RNA polymerase II (mRNA) to bind - TATA box – important in forming the initiation complex; knows where everything needs to line up ii. Once RNA polymerase binds the DNA – unwinds and transcription begins 2. Elongation a. RNA polymerase – moves along DNA template; linearizes, opens, joins nucleotides with phosphodiesterase bonds i. Bonds types - Between RNA and DNA – hydrogen bonds - Within RNA and DNA – covalent phosphodiesterase bonds b. 10-20 DNA nucleotides of the template are exposed at a time i. Nucleotides are added to the 3’ end of the RNA segment (pic of carbon primes) c. Hydrogen bonds break between DNA and new RNA – breaking bonds & anabolic; lots of energy required i. DNA the double helix reforms d. Euks – 40 nucleotides/second are transcribed i. Many RNA polymerase enzymes can work on DNA template – increases the number of mRNA transcripts produced (therefore more proteins) 3. Termination: a. Prokaryotes i. Terminator sequence in DNA – message that indicates the end of the gene; causes the RNA polymerase to detach after its been transcribed b. Eukaryotes i. RNA polymerase II transcribes a polyadenylation signal on the DNA template - AAUAAA - Occurs in pre-mRNA – mRNA that has not yet been processed ii. 10-35 nucleotides after the transcription of polyadenylation signal proteins will cut the mRNA free from the polymerase - Further processing will occur after the pre-mRNA is released – mature piece is trimmed of some excess components

Answer 38

Pre-mRNA is modified further in the eukaryotic nucleus RNA polymerase II transcribes entire DNA gene into pre mRNA o Modifications are made to the 5’ end, 3’ end, and interior prior to exportation into cytoplasm Modifications 1. Caps – added to the ends of RNA; allows recognition of the mRNA as endogenous a. 5’end – 5’ cap; modified guanine nucleotide • Guanine (base) with a ribose sugar – 3 phosphate groups between it and the rest of the RNA (nucleotides normally have 1 phosphate group) b. 3’ end – poly-adenine tail (poly A) - 50-250 adenine nucleotides will be added 2. Splicing – removes internal segments of the mRNA & remaining pieces are joined together; segment will shorten; occurs in the nucleus a. ~27,000 nucleotides in the primary mRNA (pre mRNA) • 1200 in mature mRNA – 400 amino acids in the protein product (3 nucleotides per codon) b. Introns – non coding segments that are removed prior to exportation into cytoplasm • Short nucleotide sequences at the start of each intron dictate where splicing should occur – enzymes will recognize • Nucleases – degrade the removed introns in nucleus c. Exons – coding segments; joined together & will be translated Splicing occurs via 1. Spliceosomes – snRNP and protein complex a. snRNP – small nuclear ribonucleoproteins; ~150 nucleotides o recognize splicing sites & catalyze intron removal o located in the euk nucleus b. attach at both sides of intron and remove c. RNA of spliceosome will be complimentary to mRNA 2. ribozymes – RNA with enzyme function (not all enzymes are proteins) a. intron RNA can catalyze its own removal – does not require protein or snRNPs o certain RNA bases are used in catalysis b. ssRNA (single stranded) can fold back on itself according to complimentary sequence – creates 3D structure o hydrogen bonding allows the active site of catalytic RNA to bind to complimentary portion of intron and remove mRNA is exported from nucleus to cytoplasm for translation 1. Modifications function to a. Prevent cytoplasmic degradation of the mRNA (endogenous) b. Allow the ribosome to recognize the mRNA segment and attach 2. mRNA contains 5’ & 3’ untranslated regions (UTRs) – assist in ribosome binding; not translated into protein 1 gene allows the production of 2 or more polypeptides – substitutions give multiple end products 1. Alternative splicing – removal of different exons creates protein variants from one gene a. Different from splicing o Splicing – intron removed o Alternative splicing – exon removal b. Likely responsible for different domains of a particular protein o Quaternary structure – subunits of protein can be created from alternative splicing and produce variations in final product o Different exons give rise to different domains 2. Allows fewer genes than would otherwise be required – number of proteins we can produce far outnumbers the number of genes that we encode

Answer 39

The synthesis of a polypeptide from mRNA transcript Ribosomes – facilitates translation & catalyzes peptide bonds between AA 1. Similar schematic in bacteria and eukaryotes a. Archean – not well understood • Often live in extreme conditions which create modifications b. Prokaryotes – lack nuclear membrane • Transcriptions & translation occur in the same environment – translation can begin to occur before transcription is finished c. Eukaryotes – exportation of mRNA into cytoplasm is critical for translation i. Subunits of euks o Large – 60s o Small – 40s

Answer 40

is universal; all living organisms use the same codons for AA Codons – nucleobase triplets that code for specific amino acids; smallest unit of uniform length that can encode all 20 AA 1. DNA codons is transcribed to RNA a. Only 1 or 2 DNA strands serves as template – other genes on the same DNA strand will use the opposite strand as a template • Coding strand of the DNA is the non-template strand – will match the transcribed RNA (with uracil and ribose sugar instead) b. mRNA is complimentary & antiparallel to DNA template • assembled according to base pairing rules 2. mRNA codon is translated to protein a. ribosomes read mRNA 5’ to 3’ – codons are arranged 5’ to 3’ 61/64 codons – encode amino acids 1. 3/64 – stop codons 2. AUG – universal start codon; encodes methionine AA a. Not every protein starts with AUG – most do 3. Redundant genetic code – 61 codons code for 20 AA a. Usually differ in the last nucleotide of the codon b. One codon never produces a different AA Reading frame – vital for AA sequence and protein function

Answer 41

single stranded folded RNA; translate codons of mRNA Transcription of tRNA o Occurs in the euk nucleus or prok cytoplasm according to the DNA template Function to deliver the amino acid from cytoplasm to polypeptide of ribosome a. Ribosomes add amino acids delivered by the tRNA molecules - Enzyme attaches AA to the mRNA – very specific b. Cytoplasm has a supply of the 20 amino acids – either synthesized or taken up from ECF - 9 essential – you need to get them from your diet - 11 nonessential – you can make them c. Recyclable – will collect another AA from cytoplasm after loosing initial Structure of tRNA differs depending on AA it binds – 1 tRNA with specific anticodon will translate codon of mRNA to a specific AA a. Single strand is ~80 nucleotides – hydrogen bonding within strand holds the structure tRNA (folded sides will be complimentary) - Exception – UG can form as a base pair b. Carries the AA on the 3’ end Anticodon – nucleotide sequence present on the anticodon loop of the tRNA o Complimentarily base pairs with the appropriate mRNA codon – are antiparallel (be mindful of 3’ and 5’ ends of tRNA) Amino acyl tRNA synthetases – enzyme that covalently attach AA to rRNA a. Highly specific – tRNA synthetase active site only recognizes one AA and corresponding tRNAs that code for it i. Always attaches to 3’ end of tRNA to match the anticodon of tRNA and codon of mRNA - Pair off tRNA based off codon in mRNA ii. 20 different tRNA synthetases exist -> 1 per amino acid b. Energy requiring – 1 ATP (equivalent) per attachment - ‘tase’ – indicates energy needed c. Released from active site to be used in translation The Wobble Effect – 45 tRNA molecules (45 anticodons) match with 61 codons; the third position of anticodon has flux with bases it can interact with 1. Unusual base pairs – 1 anticodon can recognize more than 1 codon a. Ex: tRNA anticodon: 3’ UCU 5’ is able to base pair with 5’ AGA 3’ or 5’ AGG 3’ codons b. Only third position by 5’ end of anticodon and third position by 3’ end of codon – watch out for this 2. Allows a single tRNA to recognize more than one codon when differences exist only in the third nucleotide position

Answer 42

Quantity depends on needs of the cell o There are 1000s present in the cell o tRNA is the most abundant RNA material Structure 1. Ribosomes consist of proteins and: a. 3 rRNA in prokaryotes (70s) b. 4 rRNA in eukaryotes (80s) 2. Synthesis of subunits occurs in the nucleolus a. Assembled in nucleolus – rRNA and proteins are processed and assembled into units • ribosomes will build ribosomal protein – imported into nucleus • DNA will be transcribed into rRNA b. Ribosomal subunits are exported via nuclear pores • Subunits are separate when not reading mRNA – assembly occurs when mRNA binds to small subunit Will have binding sites for: 1. mRNA binding site 2. 3 tRNA binding sites - P site – holds tRNA carrying the growing polypeptide chain - A site – holds tRNA carrying the next amino acid to be added to the chain - E site – holds the naked tRNA exiting from the ribosome

Answer 43

1. Initiation a. Brings together – mRNA, tRNA (usually met/AUG), and 2 ribosomal subunits i. Small ribosomal subunit binds mRNA and initiator tRNA (methionine) b. Bacteria i. Complimentary sequence in mRNA and rRNA of the small ribosomal subunit – adhere to one another & aligns reading frame on ribosome - Important in aligning reading frame c. Euks i. Formation of translation initiation complex – requires 1 GTP; assisted by initiation factor proteins - The small ribosomal subunit is already charged with initiator tRNA - Small subunit binds to 5’ cap of mRNA & scans mRNA until the AUG is located - Anticodon loop of tRNA hydrogen bonds to AUG in the mRNA sequence – completes translation initiation complex ii. Initiator tRNA is located in the P site, A site is vacant - Polypeptide assembly occurs from N terminus (NH3+) to C terminus (COO-) – new AA are added to C terminus 2. Elongation – AA are added to C terminus; requires protein elongation factors a. Energy is required for step 1 & 3 b. 3 repetitive steps occur until synthesis is complete i. Step 1 – new tRNA matches anticodon to codon and comes into A site - Codon recognition consumes 1 GTP and increases accuracy ii. Step 2 – the entire peptide chain is transferred to covalently bond to new tRNA in A site iii. Step 3 – translocation 1. Everything shuffles down A to P to E a. A site is empty for new tRNA to enter b. P has new chain – moved from A site c. E has naked tRNA – will fall off 2. Moving requires 1 GTP c. mRNA moves through the ribosome unidirectionally – read 5’ to 3’ end i. mRNA and ribosome move relative to one another 3. Termination a. Stop codon in mRNA enters the A site of the ribosome – signal translation to stop i. 3 stop codons: UAA, UAG, UGA b. Requires a release factor shaped like an amino acyl tRNA i. Binds to the stop codon in the A site & causes the addition of water ii. Hydrolyses covalent bond between polypeptide and tRNA in the P site – releases polypeptide c. Disassembly piece by piece – mRNA, subunits, release factor i. Requires protein factors & 2 GTP d. GTP requirements i. Initiation = 1 GTP ii. Addition of each subsequent AA = 2 GTP iii. Disassembly = 2 GTP - Ex. peptide 16 AA long = 1 + (2 x 15) + 2 = 33 GTP

Answer 44

multiple ribosomes may be bound to a single mRNA a. One ribosome – average polypeptide takes less than 1 min to assemble b. Multiple ribosomes speed the process – allows synthesis of many proteins very quickly o Successive ribosomes will be at successive states of synthesis o Can be seen with an electron micrograph c. Polysomes – many ribosomes accomplishing protein synthesis o Especially common in muscle cells

Answer 45

Primary, secondary, and tertiary structures a. Translation – produces a primary amino acid sequence (non-functional) b. Post translational modifications – allow folding into secondary and tertiary structure - Secondary structure – can begin during translation; will mostly occur during synthesis - Tertiary structure – will likely occur later; it needs more of the polypeptide to be present Gene sequence dictates primary protein structure -> thereby dictates protein shape (folding based on ionic and hyd bonds) o Chaperone proteins – often assist in protein folding which requires high degree of accuracy; helps proper parts of protein to come together Post translational modifications a. Addition of sugars, lipids, phosphates b. Enzyme cleavage of an amino acid from N terminus or C terminus of polypeptide c. Proteolytic activation – enzyme cleavage of the protein into 2 pieces - May form bonds with one another reassembling - May bind to another part of initial polypeptide and reattaches Formation of quaternary structure – assembly of 2+ polypeptides in tertiary structure forming protein (ex hemoglobin)

Answer 46

Location of synthesis determines location of final protein o Free ribosomes – destined to stay in the cytoplasm o RER ribosomes – exported from the cell or enter into the endomembrane system Ribosomes are structurally indistinguishable – free and bound can function as both free or bound o Cytoplasmic ribosomes always begin the synthesis of a polypeptide If destined for excretion or endomembranous system – polypeptide will contain a signal peptide a. Signal peptide – ~20 amino acid sequence at N terminus of polypeptide - Signal recognition peptide (SRP) – protein-RNA complex that recognizes signal peptide & escorts the entire ribosome, mRNA, polypeptide to the RER - Will detach once ribosome docks on RER b. Ribosome docks onto receptor on RER – polypeptide is fed through pore in RER into lumen - Signal peptide is cleaved from polypeptide - Remaining polypeptide is released into lumen of RER Other types of signal peptides o Traffic protein to the mitochondria/chloroplast o Protein will be synthesized on cytoplasmic ribosomes – proteins are sent directly to organelles

Answer 47

Always occur in DNA – changes in genetic sequences form the source of new genes Large scale mutations occur due to chromosomal rearrangement Point mutation – a change in a single nucleotide pair of a gene a. Ex. 5’ TTG ATT 3’ to 5’ TAG ATT 3’ b. Severity depends on where it occurs - Somatic cell – will only be based to daughter cell by mitosis - Gamete – will be passed along to future generations; much more serious c. Can create adverse affects & genetic disorders (ex. sickle cell disease) Mutations that affect protein function – always case by case (mutations may or may not have an affect) a. If mutation occurs in the active site of the enzyme – not likely to work properly b. May enhance protein function – may cause bacterial to be more or less transmissible c. May create non-functional or less active proteins d. May cause the insertion of a premature stop codon – polypeptide is too short - Translation stops before the protein is complete - Creates a truncated protein – almost always will be non-functional - Also always case by case

Answer 48

1) Substitutions – replaces a nucleotide in a base pair with a different one (DNA is double stranded) a. Silent mutation – change does not affect expression of gene; same AA is incorporated i. Redundant genetic code – some substitutions will not change the amino acid because multiple codons encode for the same AA ii. Ex: 3’ CCG 5’ -> 3’ CCA 5’ leads to mRNA 5’ GGC 3’ -> 5’ GGU 3’ - Both codons call for glycine amino acid b. Missense mutation – substitutions that change one amino acid to another i. May not have a noticeable effect – “personalities” of AA can be similar - Substitutes may be chemically similar to the original - Ex. both are non polar or both are changed - May occur in a region of the protein that is not important for folding or function (even if they are chemically different) 2) Insertion/deletion mutations – a nucleotide pair is added or lost from gene (removed, not only changed) a. Frameshift mutation – alters the reading frame downstream; everything will shift by 1 and it will affect all codons - More likely to significantly affect the protein

Answer 49

Can occur spontaneously as a result of DNA polymerase error during replication – usually corrected by proof-reading/repair systems a. If mutation persists – it will be passed along to daughter cells during cell division - Can become cancerous b. Responsible for 1:1010 mutations Mutagens – Physical and chemical agents that interact with DNA and cause mutation a. Mutagenic agents – has potential to cause mutation - High energy radiation (ex. X-rays/UV light) b. Chemical mutagens – bind to DNA and distort the structure - Nucleotide analogs form – result in incorrect base pairing c. Carcinogens – often mutagenic because they impact your DNA; leads to lack of regulation - Most mutagens are carcinogenic

Answer 50

Bacterial cells exhibit quorum sensing – secrete small signaling molecules that allow them to detect changes in local cellular density a. Concentration of signaling molecule will increase when there are more bacteria present b. Allows coordination of bacterial activities • Some activities will only be carried out when the bacterial population is of a sufficient size c. Allows for the formation of a biofilm – unicellular bacteria come together and form a community adhering to a surface • Better able to handle adversity and unfavorable conditions

Answer 51

Local signaling 1. Paracrine signals – local regulators & messengers secreted by the cell into immediate vicinity • Many types • Ex. growth factors – stimulates the growth of nearby cells 2. Gap junctions – allow cells in direct contact with one another to communicate; fusion of cytoplasm Long distance signaling 1. Synaptic Signaling – occurs in the animal nervous system a. Neuron – a cell of the NS i. NMJ – releases Ach into synapse; Ach binds to nAchR and initiates contraction of a muscle cell - Botox – prevents Ach from being released & muscle will not contract - Toxins can interfere with release of Ach as well – can affect diagram and respiratory system - Can signal glands to secrete as well b. AP propagated down the axon – triggers release of nt 2. Chemical signal – molecules diffuse across the synapse & triggers response in target cell by binding to receptors a. Endocrine system • Hormones – used for long distance signaling; travel through the bloodstream and bind to cells with receptors i. Travel quite quickly in the blood ii. Vary in terms of size and type - Polar vs nonpolar – dictates where receptor is located in/on cell for binding b. Only cells with receptors for the hormone will be affected – will recognize and response - Ex. epinephrine will only bind to cells with receptors and illicit responses

Answer 52

1) Reception – receptor and hormone/chemical engage a. Ligand – binds to receptor i. Lock and key fit – very specific with specific response ii. Ex. epinephrine – water soluble hormone released during fight or flight response b. Binding of ligand i. Causes conformational change in receptor – often within the cell ii. Can cause aggregation of receptors c. Polarity dictates location of receptor i. Water soluble – ligands bind to receptors located on the cell surface - A faster response ii. Hydrophobic – ligands bind to receptors located in cytoplasm or nucleus - Slower response – complex will normally affect transcription 2) Transduction – required for extracellular signals a. Multi-step activation – binding of receptor causes a cascade of responses in subsequent molecules i. Amplifies signal – activates a large number of molecules ii. More opportunity for coordination and regulation – more fine tuning iii. Fast starting and stopping of response b. Steps i. Ligand binds to the receptor – activates an intracellular molecule ii. Activates subsequent molecules – generally proteins c. The signal usually enters into the cell as a change in the shape of the receptor i. Often occurs via phosphorylation – phosphates are used often to create conformational changes because of their neg charge (will create either attraction or repulsion) 3) Response – occurs in the nucleus or cytoplasm a. Usually affects gene transcription – regulation of protein synthesis b. Activated transcription factors turn genes on i. mRNA synthesis is activated ii. Translation in the cytoplasm will occur c. One transcription factor i. May serve to regulate a number of different genes ii. May also serve to turn gene transcription off

Answer 53

Plasma membrane receptors – integral transmembrane proteins; bind to hydrophilic ligands a. Malfunction of these receptors are very important in human disease o DNA mutation – affect protein synthesis & cause malfunctions of receptors These receptors account for ~30% of all human proteins Three types of receptors: 1) G-Protein Coupled Receptors: a. G protein – loosely attached on cytoplasmic side of the membrane i. Activated – binds GTP ii. Inactivated – binds GDP b. Ligand binds to receptors & changes shape i. Receptor associates with G protein bound to GDP - G protein switches GDP for GTP – activates ii. G protein diffuses along membrane – activates adenylyl cyclase - Adenylyl cyclase – amplifier enzyme; elicits cellular response c. G protein also functions as a GTPase i. Hydrolyzing GTP to GDP – results in dissociation of G protein from adenylyl cyclase - GTP -> GDP releases energy - G protein is inactivated ii. Adenylyl cyclase enzyme is returned back to its basal inactive state iii. Allows the pathway to be turned on or off as required 2) Receptor Tyrosine Kinases – receptor with enzymatic activity a. Cytoplasmic tail functions as a tyrosine kinase (phosphorylating enzyme) – transfers phosphate group from ATP to tyrosine residue on the substrate i. Able to activate many proteins at once – through phosphorylation b. Activation i. Receptors are present as monomers before the ligand binds - Binding of ligand causes dimerization of receptors – forms a dimer & activates tyrosine tails of each monomer ii. Each activated tail phosphorylates the tail of the other receptor from ATP - Each phosphorylates the other 3x – 6 ATP required - Kinase – enzyme catalyzes c. Full receptor activation is recognized by relay proteins in the cell i. Proteins binds phosphorylated tyrosine at binding site – activates proteins via structural changes ii. The activated protein creates a cellular response 3) Ion Channel Receptors a. Binding of ligand to receptor channel regulates opening and closing of the gate – allows the passive flow of ions through when open i. Facilitates specificity in type of molecules that can cross ii. Rapid changes in ion concentration – basis of electrical activity iii. May directly affect cell activity b. Closed gate – maintains charge separation

Answer 54

may be located in the cytoplasm or the nucleus Small hydrophobic messenger must be able to diffuse through the plasma membrane a. Ex. Steroids, thyroid hormone, nitric oxide (NO) i. NO – small and e- are equally shared; causes vasodilation b. Ex: testosterone enters into the cell Activated receptor moves into nucleus a. Acts as a transcription factor – turns on the transcription of certain genes - Affects mRNA production and protein synthesis - Slower process than receptors on the cell membrane Lipid soluble hormones are only synthesized on demand because you can’t sequester it within cell – it can leave the cell when you’ve made it because it’s hydrophobic

Answer 55

acts as quick and convenient on/off switch Enzymes involved 1. Kinases – enzymes that facilitate transfer of phosphate group from ATP to molecule a. Protein + ATP -> phosphorylated protein + ADP b. 2% of all human genes are dedicated to kinase synthesis – very important 2. Phosphatases – enzymes that remove phosphates; usually inactivates protein Phosphorylation – can either activate (most often) or inactivate protein a. Phosphorylated molecules – almost always acts to phosphorylate a different protein; they themselves are kinases - Specifically on a serine or a threonine - Both have OH group present -> H is removed and phosphate group is attached

Answer 56

water soluble and non protein; able to spread through cell via diffusion cAMP – cyclic AMP Epinephrine binds to plasma membrane receptor – a GPCR 1. Ligand causes change in shape of receptor – transfers signal to G protein • G protein – exchanges GTP for GDP • Subunits of G protein (alpha, beta, gamma) diffuse throughout membrane and activates adenylyl cyclase – amplifier enzyme 2. Adenylyl cyclase converts ATP -> cAMP • Cuts of 2 phosphates (pyrophosphate) • Creates a cyclical molecule – phosphate still attached is attached to the 5’ ad 3’ of RNA • Very potent activator -> 20x increase in cytoplasmic cAMP 3. cAMP activates kinase A – phosphorylates cellular proteins and elicits cellular response 4. cAMP is degraded in order to stop response • cell is running at ‘high state’ when cAMP is present – requires a lot of energy • quickly broken down via phosphodiesterase when no longer needed

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