DNA Replication

DNA Replication in Eukaryotes Telomere Replication - As you have learned, the DNA polymerase can add nucleotides in only one direction. In the leading strand, synthesis continues until the end of the chromosome is reached. However, on the lagging strand, once the end of the chromosome is reached there is no place for an RNA primer to be added. This presents a problem for the cell because the ends remain unpaired , and over time these ends get progressively shorter as cells continue to divide. The ends of the linear chromosomes are known as telomeres. Telomeres have repetitive sequences that do not code for a gene. They are important because they prevent chromosomes from arbitrarily fusing and protect the DNA from becoming damaged. - It is the telomeres that are shortened with each round of DNA replication instead of genes. For example, in humans, a six-base-pair sequence, TTAGGG, is repeated 100 to 1000 times. The discovery of the enzyme telomerase helped explain how chromosome ends are maintained. The telomerase carries its own RNA primer which can base pair to the end of the DNA strand. The telomerase can then add DNA nucleotides to the end of the chromosome, elongating it. Once the template strand is sufficiently elongated, DNA polymerase can then add nucleotides that are complementary to the ends of the chromosome. Thus, the ends of the chromosomes are maintained in germline cells, adult stem cells, and some cancer cells. - Telomerase is not active in adult somatic cells. Adult somatic cells that undergo cell division continue to have their telomeres shortened. This essentially means that telomere shortening is associated with aging. High fidelity of DNA replication requires several proofreading mechanisms - The replication error rate is ~1 in 10 billion nucleotides added - The error rate is much lower than simply afforded by base-pair complementarity—additional mechanisms are in place - Proofreading and mismatch repair DNA polymerase 3’ to 5’ exonucleolytic proofreading Exonucleolytic : degrading nucleotide from the outside - DNA pol. Can recognize and correct a wrong base-pairing with its 3’ to 5’ exonuclease activity - Exo : from the outside, from the end - Nuclease : degrades nucleotide strands - It’s a “self-correcting” enzyme (1 in 100000 error rate) DNA mutations vs DNA damage - Mutations : wrong sequence or base pair - From errors of replication or erroneous repair of damage - If left unfixed, it would lead to a change in DNA sequence in one of the daughter cells: a mutation - DNA is structurally normal , so cannot easily detect and repair after replication - DNA damage : presence of abnormal DNA structures - If left unrepaired, DNA damage will lead to mutations - DNA repair solves DNA damage to prevent mutations - DNA mutations cannot be recognized after replication DNA damage Depurination : loss of G or A base Deamination : removal of an amino group (ex. C to U) Thymine dimers : photolesion produced by UV radiation Base excision repair (BER) - DNA glycosylases recognize altered bases and remove them from the DNA backbone (hydrolysis) - Endonucleases remove the remaining sugar-phosphate - Repair DNA polymerase + DNA ligase fill/seal the gap using the other strand as a template Nucleotide excision repair (NER) - Excision nuclease cuts around the damage - Helicase removes the cut single strand - Repair DNA polymerase + DNA ligase fill/seal the gap using the other strand as template Replication fidelity - Fidelity must be very high BUT having some mutations is good (variation, natural selection, evolution) - Fidelity is achieved by immediate proofreading + damage repair - Strand complementarity provides built-in redundancy for repair Central Dogma DNA - Much more stable to maintain genetic information - Less difficult to degrade - Less reactive (because the 2’ is lacking an O) RNA - Less stable to maintain genetic information - is more susceptible to degrading enzymes - much more reactive (because of the OH group in 2’) Transcription: produce RNA from the DNA template Both prokaryotes and eukaryotes perform fundamentally the same process of transcription, with one very important difference. In eukaryotes, transcription occurs in the membrane-bound nucleus. In prokaryotes, transcription occurs in the nucleoid region; recall that prokaryotes lack membrane-bound organelles. Once the mRNA is formed in eukaryotic cells it must be transported to the cytoplasm. Because the mRNA of prokaryotes does not need to be transported anywhere, translation can immediately follow. In both prokaryotes and eukaryotes, transcription occurs in three main stages: initiation, elongation, and termination. INITIATION : Transcription requires a small part of the DNA double helix to partially unwind. The DNA must unwind to allow enzymes and additional proteins to access specific genes which will then be used to make mRNA. The region of the DNA that is unwound is called the transcription bubble Several proteins and enzymes bind to a region at the beginning of the gene called a promoter , a particular sequence of DNA nucleotides that trigger the start of transcription. In most cases, promoters exist upstream, or in front of, the genes they regulate. The specific sequence of a promoter is very important because it determines whether the corresponding gene is transcribed all of the time, some of the time, or hardly at all. ELONGATION : Transcription always proceeds from one of the two DNA strands, which is called the template strand. The mRNA is complementary to the template strand and is almost identical to the other DNA strand, called the non-template strand. The two big exceptions are that RNA nucleotides contain the sugar ribose while DNA nucleotides contain the sugar deoxyribose and that RNA contains the nitrogenous base uracil (U) instead of the thymine (T) found in DNA. During elongation, an enzyme called RNA polymerase proceeds along the DNA template adding RNA nucleotides by base pairing with the DNA template in a manner like DNA replication. As elongation proceeds, the DNA is continuously unwound ahead of the enzyme and then rewound behind it. TERMINATION : When the polymerase has reached the end of the gene, the RNA polymerase needs to be instructed to dissociate, or separate, from the DNA template strand. Once the RNA polymerase dissociates, the newly made mRNA transcript is released. Depending on the gene being transcribed, there are two kinds of termination signals, but both involve repeated nucleotide sequences in the DNA template. These repeated sequences cause the RNA polymerase to stall, separate from the DNA template, and free the newly synthesized mRNA. At the end of the termination, the process of transcription is complete. In a prokaryotic cell, by the time termination occurs, the mRNA is already being used to synthesize numerous copies of the encoded protein. This is possible because prokaryotic cells do not have their DNA enclosed in membrane-bound nuclei. As soon as the mRNA is partially synthesized, ribosomes attach and begin generating the protein. Because of their nucleus, this is not possible for eukaryotic cells. Once the mRNA has been synthesized and undergoes modifications it must first be moved out of the nucleus and into the cytoplasm before translation can begin. This prevents simultaneous transcription and translation in eukaryotic cells. Eukaryotic RNA Processing - The newly transcribed eukaryotic mRNAs are referred to as primary transcripts. These primary transcripts must undergo several processing steps before they can be transferred from the nucleus to the cytoplasm and then translated into a protein. The additional steps involved in eukaryotic mRNA maturation create a molecule that is much more stable than a prokaryotic mRNA. For example, eukaryotic mRNAs last for several hours, whereas the typical prokaryotic mRNA lasts no more than five seconds. - The mRNA transcript is first coated in RNA-stabilizing proteins to prevent it from degrading while it is processed and exported out of the nucleus. This occurs while the mRNA transcript is still being synthesized and involves adding a special nucleotide “cap” to the 5' end of the growing transcript. In addition to preventing degradation, factors involved in protein synthesis recognize the cap to help initiate translation by ribosomes. - Once elongation is complete, an enzyme then adds a string of approximately 200 adenine nucleotides to the 3' end, called the poly-A tail . This modification further protects the mRNA transcript from degradation and signals that the mRNA transcript is ready to be exported to the cytoplasm. - Eukaryotic DNA, and thus complementary mRNA, contains long non-coding regions that do not code for amino acids. Their function is still not well understood, but the process called splicing removes these non-coding regions, called introns , from the mRNA transcript. The non-coding regions are called introns because they are intervening sequences. The coding regions are called exons ; exon signifies that they are expressed. - A spliceosome , a structure composed of various proteins and other molecules, attaches to that mRNA transcript and “splices” or cuts out the non-coding , introns . The remaining exons are pasted together to form the mature mRNA which will then be transported to the cytoplasm. - Some of the segments that are removed from mRNA during splicing are not always non-coding. When different coding regions of mRNA are alternatively spliced out, different variations of the protein will result, with differences in structure and function. This process results in a much larger variety of possible proteins and protein functions from a given genome. Humans, for example, have just over 20,000 genes, yet the human body produces over 80,000 different proteins. TRANSLATION Silent Mutation : the new codon codes for the same amino acid (redundancy). Though we had a mutation, the resulting amino acid is still the same. Missense Mutation : the new codon codes for a different amino acid (changes the meaning/sense) Nonsense Mutation : the new codon codes for a stop codon (there is no amino acid, no meaning/sense). End of translation. AN RNA sequence can in principle be translated into three different reading frames - 3 reading frames, based on where the decoding process begins - This would result in 3 very different proteins - The cell must be able to determine the correct frame for translation initiation - The deletion could cause encodings or subsequent amino acids to be completely different from what they should have been - Choosing a reading frame is crucial for the organism tRNA: adaptors between codons and amino acids - RNA-based adaptor - tRNA binds a specific (“cognate”) amino acid on one end - has an anti-codon on another end - Anti-codon base pairs with the codon on the mRNA, placing the aa in line for polymerization tRNA synthetases: adaptors between tRNAs and their amino acids - For the code to work, crucial step is to properly attach tRNA to its amino acid - Aminoacyl-tRNA synthetases covalently attach the correct amino acid to the end of each tRNA - Tri-partite system : enzyme + tRNA + aa 1. The enzyme activates the aa by covalently attaching AMP (from ATP, releasing PPi) 2. The enzyme covalently attaches the aa to the tRNA (and releases AMP) - Translation has 2 adaptor steps: - tRNA synthetases (tRNA, aa) - tRNA (aa, codon) - an error in either adaptor would cause the wrong aa to be incorporated into the protein) - tRNA synthetases have proof-reading capabilities Polypeptide formation (1): Chemical bond - Protein chain (polypeptide) grows by stepwise addition of each amino acid (polymer) - Amino acids are linked by peptide bonds between amino and carboxyl ends of aa - Peptides grow from the N to the C terminus Polypeptide formation (2): charged tRNAs - The aa and growing polypeptides are not “floating”, but charged on tRNAs - The N terminus of aa on tRNA n+1 attacks the C terminus of the growing polypeptide on tRNA n - Peptide bond is formed, tRNA n is released and the extended polypeptide is now attached to the tRNA n+1 Polypeptide formation (3): ribosomes (translation machinery) - The charged tRNAs that undergo peptide bond formation are not “floating” either: peptide formation occurs in the ribosome (in the cytoplasm) - Ribosomes are large molecular assemblies of multiple proteins and ribosomal RNA (rRNA) - Overall structure: “small subunit” and “large subunit” - Small subunit : responsible for binding directly to the mRNA; for codon and anti-codon pairing - Large subunit : sequentially binds transfer RNAs; for peptide bond formation - tRNA : is a type of RNA molecule that brings amino acids to the growing polypeptide chain - 4 sites for RNA binding - 1 for mRNA - 3 for tRNA (A, P, E sites) - These sites are mostly formed by rRNA - Peptidyl transferase activity also catalyzed by rRNA - A (aminoacyl site) : tRNA n+1 with incoming aa; where new incoming tRNA will come in - P (peptidyl site) : tRNA n with growing chain; where the tRNA that is bound to the growing peptide chain will bind - E (exit) site : emptied tRNA n-1 ; where the tRNA moves after the reaction has occurred - Ribosome catalyzes bond formation and moves over to the next codon - Peptide bond formation (peptidyl transfer) is happening in the ribosome - It is catalyzed by the rRNA in the large subunit - Efficient and accurate translation is aided by elongation factors (GTP hydrolysis) The ribosome is a ribozyme! - Catalytic activity for peptide bond formation is provided by the rRNA, not the protein subunits - Bacterial ~20 aa/s; eukaryotic ~4 aa/s Translation initiation in prokaryotes - Initiator tRNA-Met (tRNAi-Met) is loaded in the P site (initiation factors involved) - A small subunit is recruited to mRNA and scans for the Shine-Dalgarno sequence (will guide the small subunit to the start codon) - Large subunit then recruited - The initiator tRNA is the only tRNA that can go directly into the P site . Otherwise, every other amino acid goes into the A site . - Small subunit recognizes “ Shine-Dalgarno sequence ” (AGGAGGU) upstream of AUG - Recognized by base-pairing of 16S rRNA - This positions the small subunit/ribosome close to the start codon so translation can begin - No mRNA cap (important for recognition in eukaryotes) Translation initiation in eukaryotes - Similar but more complex - Initiator tRNA loads up with methionine - Eukaryotic initiation factor2 (eIF2) rings tRNA i -Met to P site - Initiation factors bound to mRNA’s cap and polyA tail bring mRNA to small subunit - The 5’ caps play a crucial role in recruiting and positioning the small subunit - Small subunit with tRNA i scans for the first AUG - AUG must be in the context of a consensus sequence (“Marilyn Kozak sequence ”: ACCAUG G) for efficient recognition - Small subunit rRNA base pairs with Kozak sequence and stops there - Large subunit is recruited-full ribosome is in place - First peptide bond between aa in A an P sites - Translation proceeds Multiple SD sequences (and the lack of a 5’ cap) allow for translation of polycistronic mRNAs - Polycistronic : a single mRNA can contain the information it can encode for multiple proteins - Ribosomes must know where to start, but also where to stop - Proteins on the operon might be encoded on different reading frames

Termination - End of encoded protein signaled by stop codons : UAA, UAG, UGA - Not recognized by a tRNAs - Instead, recognized by released factors - Release factor binds to stop in A site - Then moves to the P site (GTP) - Catalysis with H 2 O, releasing peptide from tRNA in E site - Ribosome-released mRNA separates Proteins are made on polyribosomes - Multiple ribosomes can latch on to a single mRNA at the same tie - >80 nt apart - Leads to faster protein production - In humans, actively translated mRNAs have average 10-2- ribosomes 1. Replication of linear presents a problem that is not experienced by circular chromosomes. This problem is solved by telomerase. What’s the problem experienced by linear chromosomes? - Replication cannot be completed at one end of each daughter chromatid 2.Telomerase has an enzymatic activity. What does it catalyze? - The extension of a DNA polymer, using an RNA template 3. Nucleotide excision repair results in error-free repair of UV-induced damage in humans. How does this repair process manage to remove damaged DNA and replace it with undamaged DNA with the same sequence? - It employs the undamaged strand as a template for new synthesis 4. Cytosine can spontaneously deaminate to form uracil. This is a potentially mutagenic form of DNA damage, because U base pairs with A, while the original base, C, base pairs with G. To prevent mutagenesis, the cell produces an enzyme called uracil glycosylase, which catalyzes the first step in an excision repair pathway to remove and replace the damaged base. However, no similar enzyme exists to recognize and remove uracil from RNA. Why is this? - U’s are supposed to be in RNA, it would be bad to remove them 5. DNA double-strand breaks are intentionally generated by the cell during _____ and are repaired ____ via _____. - Meiosis, homologous recombination 6. A human gene encodes insulin. You would like to express this gene in the bacterium E. coli. You replace the human promoter sequence with an E coli promoter sequence. What else needs to be changed to get an expression of this protein in E. coli? - Any intron sequences will have to be removed, if present & You need to add a Shine-Dalgarno sequence just 5’ of the AUG 7. Errors in transcription are not as potentially harmful as errors in replication because: - RNA is not copied into more RNA molecules & RNA strands have a relatively short life span 8. An operon encoding an anabolic (synthetic, rather than degradative) pathway is generally regulated as follows: - The small molecule that is the product of the entire pathway represses transcription of the operon 9. A 2-nucleotide deletion mutation is generated in the center of a eukaryotic intron sequence. Such a mutation would cause… - No change in the protein sequence 10. The histidine biosynthetic operon would be expected to be regulated as follows: - Lots of histidine -> repressor -> histidine biosynthetic genes - Lots of histidine – I activator -> histidine biosynthetic genes - Either one would work 11. What is the biological function of the lactose regulatory system? - It prevents the cell from wasting energy making lactose-degrading enzymes when lactose is not present, or a better option is available 12. A signal transduction pathway is illustrated in Figure below. It outlines the series of stimuli and regulatory processes that affect a student’s decision to go to the movies and/or eat at a Thai restaurant. In the presence of the signal (=boredom), what is the student’s response? - Go to the movies, but no Thai food 13. A mutant is isolated with a nonsense mutation in the 5’ end of the coding region of the gene encoding the lac repressor. Such a mutant would be expected to: - Strongly express the lac operon only in the presence of high concentrations of cAMP 14. Which of the following can be used to increase a protein’s activity quickly within seconds? - The reversible binding of a small molecule - Phosphorylation of the protein 15. The coding region for the lacZYA operon is fused to a strong, constitutively expressed eukaryotic promoter (the operon’s usual promoter and regulatory regions are removed). This hybrid DNA is then transformed into a human cell, where it integrates into the genome and is successfully transcribed. An assay for beta-galactosidase (lacZ gene product) is performed, and the activity is detected. However, neither lacY (transporter) or lacA (acetylase) proteins are observed. What’s the problem? Why is only one protein expressed? - Eukaryotic ribosomes initiate synthesis at the first AUG codon on the message 16. The most costly way (requires the most ATP) to negatively regulate gene expression is: - Degrade (destroy) the protein made by the genes 17. Eukaryotic genes are rarely clustered into operons, but bacterial genes often are. Why are operons selectively advantageous in bacteria but not eukaryotes? - Because segments of the bacterial genome are often transmitted “horizontally” (between species) as well as “vertically” (from mother to daughter) 18. In eukaryotes, transcription takes place in the nucleus. Before the mRNA leaves the nucleus, it must be: - Capped, polyadenylated, and spliced 19. The activity of factor S (shown in Figure Y) is consistent with its being a… - A repressor protein 20. How can harmful mutations be eliminated? - Through natural selection 21. Which of the following activities does tRNA aminoacyl synthetase perform? - Addition of amino acids to tRNAs 22. Figure X depicts polyribosomal mRNA. Polyribosomal RNA is found in… - Eukaryotes and Prokaryotes 23. Which of the following chemical reactions does the ribosome perform? - Formation of peptide bonds 24. In the initiation stage of transcription what does the promoter specify: - The site of transcription initiation - The direction the polymerase will travel 25. Eyes are useless in a lightless environment, and many cave-dwelling species have lost their eyes. This loss might be due to… - Natural selection for some beneficial aspect of eyelessness 26. Men have on eX chromosome, while women have two. You might think that because women have two doses of the genes on X, they’d make twice as much protein (compared with men) from these X-linked genes-but they don’t. “Gene dosage” in males vs. females is balanced because: - Women transcribe the genes on only one of the X chromosomes in any cell 27. While out taking a hike you come across a plant with floral organs in the following order: sepal, sepal, carpel, carpel. One explanation is that this plant has: - A mutation causing loss of B class function 28. Only 10% of the maize (corn) genome encodes for protein. The majority of the rest of the genome is made up of: - Retroviral and transposable element sequences 29. The information encoded in DNA is very stable genes are still functional in spite of the fact that they are billions of years old. The stability is achieved through… - The remarkable accuracy of DNA replication - The ability of the cell to use an undamaged strand as a template for repair - (Both b and d contribute to the longevity of DNA sequences) 30. A retroviral genome carries a mutation that results in a stop codon early in the coding sequence for the viral coat protein (=capsid protein). Such a mutation would make the virus unable to: - Infect other cells 31. In lab, you engineer an Arabidopsis plant so that B class function is expanded into whorl 4. What is the expected pattern of floral organs? - Sepal, petal, stamen, stamen 32. The majority of the DNA in a bacterial chromosome has been synthesized by - DNA polymerase III 33. Adding another nucleotide to the 3’ of a growing DNA chain is energetically favorable because… - You are releasing two inorganic phosphates 34. Mismatched bases that escape the notice of DNA polymerase can still be fixed via “mismatch repair”. This process works by… - Recognizing the mismatched bases and ripping out the younger strand 35. Mutations can be repaired by… - Mutations can’t be repaired—there’s no way to tell which base pair is incorrect 36. Chromosomes are made of both DNA and protein and were known to contain the heritable material. However, no one knew if the DNA or the protein was actually the chemical that encoded “inheritance”. It was very hard to convince people that DNA, rather protein made up the genetic code because: - Proteins have 20 different “doo-dads” (R groups) while DNA only has 4 bases - DNA has no enzymatic activities, while proteins do 37. Transcription is like DNA replication in that: - Synthesis is antiparallel to the template strand 38. Which of the following activities does tRNA aminoacyl synthase perform? - Error-checking for correctly charged tRNAs - Charging of tRNAs with amino acids 39. In eukaryotes, the correct reading frame for translation is precisely determined by: - The position of the most 5’ methionine codon 40. In prokaryotes, the correct reading frame for translation is precisely determined by: - The position of the methionine codon preceded by the Shine Dalgarno sequence 41. DNA polymerase proofreads its work. Which of the following components of gene expression also proofread? - tRNA synthetases 42. In eukaryotes, transcription and processing of mRNAs occurs in the _____, while translation occurs in the _____. - Nucleus, cytoplasm 43. The Lac operator - Is a DNA sequence located between the RNA polymerase binding site and the first coding sequence - Is a DNA sequence that binds the repressor 44. Enhancers are frequently found in eukaryotic genes, but no in prokaryotic genes. One possible explanation for this difference is: - The density of genes (# genes per kilobase) is too high for enhancers to specifically affect a single operon 45. Bacteria tend to locate their genes in functional clusters (operons) – where the genes involved in a single pathway involving several enzymes will be co-located. In eukaryotes, however, the genes involved in a biosynthetic pathway are randomly distributed throughout the genome. Why does this discrepancy exist? - DNA is often inherited via horizontal, as well as vertical, transmission in prokaryotes 46. A mutant is isolated with a nonsense mutation in the 5’ end of the gene encoding the trp repressor. Such a mutant would be expected to: - Strongly and constitutively express the trp operon 47. A mutant is isolated in the trp repressor protein that prevents binding of trp to the repressor. Such a mutant would be expected to: - Strongly and constitutively express the trp biosynthesis genes 48. The plant Coleus sometimes turns its leaves red. A hormone (here called “signal”) is involved in the regulation of flowering and leaf color. A pathway (see Figure above) is hypothesized to explain its action. All members of the signal transduction pathway are shown. The T-bar symbol depicted between M and K is consistent with the action of a protein that: - Degrades protein K - Binds a regulatory region near the promoters of its target genes, inhibiting RNA polymerase II binding 49. “Signal” is a small molecule. Predict the correct outcome when “signal” is applied to the plant, according to the pathway shown in the Figure above, both in wilde-type (wt) and in mutants that do not express protein “M”. - wt: Flowers and green leaves, mutant: flowers and green leaves 50. Generally speaking, the genes for regulatory proteins that directly detect environmental signals (for example, the Lac Repressor protein) are transcribed… - All the time (constitutively) 51. The fastest way to negatively regulate an enzymatic activity would be: - Allosteric inhibition of the protein 52. The least costly (least ATPs!) way to negatively regulate an enzymatic activity would be: - Turn off transcription of its gene 53. Which of the following eukaryotic processes reduce the effect of the relentless accumulation of mutations? - Crossing over 54. The eukaryotic RNA polymerase II, required for mRNA production, cannot recognize promoters on purified DNA without assistance. What additional proteins/events are required for RNA pol II to bind to the promoter? - Transcription factor II D must be bound to the TATA box 55. In cells, which express telomerase, telomerase will add telomeric sequences to: - The 3’ ends of chromosomes 56. Photolyase repairs: - UV-induced DNA damage 57. Transformed mice are created by injecting DNA into embryonic stem cells, and then injecting the stem cells into a very early mouse embryo (a “blastocyst”), and then injecting that embryo into the uterus of a host mouse (see. Figure below). The resulting mouse is expected to be: - A chimera of the cells of the blastocyst and the transformed embryonic stem cells 58. All eukaryotic nuclearly-encoded messenger RNAs begin translation in the cytoplasm. However, some complete translation elsewhere. Which types of proteins complete translation in a different compartment, and where does this occur? - Proteins destined for secretion out of the cell, at the rough endoplasmic reticulum 59. The ribosome performs many functions that assure accurate translation of mRNA. Which of the functions below is NOT performed by the ribosome? - Proofreading for the placement of the correct amino acids on the correct tRNAs 60. Replicators are molecules that evolve, each new allele being more or less likely to increase in copy number in a population. - Retrotransposons 61. Lambda phage that have integrated into the host genome are referred to a “prophage”. - The prophage will not excise in response to UV 62. Lambda has infected a host that carries a mutation in lexA that makes the LexA protein resistant to cleavage by recA protease. - The prophase will still excise in response to UV 63. In the figure below the line running through horizontally through the figure is: - A messenger RNA. Its 5’ end is on the left 64. Bacterial genes with related functions are often clustered. - CRP regulates a lot of other operons, it is not specialized for lactose metabolism 65. The pathway below is subject to allosteric feedback regulation. Number indicates enzymes. If the concentration of AMP is too high, and the concentration of GMP is too low, a good way to adjust the concentration would be to have: - AMP act as a negative regulator of enzyme 5 66. The hormone estrogen moves directly across the cell membrane without the assistance of a transporter protein. The process by which estrogen crosses the membrane is called: - Diffusion 67. A phospholipid is shown in the attached figure. If these molecules are mixed with water, they will form a micelle or vesicle because: - the hydrophobic tails are excluded from water - hydrogen bonds stabilize the interaction of the phospholipid polar head groups with water 68. “Pumps” push solutes up/against a gradient. Because this activity increases the potential energy stored in the gradient, the cell might “pay” for this through… -hydrolysis of ATP (a), coupling to flow of another solute down a gradient (b) - either a or b might power a biological “pump” 69. Enzymes allow a reaction to happen, providing a low potential energy catalytic pathway. An enzyme is found that catalyzes the reaction A  B. The direction of this reaction is determined by: - The ∆G of the reaction 70. Which of the following types of tubulin dimers preferentially assembles onto the plus end of a microtubule - ⍺ -GTP/β-GTP 71. Cytoskeleton filaments are assembled from small subunits. Which of the following statements is FALSE regarding cytoskeleton filament assembly? - While tubulin subunits assembly head-to-tail, actin subunits assemble head-to-head 72. Which statement about motor proteins is true? - Kinesins and dyneins interact with microtubule-associated proteins to regulate microtubule dynamics 73. A cell cycle checkpoint is best described as: - A point in the cell cycle where further progress along the cycle can be stopped 74. A diploid cell has a DNA content of 10 mg (micrograms). After it completes a meiotic cell cycle, each of the resulting cells will have a DNA content of: - 5 mg 75. Meiosis II resembles mitosis in that _____ but differs from mitosis in that _____ - sister chromatids separate, the sister chromatids are not identical 76. Consider a somatic animal cell with 10 chromosomes. In mitotic metaphase, this cell contains: - 2 centrosomes, 10 chromosomes, 20 kinetochores 77. Consider a somatic animal cell with 10 chromosomes. In mitotic metaphase, this cell contains: