Mastering Bio #6
The operon model describes how bacteria control the production of groups of enzymes. In this model, synthesis of the messenger RNA coding for these enzymes is switched on or off by regulatory proteins. a.) Can you match terms related to operons to their definitions? b.) The trp and lac operons are regulated in various ways. How do bacteria regulate transcription of these operons? Sort the statements into the appropriate bins depending on whether or not each operon would be transcribed under the stated conditions. c.) You are studying a bacterium that utilizes a sugar called athelose. This sugar can be used as an energy source when necessary. Metabolism of athelose is controlled by the ath operon. The genes of the ath operon code for the enzymes necessary to use athelose as an energy source. You have found the following: The genes of the ath operon are expressed only when the concentration of athelose in the bacterium is high. When glucose is absent, the bacterium needs to metabolize athelose as an energy source as much as possible. The same catabolite activator protein (CAP) involved with the lac operon interacts with the ath operon. Based on this information, how is the ath operon most likely controlled?
a.) 1. A(n) operon is a stretch of DNA consisting of an operator, a promoter, and genes for a related set of proteins, usually making up an entire metabolic pathway. 2. The genes of an operon is/are arranged sequentially after the promoter. 3. A(n) promoter is a specific nucleotide sequence in DNA that binds RNA polymerase, positioning it to start transcribing RNA at the appropriate place. 4. A(n) regulatory gene codes for a protein, such as a repressor, that controls the transcription of another gene or group of genes. 5. Regulatory proteins bind to the operator to control the expression of the operon. 6. A(n) repressor is a protein that inhibits gene transcription. In prokaryotes, this protein binds to the DNA in or near the promoter. 7. A(n) inducer is a specific small molecule that binds to a bacterial regulatory protein and changes its shape so that it cannot bind to an operator, thus switching an operon on. explanation: An operon is made up of a promoter and the genes of the operon. The promoter, which includes an operator, is the stretch of DNA where RNA polymerase binds. Regulatory proteins bind to the operator. The genes of the operon code for a related set of proteins. A regulatory gene located away from the operon codes for a protein that controls the operon. ----------------------------------------------------- b.) 1.operon is not transcribed->lac operon:lactose absent,trp operon:tryptophan present 2.operon is transcribed but not sped up through positive control->lac operon:lactose present,glucose present,trp operon: tryptophan absent 3.operon is transcribed quickly through positive control->lac operon: lactose present, glucose absent The trp operon is regulated through negative control only. When tryptophan is present, the operon genes are not transcribed. The lac operon is regulated through both negative control and positive control. Negative control: When lactose is absent, the repressor protein is active, and transcription is turned off. When lactose is present, the repressor protein is inactivated, and transcription is turned on. Positive control: When glucose is absent, another regulatory protein (CAP) binds to the promoter of the lac operon, increasing the rate of transcription if lactose is present. ----------------------------------------------------- c.) a)inactive activator b)cAMP c)Active activator d)active repressor(1st box) e)athelose f)inactive repressor(4th box) explanation: Metabolism of the sugar athelose in this hypothetical system is controlled by an operon that exhibits both positive control and negative control. Transcription of the ath operon is turned on when athelose is present (negative control), and sped up when the bacterium runs out of glucose and must rely on athelose for energy (positive control).
a.) You are given two test tubes, one containing cells from a human cancer and the other containing normal cells. Before you begin your studies, the labels fall off the test tubes How could you determine which sample contains the cancer cells?
normal cells: -need anchorage for growth -don't divide more than 50-60 times in culture cancer cells: -produce tumors being injected into nude mice -have mechanisms for replenishing their telomeres in culture -grow well in culture -grow to higher population densities in culture
Indicate whether each of the following descriptions applies to an oncogene, a proto-oncogene, or a tumor suppressor gene. Some descriptions may apply to more than one of these gene types.
Oncogene and proto-oncogene - gene(s) that could code for a normal growth factor. Oncogene - type of gene(s) found only in cancer cells. Proto-oncogene and tumor suppressor gene - type of gene(s) found in normal cells and cancer cells. Oncogene - gene(s) whose presence can cause cancer. Proto-oncogene and tumor suppressor gene - type of gene(s) found in normal cells. Oncogene, proto-oncogene, and tumor suppressor gene - type of gene(s) found in cancer cells. Tumor suppressor gene - gene(s) whose absence can cause cancer.
The central paradigm of biochemistry holds that information flows from DNA to RNA to protein. The process of making protein from the mRNA is called translation. Translation is carried out by the ribosome, which binds to the mRNA and binds tRNA, which recognizes the codons on the mRNA and brings the appropriate amino acid with it. The ribosome forms the peptide bond between the new amino acid and the growing peptide chain. a.) Complete the following vocabulary exercise related to the process of translation of mRNA to protein by the ribosome. Match the words in the left-hand column with the appropriate blank in the sentences in the right-hand column.
a.) 1. Initiationof translation always happens at the start codon of the mRNA. 2. The RNA that has an amino acid attached to it, and that binds to the codon on the mRNA, is called atRNA 3. Amino acids are attached to tRNA by enzymes called aminoacyl-tRNA synthetase. 4. Terminationof translation happens when the ribosome hits a stop codon on the mRNA. 5. The process, performed by the ribosome, of reading mRNA and synthesizing a protein is called translation. explanation: The process of translation, or protein synthesis, is a crucial part of the maintenance of living organisms. Proteins are constantly in use and will break down eventually, so new ones must always be available. If protein synthesis breaks down or stops, then the organism dies.
Chromosomal mutations are changes in the normal structure or number of chromosomes. Changes in chromosome structure can result from errors in meiosis or from exposure to radiation or other damaging agents. Certain changes in chromosome number can result from nondisjunction during either meiosis or mitosis. Both structural mutations and nondisjunction can play a role in trisomy 21, commonly known as Down syndrome. a.) The diagram below shows two normal chromosomes in a cell. Letters represent major segments of the chromosomes. The following table illustrates some structural mutations that involve one or both of these chromosomes. Identify the type of mutation that has led to each result shown. Drag one label into the space to the right of each chromosome or pair of chromosomes. You can use a label once, more than once, or not at all. b.) Suppose a diploid cell with three pairs of homologous chromosomes (2n = 6) enters meiosis. How many chromosomes will the resulting gametes have in each of the following cases? Drag one label into each space at the right of the table. Labels can be used once, more than once, or not at all. c.) Down syndrome is caused by trisomy 21, the presence of three copies of chromosome 21. The extra copy usually results from nondisjunction during meiosis. In some cases, however, the extra copy results from a translocation of most of chromosome 21 onto chromosome 14. A person who has had such a translocation in his or her gamete-producing cells is a carrier of familial Down syndrome. The carrier is normal because he or she still has two copies of all the essential genes on chromosome 21, despite the translocation. However, the same may not be true for the carrier's offspring. The diagram shows the six possible gametes that a carrier of familial Down syndrome could produce. Suppose that a carrier of familial Down syndrome mated with a person with a normal karyotype. Which gamete from the carrier parent could fuse with a gamete from the normal parent to produce a trisomy-21 zygote? Drag one of the white cells (representing gametes) to the white target in the diagram. Drag one of the pink cells (representing zygotes) to the pink target.
a.) 1. deletion 2. duplication 3. Robertsonian translocation 4. Pericentric inversion 5.duplication 6. unequal translocation 7. paracentric inversion explanation: -A deletion is the loss of part of a chromosomal segment. -A duplication is the repetition of a segment. The repeated segment may be located next to the original or at a different location, and its orientation may be the same as the original or the reverse. -An inversion is the removal of a segment followed by its reinsertion into the same chromosome in the reverse orientation. -A translocation is the transfer of a segment to a nonhomologous chromosome. Translocations may be reciprocal (two nonhomologous chromosomes exchange segments) or nonreciprocal (one chromosome transfers a segment without receiving one). -------------------------------------------------------- b.) Meiosis occurs normally: When regular meiosis takes place then 2n gives rise to n, so each daughter cell has 3 chromosomes Nondisjunction of one chromosome pair in meiosis After the process of meiosis ! one of the daughter cells will have 2 chromosomes and the other will have 4 chromosomes. If there is a normal disjunction during the process two daughter cells will have 4 chromosomes and two daughter cells will have 2 chromosomes. Nondisjunction of all three chromosome pairs in meiosis 1: After the process of meiosis1 one daughter cell has 6 chromosomes and the other does not have any chromosomes. After the process of meiosis 2, two daughter cells have 6 chromosomes and two daughter cells have 0 chromosomes (6, 6, 0, 0). Nondisjunction of one chromosome in one daughter cell in meiosis 2: After the process of meiosis1 both daughter cells have 3 chromosomes. After the process of meiosis 2, two daughter cells have 3 chromosomes and one daughter cell has 4 chromosomes and the other has 2 chromosomes (3, 3, 4 ,2). Nondisjunction of all three chromosomes in one daughter cell in meiosis 2: After the process of meiosis 1 both daughter cells have 3 chromosomes. Non-disjunction in all three chromosomes in meiosis 2 in one daughter cell has 6 chromosome, 3 chromosomes and no chromosomes in other cells. explanation: If one chromosome pair undergoes nondisjunction in meiosis I, half the gametes will have an extra chromosome (n +1), and half will be missing a chromosome (n - 1). If all chromosome pairs undergo nondisjunction in meiosis I, half the gametes will have twice the normal haploid number of chromosomes (2n), and half will have no chromosomes. If one chromosome undergoes nondisjunction in meiosis II, half the gametes will have the normal haploid number of chromosomes (n), one-quarter will have an extra chromosome (n +1), and one-quarter will be missing a chromosome (n - 1). If all chromosomes undergo nondisjunction in meiosis II, half the gametes will have the normal haploid number of chromosomes (n), one-quarter will have twice the haploid number (2n), and one-quarter will have no chromosomes. ------------------------------------------------------- c.) The correct answer contains three copies of chromosome 21 (hence, Trisomy 21), one of those copies of 21 coming from a disjunction pairing between chromosomes 14 and 21 (thus, a familial down syndrome zygote, not simply a down syndrome zygote) A carrier of familial Down syndrome has two copies of chromosome 21 and a normal phenotype. However, one of those copies has been translocated to another chromosome, often chromosome 14. Some of the carrier's gametes will contain both the normal and the translocated chromosome 21. If one of those gametes fuses with a gamete from a person with a normal karyotype, a zygote with trisomy 21 will result.
The three major classes of RNA found in the cytoplasm of a typical eukaryotic cell are rRNA, tRNA, and mRNA. a.) For each, indicate two kinds of processing that the RNA has almost certainly been subjected to. Between the two events choose the processing event which is unique to that RNA species. Drag the appropriate items to their respective bins. b.) Indicate processing events that you would also expect to find for the same species of RNA from a bacterial cell. Drag the appropriate items to their respective bins.
a.) Processing Event: Unique Processing Event: rRNA: Clevage of large Clevage of large precursor into several precursor into molecules; Degradation several molecules of transcribed spacers RNA: Addition of CCA sequence Addition of at 3' end (if not already added); CCA sequence methylation of bases at 3' end mRNA: Elimination of Introns; Addition of Addition of poly(A) tail at Poly(A) tail 3' end. -------------------------------------------------------- b.) rRNA: degradation of transcribed spacers, generation of one large molecule from one large precursor. tRNA: addition of CCA sequence if necessary, Methylation of bases mRNA: none in common Not found in bacterial cells: Addition of poly-A tail on 3' end, capping of 5' end, removal of leader sequence at 5' end.
In eukaryotic cells, the nuclear DNA codes for the synthesis of most of the cell's proteins. Each step of protein synthesis occurs in a specific part of the cell. In addition, various forms of RNA play key roles in the processes of protein synthesis. a.) In eukaryotic cells, the processes of protein synthesis occur in different cellular locations. b.)RNA plays important roles in many cellular processes, particularly those associated with protein synthesis: transcription, RNA processing, and translation. Drag the labels to the appropriate bins to identify the step in protein synthesis where each type of RNA first plays a role. If an RNA does not play a role in protein synthesis, drag it to the "not used in protein synthesis" bin c.) Life as we know it depends on the genetic code: a set of codons, each made up of three bases in a DNA sequence and corresponding mRNA sequence, that specifies which of the 20 amino acids will be added to the protein during translation. Imagine that a prokaryote-like organism has been discovered in the polar ice on Mars. Interestingly, these Martian organisms use the same DNA → RNA → protein system as life on Earth, except that there are only 2 bases (A and T) in the Martian DNA, and there are only 17 amino acids found in Martian proteins. Based on this information, what is the minimum size of a codon for these hypothetical Martian life-forms?
a.) a.formation of ribosomal subunits. b.attachment of an amino acid to tRNA. c.translation of cytoplasmic proteins. d.transcription and RNA processing. e.translation of secreted proteins. The above are the respected locations indicated in the diagram in the eukaryotic cell, where the process of protein synthesis occurs in different cellular locations. ------------------------------------------------ b.) Transcription/ RNA processing: Pre-mRNA, snRNA, mRNA translation: rRNA, tRNA Not used in protein synthesis: RNA primers explanation: In eukaryotes, pre-mRNA is produced by the direct transcription of the DNA sequence of a gene into a sequence of RNA nucleotides. Before this RNA transcript can be used as a template for protein synthesis, it is processed by modification of both the 5' and 3' ends. In addition, introns are removed from the pre-mRNA by a splicing process that is catalyzed by snRNAs (small nuclear RNAs) complexed with proteins. The product of RNA processing, mRNA (messenger RNA), exits the nucleus. Outside the nucleus, the mRNA serves as a template for protein synthesis on the ribosomes, which consist of catalytic rRNA (ribosomal RNA) molecules bound to ribosomal proteins. During translation, tRNA (transfer RNA) molecules match a sequence of three nucleotides in the mRNA to a specific amino acid, which is added to the growing polypeptide chain. RNA primers are not used in protein synthesis. RNA primers are only needed to initiate a new strand of DNA during DNA replication. -------------------------------------------------- c.) Answer: 5 In the most general case of x bases and y bases per codon, the total number of possible codons is equal to x^y . In the case of the hypothetical Martian life-forms, is the minimum codon length needed to specify 17 amino acids is 5 (25 = 32), with some redundancy (meaning that more than one codon could code for the same amino acid). For life on Earth, x = 4 and y = 3; thus the number of codons is 43, or 64. Because there are only 20 amino acids, there is a lot of redundancy in the code (there are several codons for each amino acid).
Living organisms make and use three main types of ribonucleic acids (RNA) for their biological functions: ribosomal RNA (rRNA) messenger RNA (mRNA) transfer RNA (tRNA) a.) Sort each description by the type of RNA it describes. Drag each item to the appropriate bin. b.) Indicate at which step of the replication-transcription-translation process each type of RNA first plays a role. During which step of the replication-transcription-translation process does each type of RNA first play a role? Drag each item to the appropriate bin.
a.) tRNA: contains an anticodon, has amino acid covalently attached mRNA: contains a exon, specifies the amino acid sequence for a protein rRNA: is a component of ribosome, is the most abundant form of RNA explanation: The cell uses three different types of RNAs to build proteins. rRNA is part of the ribosome, which is the site of protein synthesis. mRNA carries the genetic information from the DNA; the information specifies the sequence of amino acids in the new protein. tRNA interprets the information from the mRNA and brings the appropriate amino acids to the ribosome. -------------------------------------------------- b.) Replication: Transcription/ RNA processing: mRNA Translation: rRNA, tRNA explanation: The understanding that genetic information flows in one direction, from DNA to RNA to protein, became known as the central dogma of biology. The replication-transcription-translation pathway is connected with the DNA→RNA→protein flow of information. During replication, a faithful copy of a DNA molecule is made. During transcription, the DNA "message" is copied onto a molecule of mRNA. During translation, the information carried in the mRNA is transferred to molecules of tRNA to build a protein on the ribosomes.
The following is the actual sequence of a small stretch of human DNA: 3′ AAT TAT ACA CGA TGA AGC TTG TGA CAG GGT TTC CAA TCA TTA A 5′ 5′ TTA ATA TGT GCT ACT TCG AAC ACT GTC CCA AAG GTT AGT AAT T 3′ a.) What are the two possible RNA molecules that could be transcribed from this DNA? b.) Only one of these two RNA molecules can actually be translated. Explain why. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Terms can be used once, more than once, or not at all. c.) The RNA molecule that can be translated is the mRNA for the hormone vasopressin. What is the apparent amino acid sequence for vasopressin? d.) In its active form, vasopressin is a nonapeptide (that is, it has nine amino acids) with cysteine at the N-terminus. How can you explain this in light of your answer to the previous part? e.) A related hormone, oxytocin, has the following amino acid sequence: Cys-Tyr-Ile-Glu-Asp-Cys-Pro-Leu-Gly Where and how would you change the DNA that codes for vasopressin so that it would code for oxytocin instead?
a.) 5' UUA AUA UGU GCU ACU UCG AAC ACU GUC CCA AAG GUU AGU AAU U 3' 3' AAU UAU ACA CGA UGA AGC UUG UGA CAG GGU UUC CAA UCA UUA A 5' b.) Only the first of the two R N A sequences has an initiation codon (AUG) in the correct (5'-3') direction, so it must be the messenger. Notice also that the same R N A sequence has two stop codons in the correct reading frame near the 3' end (UAG and UAA). c.) Met-Cys-Tyr-Phe-Glu-His-Cys-Pro-Lys-Gly-(stop) d.) The methionine at the N-terminal end is apparently cleaved to generate the mature polypeptide. e.) The most conservative DNA sequence would be: 3' AAT TAT ACA CGA TGT AGC TTC TGA CAG GGA ATC CAA TCA TTA A 5'
A portion of a polypeptide produced by a mammalian cell was found to have the following sequence of amino acids: ...Lys-Ser-Pro-Ser-Leu-Asn-Ala... - In a normal cell ...Lys-Val-His-His-Leu-Met-Ala...- In a mutant cell a.) What was the nucleotide sequence of the mRNA segment that encoded this portion of the original polypeptide? b.) What was the nucleotide sequence of the mRNA encoding this portion of the mutant polypeptide? c.) Can you determine which nucleotide was deleted and which was inserted? d.) e.) If the insertion of histidines in the sequence results in abnormal activity of the protein, what changes would you suggest in the DNA that can change the insertion of histidines to the normal protein?
a.) 5'-AAA(or G) AGU CCA UCA CUU AAU GCN-3'(where N is any nucleotide) b.) 5'-AAA(or G) GUC CAU CAC UUA AUG GCN-3'(where N is any nucleotide) c.) In terms of the mRNA, we know that a(n) A was deleted and a(n) G was inserted. In terms of the DNA template strand, these changes would correspond to a deleted T and an inserted C. d.) Speaking again in terms of the mRNA, if the Lys codon at the 5'end of the original sequence was AAA, making the four nucleotides AAAA, any of the A nucleotides could have been deleted. On the other hand, if the Lys codon was AAG, the deleted A must have been the next one in the sequence. The new G nucleotide could have been inserted on either side of the G nucleotide that is the third nucleotide from the 3'end. e.) the substitution of T with A of the second histidine sequence (the one further from lysine)
Translation is the mRNA-directed synthesis of polypeptides. In translation, the information encoded in a sequence of RNA nucleotides is converted into a sequence of amino acids according to the genetic code. Translation also includes the first stage of targeting proteins to their eventual cellular location. a.) Ribosomes provide the scaffolding on which tRNAs interact with mRNA during translation of an mRNA sequence to a chain of amino acids. A ribosome has three binding sites, each of which has a distinct function in the tRNA-mRNA interactions. Drag the appropriate tRNAs to the binding sites on the ribosome to show the configuration immediately before a new peptide bond forms. If no tRNA is bound to a site at that time, leave that binding site empty. b.) The diagram below shows an mRNA molecule that encodes a protein with 202 amino acids. The start and stop codons are highlighted, and a portion of the nucleotide sequence in the early part of the molecule is shown in detail. At position 35, a single base-pair substitution in the DNA has changed what would have been a uracil (U) in the mRNA to an adenine (A). Based on the genetic code chart above, which of the following would be the result of this single base-pair substitution? c.) The DNA in a cell's nucleus encodes proteins that are eventually targeted to every membrane and compartment in the cell, as well as proteins that are targeted for secretion from the cell. For example, consider these two proteins: Phosphofructokinase (PFK) is an enzyme that functions in the cytoplasm during glycolysis. Insulin, a protein that regulates blood sugar levels, is secreted from specialized pancreatic cells. Assume that you can track the cellular locations of these two proteins from the time that translation is complete until the proteins reach their final destinations. For each protein, identify its targeting pathway: the sequence of cellular locations in which the protein is found from when translation is complete until it reaches its final (functional) destination. (Note that if an organelle is listed in a pathway, the location implied is inside the organelle, not in the membrane that surrounds the organelle.)
a.) During translation, new amino acids are added one at a time to the growing polypeptide chain. The addition of each new amino acid involves three steps: Binding of the charged tRNA to the A site. This step requires correct base-pairing between the codon on the mRNA and the anticodon on the tRNA. Formation of the new peptide bond. In the process, the polypeptide chain is transferred from the tRNA in the P site to the amino acid on the tRNA in the A site. Movement of the mRNA through the ribosome. In this step, the discharged tRNA shifts to the E site (where it is released) and the tRNA carrying the growing polypeptide shifts to the P site. b.) a nonsense mutation resulting in early termination of translation The effect of a single base substitution depends on the new codon formed by the substitution. To identify the new codon, it is first necessary to determine the reading frame for the amino acid sequence. The first codon starts with base 1, the second codon with base 4, the third with base 7, and so on. In this problem, the codon that contains the single base substitution begins with base 34. The original codon (UUA, which encodes the amino acid leucine) is converted by the single base substitution to UAA, which is a stop codon. This will cause premature termination of translation, also called a nonsense mutation -------------------------------------------------------- c.) PFK = Cytoplasm Only insulin = ER ->Golgi -> Outside of cell explanation: There are two general targeting pathways for nuclear-encoded proteins in eukaryotic cells. Proteins that will ultimately function in the cytoplasm (PFK, for example) are translated on free cytoplasmic ribosomes and released directly into the cytoplasm. Proteins that are destined for the membranes or compartments of the endomembrane system, as well as proteins that will be secreted from the cell (insulin, for example), are translated on ribosomes that are bound to the rough ER. For proteins translated on rough ER, the proteins are found in one of two places at the end of translation. If a protein is targeted to a membrane of the endomembrane system, it will be in the ER membrane. If a protein is targeted to the interior of an organelle in the endomembrane system or to the exterior of the cell, it will be in the lumen of the rough ER. From the rough ER (membrane or lumen), these non-cytoplasmic proteins move to the Golgi apparatus for processing and sorting before being sent to their final destinations.
You perform a series of experiments on the synthesis of the pituitary hormone prolactin, which is a single polypeptide chain 199 amino acids long. The mRNA coding for prolactin is translated in a cell-free protein synthesizing system containing ribosomes, amino acids, tRNAs, aminoacyl-tRNA synthetases, ATP, GTP, and the appropriate initiation, elongation, and termination factors. Under these conditions, a polypeptide chain 227 amino acids long is produced. a.) How might you explain the discrepancy between the normal length of prolactin (199 amino acids) and the length of the polypeptide synthesized in your experiment (227 amino acids)? b.) You perform a second experiment, in which you add SRP to your cell-free protein-synthesizing system, and find that translation stops after a polypeptide about 70 amino acids long has been produced. How can you explain this result? Can you think of any purpose this phenomenon might serve for the cell? Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all items will be used. c.) You perform a third experiment, in which you add both SRP and ER membrane vesicles to your protein-synthesizing system, and find that translation of the prolactin mRNA now produces a polypeptide 199 amino acids long. What is the explanation of the results? Where would you expect to find this polypeptide?
a.) In the absence of SRP and ER membranes, a prolactin precursor molecule is synthesized. b.) SRP binds to the ER signal sequence of the newly forming polypeptide chain and halts protein synthesis, thereby preventing the chain from growing beyond 70 amino acids. Inside cells, this blockage is normally maintained until the SRP binds the Ribosome to the ER membrane. Without this blockage, a complete preprolactin chain might be produced by cells and released into the cytosol, rather than being transported into the ER lumen. c.) in the presence of ER membrane vesicles, the SRP binds the mRNA-ribosome complex to the ER membrane, where signal peptidase cleaves the ER signal sequence from the newly forming prolactin chain as it moves across the ER membrane. Removal of the ER signal sequence reduces the length of the final polypeptide from 227 amino acids to 199 amino acids. The final polypeptide chain is released into the ER lumen
Steroid hormones are known to increase the expression of specific genes in selected target cell types. For example, testosterone increases the production of a protein called α2-microglobulin in the liver, and hydrocortisone (a type of glucocorticoid) increases the production of the enzyme tyrosine aminotransferase in the liver. a.) Based on your general knowledge of steroid hormone action, explain how these two steroid hormones are able to selectively influence the production of two different proteins in the same tissue. b.) Prior to the administration of testosterone or hydrocortisone, liver cells are exposed to either puromycin (an inhibitor of protein synthesis) or α-amanitin (an inhibitor of RNA polymerase II). How would you expect such treatments to affect the actions of testosterone and hydrocortisone? c.) Suppose you carry out a domain-swap experiment in which you use recombinant DNA techniques to exchange the zinc finger domains of the testosterone receptor and glucocorticoid receptors with each other. What effects would you now expect testosterone and hydrocortisone to have in cells containing these altered receptors? d.) If recombinant DNA techniques are used to substitute a testosterone response element for the glucocorticoid response element that is normally located adjacent to the tyrosine aminotransferase gene in liver cells, what effects would you expect testosterone and hydrocortisone to have in such genetically altered cells?
a.) Testosterone and hydrocortisone bind to different hormone receptors present in liver cells. --------------------------------------------------- b.) Inhibitors of either RNA synthesis or protein synthesis would be expected to block the ability of testosterone and hydrocortisone to stimulate the production of α2-microglobulin and tyrosine aminotransferase, respectively. --------------------------------------------- c.) The testosterone receptor would bind to glucocorticoid response elements, and the hydrocortisone receptor would bind to testosterone response elements. The binding of testosterone to its receptor would now be expected to increase the production of tyrosine aminotransferase, and the binding of hydrocortisone to its receptor would be expected to increase the production of α2-microglobulin. ------------------------------------------------- d.) Testosterone would increase the production of both of these proteins. Hydrocortisone would increase the production of neither of these proteins
DNA is transcribed to messenger RNA (mRNA), and the mRNA is translated to proteins on the ribosomes. A sequence of three nucleotides on an mRNA molecule is called a codon. As you can see in the table, most codons specify a particular amino acid to be added to the growing protein chain. In addition, one codon (shown in blue) codes for the amino acid methionine and functions as a "start" signal. Three codons (shown in red) do not code for amino acids, but instead function as "stop" signals a.) Use the table to sort the following ten codons into one of the three bins, according to whether they code for a start codon, an in-sequence amino acid, or a stop codon. Drag each item to the appropriate bin. b.) During translation, nucleotide base triplets (codons) in mRNA are read in sequence in the 5' → 3' direction along the mRNA. Amino acids are specified by the string of codons. What amino acid sequence does the following mRNA nucleotide sequence specify? 5′−AUGGCAAGAAAA−3′ Express the sequence of amino acids using the three-letter abbreviations, separated by hyphens (e.g., Met-Ser-Thr-Lys-Gly). c.) Before a molecule of mRNA can be translated into a protein on the ribosome, the mRNA must first be transcribed from a sequence of DNA. What amino acid sequence does the following DNA template sequence specify? 3′−TACAGAACGGTA−5′ Express the sequence of amino acids using the three-letter abbreviations, separated by hyphens (e.g., Met-Ser-His-Lys-Gly).
a.) start/methionine = AUG stop codon = UGA, UAG, UAA amino acid = AAA, UGC, CAC, ACU, GCA, AUC Nearly every mRNA gene that codes for a protein begins with the start codon, AUG, and thus begins with a methionine. Nearly every protein-coding sequence ends with one of the three stop codons (UAA, UAG, and UGA), which do not code for amino acids but signal the end of translation. -------------------------------------------------------- b.) Met-Ala-Arg-Lys An amino acid sequence is determined by strings of three-letter codons on the mRNA, each of which codes for a specific amino acid or a stop signal. The mRNA is translated in a 5' → 3' direction. --------------------------------------------------- c.) The transcription of DNA molecules forms mRNA. In RNA, thymine base pairs with adenine and adenine base pairs with uracil instead of thymine. Guanine base pairs with cytosine and vice-versa. Amino acids are encoded by a triplet of nucleotides known as a codon. DNA nucleotide sequence: 3'- TACAGAACGGTA - 5' -mRNA sequence: 5'--AUGUCUUGCCAU--3' -amino acid sequence: AUG UCU UGC CAU -Met Ser Cys His The codon AUG codes for methionine (Met), UCU codes for serine (Ser), UGC codes for cysteine (Cys) and CAU codes for histidine (His). remember: Before mRNA can be translated into an amino acid sequence, the mRNA must first be synthesized from DNA through transcription. Base pairing in mRNA synthesis follows slightly different rules than in DNA synthesis: uracil (U) replaces thymine (T) in pairing with adenine (A). The codons specified by the mRNA are then translated into a string of amino acids.