Discovery of the operon- Jacob and Monod In study of induction of lact.pdf

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Discovery of the operon: Jacob and Monod In study of induction of lactose metabolism, Jacob and Monod isolated phenotypic mutants of two kinds. We now know the molecular basis for these mutant phenotypes. 1. uninducible- I 3 mutant repressor cannot bind inducer, P - mutant promoter (-10/-35 sequences) 2. constitutive- I- lacI mutant (no protein made); O c mutant operator sequence w/lower affinity for repressor, I d mutant repressor can t bind to the operator. They did not have operon sequence or protein structural data, but they could assay the phenotype of each mutant. In hindsight, we can fill in this table to test our understanding of the system, assuming a single O sequence. (In E.coli grown on glycerol, IPTG is used as the inducer b/c it is not metabolized.) Fill in the table below. A " + " indicates that the protein listed would be expressed in the given conditions; a " indicates it would not be expressed. Jacob and Monod had a model for two classes of regulatory factors-cis (DNA elements) and trans (proteins). But without knowledge of the operon structure and the nature of the proteins, how could they distinguish O C from I I 4 ( 5 vs. 4,6 above), or between I 3 and P ( 7 vs. 8 above)? A genetic cis trans test (meridiploids) was used to form a model of the operon. They made partial diploids by introducing the operon on a phage and determined the phenotype of these strains. Fill in the table, using the same symbols as above. In the column labeled "genetic" indicate if the I. P or O mutation is cis or trans and recessive or dominant. For 13 and 14, diagram the lac operon and draw the possible repressor molecules formed..

Education

1 At least 2 questions from this section will be on the final exam SAMPLE QUESTIONS FOR THE FINAL EXAM Question 1. Ferritin is a protein involved in the storage of iron inside cells. To prevent toxic accumulation of too much iron inside cells, the intracellular level of ferritin is tightly regulated. To study the regulation of ferritin synthesis, mammalian cells are grown with or without iron in the culture medium. Note that iron in the culture medium is rapidly transported inside cells. a) Upon addition of iron to the culture medium, the intracellular concentration of ferritin mRNA is unchanged but the concentration of ferritin protein increases. How do you think ferritin expression is regulated? Briefly explain. The regulatory sequence given below is found in the ferritin mRNA between the cap structure and the start codon. 5’-GGGUUUCCGUUCAACAGUGCUUGGACGGAAACCC-3’ Mutations within in this sequence are used to study the regulation of ferritin expression. The following observation are made: • ferritin expression is high, independent of the iron concentration, when (i) the entire region is deleted, or (ii) the region located upstream of the underlined sequence is deleted or (iii) the underlined sequence is replaced with a random sequence. • ferritin expression remains iron-dependent when this region is replaced by the following sequence: 5’-GGGCUCAGGUUCAACAGUGCUUGGACCUGAGCCC-3’. Note that the sequence differences are indicated in bold. b) Explain why these observations suggest that both sequence and structure of the 5’ end of ferritin mRNA are important for the regulation of ferritin expression. c) Ferritin translation becomes iron-independent when the regulatory sequence is moved from the 5’ side (upstream of the open reading frame) to the 3’ side (downstream of the open reading frame) of ferritin mRNA. Which step of ferritin translation do you think is affected by the intracellular level of iron? d) IRP is a protein involved in the regulation of ferritin expression. Anti-IRP antibodies attached to sepharose beads are added to a cell extract, then the extract is centrifuged to separate the pellet fraction (containing the sepharose beads ) from the supernatant fraction. If the cells are cultured in the absence of iron, ferritin mRNA is found together with IRP in the pellet. In contrast when cells are cultured in the presence of iron ferritin mRNA remains in the supernatant fraction while IRP alone is found in the pellet. Briefly explain the likely role of IRP in the regulation of ferritin expression. Question 2. You are studying the development of a newly discovered insect. Like drosophila, it undergoes a stage in early larval development where the eve gene is expressed in a pattern of 7 stripes. You are particularly interested in stripes 2 and 5. The following figures show the organization of the cis-acting elements that control the expression o ...

Zoology Second Year Important Question | Exam Tips and Tricks

PreethyKs

Lecture Merodiploids -lac genes_slideshare.pdf

Kristu Jayanti College

The document discusses the use of merodiploids, or partially diploid prokaryotes, in studying gene mutations in the lac operon in E. coli. Merodiploids contain two copies of certain genes, which allows researchers to examine how wild-type and mutated alleles interact. A series of examples are provided to illustrate how different mutations in the lac operon's operator, repressor, and promoter regions would affect the production of beta-galactosidase and permease in merodiploid E. coli under varying conditions.

You have identified two genes Parl and Mst5 which you susp.pdf

adnankhan605720

You have identified two genes, Parl and Mst5, which you suspect are involved in the embryonic development of the mouse. To study these genes, you isolate total DNA, total RNA, and total protein samples from mice at the following developmental stages: 1) sarly embryos ( 8 days post fertilization, or 8dpf ) 2) late embryos ( 15 days post fertilization, or 15dpf ) 3) newborn pups (NB) 4) 3-month-old adults (A) The expression of Parl and Mst5 during development may be controlled at the RNA level (by regulating either transcription or mRNA stability) or at the protein level (by regulating translation or protein stability). To begin to understand Parl's regulation, you decide to perform Northern and Western blots on your samples. a. Compare and contrast theYou have isolated two different yeast strains, strain 1 and strain 2, each of which fils to grow in the absence of arginine. You want to clone the wild-type copy of the gene or genes that are mutated in strain 1 and strain 2 . To do so you plan to: i. Obtain fragments of the entire yeast genomic DNA ii. Cut the chosen vector and ligate each fragment into a vector iii. Use this pool of original and recombinant vectors to transform E. coll cells iv. Select for E coli cells that have obtained an original or recombinant vectors v. Screen for E. coli transformed with a recombinant plasmid vi. Obtain recombinant plasmids from the library and transform yeast vii. Clone by complementation the gene that can restore the yeast to arginine prototrophy. a. To construct a yeast genomic library in E. coli that will allow you to successfully complete step 7 above, what would be the phenotype of the yeast you would choose as the donor for the genomic DNA? ( 3 marks) b. You choose the vector pBlue, shown below. Note that the cloning site lies within lacZ, the coding region of the gene that encodes -galactosidase. A cell that expresses -galactosidase can take a substrate called X-gal and cleave the -1,6 linkage to form a product that is bright blue. For each of the following sequences found on PB Bue, list the step or steps (1-7 above) for which that sequence is needed and explain the role that sequence plays. (7 marks).

Part I Genetics – From Genes to Proteins, Mutations Background .docx

bridgelandying

Part I: Genetics – From Genes to Proteins, Mutations Background: DNA represents the architectural blueprint for all living systems, encoding specific instructions in the sequence of its four nitrogen-containing base pairs that are necessary for building the organism. During the process of transcription, the information in the DNA codons of a gene is transcribed into RNA. A change in the DNA sequence, for example as a result of a "mistake" during DNA replication, is defined as a mutation. Mutations may result in a change in the "blueprint," which may then change the resulting protein product. Assignment details for part one: Transcribe and translate EACH of the three following DNA gene sequences. Turn these in using the graphic organizer. The letters represent the DNA N-base sequences of the genes. The first is the original gene. The next two are mutations of the original. Note that the changes (mutations) are shown in red. Original Gene sequence 3'-T A C C C T T T A G T A G C C A C T-5 Mutated gene sequence 1 3’-T A C G C T T T A G T A G C C A T T-5' Mutated gene sequence 2 3’-T A A C C T T T A C T A G G C A C T-5’ Also, answer the following questions: What is the significance of the first and last codons of an mRNA transcript? What meaning do these mRNA codons have for protein synthesis? Did the mutations result in a change in the final proteins? If so, describe the change. In general, why might a change in amino acid sequence affect protein function? Part II: Inheritance of Traits or Genetic Disorders Background : Cystic fibrosis is an inherited disease that affects the respiratory, digestive, or reproductive systems of the body, with symptoms ranging from mild to severe. Bob and Sally are recently married. Upon deciding to plan a family, both Sally and Bob find out that they are both heterozygous for cystic fibrosis , but neither of them has symptoms of the disorder. Assignment details: Complete a Punnett Square for cystic fibrosis for this couple using the following alleles: C = normal allele ; and c = allele for cystic fibrosis . Complete the Punnett Square in the graphic organizer. Also answer the following questions: Based on the Punnett square, calculate chances (percentages) for the following: for having a healthy child (not a carrier) a child that is a carrier for the cystic fibrosis trait a child with cystic fibrosis Be sure to submit these percentages as part of your assignment in the Graphic Organizer. Part III: Cell division, sexual reproduction and genetic variability Background : Eukaryotic cells can divide by mitosis or meiosis. In humans, mitosis produces new cells for growth and repair; meiosis produces sex cells (gametes) called sperm and eggs. Although mutations are the ultimate source of genetic variability, both meiosis and sexual reproduction also can contribute to new genetic combinations in offspring. Assignment Essay Question: How do both meiosis and sexual reproduction (fertilizat.

Genetic Dna And Bioinformatics ( Accession No. Xp Essay

Jessica Deakin

Question 11 of 24 In E. coli, three structural Deand enzymes A. D, an.pdf

fathimafancyjeweller

Question 1. Dscribe the change you observed when you added 1 mL of 0.1 M KSCN to the 2 mL portion of the diluted solution. Copy the equation from the procedure and explain your observations in terms of LeChatelier’s Principle. Choose the best answer. The additional thiocyantate ion shifted the equilibrium toward the products, producing morehexathiocyanatoferrate(III) lightening the color. Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq). The additional thiocyantate ion shifted the equilibrium toward the reactants, producing lesshexathiocyanatoferrate(III) and deepening the color. Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq). The additional thiocyantate ion shifted the equilibrium toward the reactants, producing morehexathiocyanatoferrate(III) lightening the color. Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq). The additional thiocyantate ion shifted the equilibrium toward the products, producing morehexathiocyanatoferrate(III) and deepening the color. Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq). Question 2. Show your calculations to determine the final volume of the 0.001 M FeCl3 solution. Choose the best answer. M1V1 = M2V2. (0.001 M)V1=(0.1 M)(0.05 L) V1=0.050 L = 50 mL M1V1 = M2V2. (0.001 M)V1=(0.1 M)(0.005 L) V1=0.500 L = 500 mL M1V1 = M2V2. (0.1 M)(0.005 L)=(0.001 M)V2 V2=0.500 L = 500 mL M1V1 = M2V2. (0.1 M)(0.0005 L)=(0.001 M)V2 V2=0.050 L = 50 mL Question 3. Describe the change you observed when you added 1 mL of 0.1 M FeCl3 to the diluted mixture of 0.001 M FeCl3(aq) and 0.1 M KSCN(aq). Refer to the equation in step 1 and explain your observations in terms of LeChatelier’s Principle. Choose the best answer. Additional iron(III) ion shifted the equilibrium toward the products to form the red-colored hexathiocyanatoferrate(III) complex and the intensity of color increased.color. Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq). Additional iron(III) ion shifted the equilibrium toward the products to form the red-colored hexathiocyanatoferrate(III) chloride precipitate. Fe3+(aq) + 6SCN-(aq) Fe(SCN)6Cl3(s). Additional iron(III) ion caused the equilibrium to shift to the reactants to compensate for the additional added iron(III), which increased the intensity of the red color. Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq). Additional iron(III) ion increased the concentration of the red-colored iron(III) ion. Fe3+(aq) + 6SCN-(aq) Fe(SCN)63-(aq). Question 4. Predict and then describe what happened when you added HCl(aq) to the copper nitrate solution. Write down the equation for the reaction. Choose the best answer. The added hydrogen ion shifted the equilibrium toward the products so that more CuCl42-(aq) was createdand the solution changed from blue to green. Cu(H2O)42+(aq) + 4Cl-(aq) CuCl42-(aq) + 4H2O(l). The added chloride ion shifted the equilibrium toward the products so that more CuCl42-(aq) was created and the solution changed from blue to green. Cu(H2O)42+(aq) + 4Cl-(aq) CuCl42-(aq) + 4H2O(l). The added chloride ion caused the precipitation of CuCl42-(s), which shifted the equilibrium toward t.

Online Molecular Biology Homework Help

Nursing Assignment Help

I am Monica A. I am a Molecular Biology Assignment Expert at nursingassignmenthelp.com. I hold a Masters’ in Medical Biotechnology, from University of Essex, UK. I have been helping students with their assignments for the past 15 years. I solve assignments related to Molecular Biology. Visit nursingassignmenthelp.com or email info@nursingassignmenthelp.com. You can also call on +1 678 648 4277 for any assistance with Molecular Biology Assignments.

Reprogramming refers to erasure and remodelling of epigenetic marks, such as DNA methylation, during mammalian development. Exposure of a differentiated cell nucleus to the cytoplasm of less differentiated cell leads to erasure of the stable epigenetic code that maintains the differentiated cell’s phenotype. Gradually, the nucleus acquires a new epigenetic code that is characteristic of the dedifferentiated cell donating the cytoplasm, a process termed cellular reprogramming.

REGULATION OF GENE EXPRESSION

YESANNA

DNA contains genetic information in the form of genes. Gene expression is regulated through positive and negative regulation, and this regulation is essential for organism growth and development. The lac operon in E. coli regulates the expression of genes involved in lactose metabolism. It consists of structural and regulatory genes. In the absence of lactose, a repressor protein binds to the operator gene and prevents transcription. In the presence of lactose, it binds to the repressor and induces a conformational change, allowing transcription and expression of the structural genes.

Please see link for image of the problemhttpsi.imgsafe.orgacb.pdf

amrishinda

Please see link for image of the problem: https://i.imgsafe.org/acb03acb2a.png a. A geneticist has devised a strategy to study protein translocation into the endoplasmic reticulum (ER) in yeast cells. She is interested in two different signal sequences that are thought to operate via slightly different translocation mechanisms. Using genetic engineering, she has fused the first signal sequence to a protein whose cytosolic expression is absolutely necessary for cell survival in the selective medium, but is inactive when in the ER. In the same cell, she has also fused the second signal sequence to a toxic protein whose cytosolic expression leads to cell lysis but is harmless when in the Whereas wild-type cells undergo lysis upon the expression ofthese fusion proteins, she has been able to identify viable mutants, each of which has a loss-of-function mutation in a gene encoding a protein involved in membrane translocation. The products of these genes are probably involved In the general transport of protelns Into the ER regardless of the type of signal sequence. involved in the transport of proteins with the first signal sequence but not the second one. Involved In the transport of protelns with tne second signal sequence but not the first one Involved in the transport of protelns with a novelsignal sequence (l.e.nether the first signal sequence nor the second one- 9. In the following graph, the magnitude of concentration difference across the nuclear pore complexes (NPCs) is plotted for four molecules (A to D) as a function of time, starting from an arbitrarily chosen initial concentration difference. Which of the following curves describes: 1) a large protein that is being actively transported across the NPC and 2) a large protein that is NOT actively transported into or out of the nucleus 10 Time (min) 1A, 2C 10. Considering the following hydropathy plot for a protein with 4 transmembrane domain. This protein functions in the plasma membrane. The position of positively charged residues near the first transmembrane helixis indicated by an asterisk on the plot. Are the protein termini expected to be located inside (l) or outside (O of the cell? Which of the regions indicated as 2 and 3 in the plot is expected to have more positively charged residues? Hydropathy index 1000 200 Residue number O, 2 Solution 8. A geneticist has devised a strategy to study protein translocation into the endoplasmic reticulum (ER) in yeast cells. She is interested in two different signal sequences that are thought to operate via slightly different translocation mechanisms. Using genetic engineering, she has fused the first signal sequence to a protein whose cytosolic expression is absolutely necessary for cell survival in the selective medium, but is inactive when in the ER. In the same cell, she has also fused the second signal sequence to a toxic protein whose cytosolic expression leads to cell lysis but is harmless when in the ER. Whereas wild-type cells undergo l.

this is all of the information that I have please help Lab 5 In.pdf

ambikacomputer4301

this is all of the information that I have please help Lab 5 Introduction - Genetic mapping See figure 5.1 for a schematic of the fly the cross you initially started with, you'll either crosses you have been working on. Two labs ago, map the distance between the w gene and the m you set up a pair of reciprocal parental crosses, gene, or between the w gene and the y gene. between mutant and wild type flies (fig. 5.1a). You had one of two different mutant strains, each with two mutant phenotypes - either white eyes all F2 individuals will receive only recessive (w) and miniature wings (m), or white eyes (w) alleles from this parent (fig. 5.1d, orange and and yellow body (y). The phenotypes of the F1 yellow chromosomes). Because of this, the flies should have indicated to you that all mutant phenotype of each F2 fly will tell you which phenotypes in question are x- linked recessive alleles (mutant or WT) were inherited from the (fig. 5.1b). heterozygous female F1 parent (fig. 5.1d, dark and light blue chromosomes). The first F2 fly Last lab, you used the F1 flies from one of shown in figure 5.1d inherited 'a B' from the your parental crosses to set up an F1 cross (fig. heterozygous parent and will end up with the consequence. After crossing two pure breeding this F2, you observe a ABphenotype and parents, F1 offspring will be heterozygous for therefore know that this fly received 'A B' from nearly all genes in question - the exception is X - the heterozygous parent. linked genes in the male offspring. Since the Y chromosome is equivalent to recessive alleles for The goal of genetic mapping is to X-linked genes, these F1 males are recessive for determine the likelihood of cross over between all X-linked genes and act as a test cross. The two loci/genes. If we score the phenotypes of a heterozygous F1 females and recessive F1 males large F2 population from our crosses, we can (test cross) can be used to map the distance determine the recombination frequency of your between the genes causing the two phenotypes two genes. of your parental mutant female. Depending on e) F2 phenotype scoring: f) Recombination frequency: Eigure 5.1: Schematic of your Drosophila crosses, See text of lab 5 intro for description. For a given F1 gamete for the F2 individual it number of flies, it is easy to calculate the creates), if no cross over occurs between the two recombinant frequency between your two genes genes in question, the F2 phenotype will be the (fig. 5.1ef ). same as one of the original parents - either fully WT or double mutant in this case. In figure 5.1d Recall from last time that a lower recombinant these have blue chromosomes of a single colour. frequency is observed when genetic map If a crossover does occur between the two genes, distances are small. When genes are close to the F2 fly will have a phenotype unlike either of each other, there's a narrow range on the the parents - a recombinant phenotype (shown chromosome for a random crossover to land.

Exam Questions and Solutions of Molecular Biology

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Unit 5 -chapters_16_-20

sbarkanic

This document provides an overview of the content covered in Unit 5 - Molecular Genetics, including Chapters 16-20. Chapter 16 focuses on DNA as the genetic material and DNA replication and repair. Chapter 17 examines the connection between genes and proteins, including transcription, translation, and RNA/protein synthesis. Chapter 18 covers the genetics of viruses and bacteria, including viral replication cycles and bacterial genetic variation. Chapter 19 discusses the organization and control of eukaryotic genomes, including chromatin structure, genome organization, and gene expression control. Finally, Chapter 20 reviews DNA technology and genomics, such as DNA cloning, analysis techniques, the Human Genome Project, and applications of DNA technology.

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I am Robert D. I am a Molecular Biology Assignment Expert at nursingassignmenthelp.com. I hold a Masters’ in Medical Biotechnology, from University Of Portsmouth, UK. I have been helping students with their assignments for the past 10 years. I solve assignments related to Molecular Biology. Visit nursingassignmenthelp.com or email info@nursingassignmenthelp.com. You can also call on +1 678 648 4277 for any assistance with Molecular Biology Assignments.

R1 research paper

Karen Amor

1) The document discusses the core oligosaccharide structures of E. coli lipopolysaccharides. There are 5 known core types (R1-R4 and K-12), with R1 being the most prevalent among clinical isolates. 2) It describes the organization of the waa locus responsible for core biosynthesis, including 3 main operons (gmhD, waaQ, waaA). The central waaQ operon contains both conserved and core type-specific genes involved in outer core assembly. 3) Recent work characterized the functions of 9 genes in the waaQ operon of the predominant R1 core type. This helped define the pathway for R1 core assembly and assigned

Please help with an explanation method on how to approach these. I .pdf

archiesgallery

Please help with an explanation/ method on how to approach these. I get so confused. Complete the table below using + to indicate if the enzyme is synthesized and - to indicate if the enzyme will not be synthesized. Genotype symbols l = acl (the repressor gene) P = P_lac (the promoter), O = lacO (the operator), Z = lacZ (the beta = -galactosidase gene), and Y = lacY (the permease gene). The minus sign accompanying the symbol (e.g., I^1) indicates a mutant gene, the plus sign indicates wild-type/functional gene. O^c is the constituitive operator mutation (the same as O-) and l^-d genes make mutant subunits of the repressor protein (as opposed to I^1 which you can think of as not making a product) and I^s codes for the super-repressor protein, which will not release from the operator. Solution It is related to Lac operon and Here I, P & O are regulatory genes and Z & Y are structural coding genes. The expression of coding genes depends on regulatory genes. Generally O & P promotes expression of Z & Y. If I is functional, O & P cannot promote the expression of Z & Y. Inducer can remove the I effect and promotes Z & Y expression Based on the above words, we can tell weather the Z and Y are exprssed or Not. I have give some in below. Please find Genotype Inducer Absent Inducer Present B-Galactosidase or Lac-Z Permease or Lac-Y B-Galactosidase or Lac-Z Permease or Lac-Y A - - + + B - - - + C - - + - D + + + + E - - - - F - - + + G - - - - Genotype Inducer Absent Inducer Present B-Galactosidase or Lac-Z Permease or Lac-Y B-Galactosidase or Lac-Z Permease or Lac-Y A - - + + B - - - + C - - + - D + + + + E - - - - F - - + + G - - - -.

Gene expression and regulation

Tapeshwar Yadav

This document discusses gene expression and regulation. It begins by defining key terms like genes, genome, and gene expression. It then explains that organisms adapt to their environment by altering gene expression. There are two main types of gene expression: positive regulation which increases expression, and negative regulation which decreases it. Gene expression is studied in detail in prokaryotes using the lactose operon in E. coli as an example. Gene expression in eukaryotes is more complex due to larger genomes and separation of DNA and protein synthesis by the nuclear membrane.

Gene expression & its regulation.

Mohd Uzair Ansari

Nuclear Transport And Its Effect On Breast Cancer Tumor Cells

Stephanie Clark

The document discusses nuclear transport and its role in breast cancer tumor cells. It explains that nuclear transport involves the movement of large molecules in and out of the cell nucleus through nuclear pore complexes. It notes that various nuclear transport pathways play an important role in the progression and suppression of breast cancer tumor cells. Disrupting nuclear transport sequences can alter protein localization and functionality, as seen with spleen tyrosine kinase, which is a candidate tumor suppressor for breast cancer.

JG_ResearchPoster-fe

Jerry Gomez

This research analyzes the conformations of proteins related to prion and Parkinson's diseases through steered molecular dynamics simulations. Simulations of the human prion protein and alpha-synuclein peptides show transitions between conformations as hydrogen bonds break during pulling. Certain conformations are preferred at initial and final pulling stages. Understanding structural transitions provides insight into how protein misfolding occurs in disease. Future work includes varying pulling rates and simulating mutations to further study protein stability and misfolded states.

mitochondria poster

Abhinav Suri

1. The document discusses complex I, the first enzyme in the mitochondrial electron transport chain. It has 45 subunits encoded by both nuclear and mitochondrial DNA. Deficiencies in complex I can cause various mitochondrial diseases. 2. The lab studies complex I structure and assembly using three mouse fibroblast cell lines: A9 (normal complex I), 4A (mutated ND6 gene impairing complex I), and 4AR (restored complex I despite ND6 mutation). They use gel electrophoresis, western blotting, and co-immunoprecipitation to analyze complex I proteins. 3. Defects in mitochondrial complex I subunits can lead to diseases like Parkinson's, cancers, and neurological/my

Gene Regulation Translation Russell.ppt

Pudhuvai Baveesh

The document describes the process of translation, where mRNA is translated into proteins with the help of tRNAs and ribosomes. It discusses the key findings that led to the determination that the genetic code is read in triplets and is almost universal across organisms. The genetic code is described as being triplet, comma-free, non-overlapping, and degenerate. The process of translation involves tRNAs becoming charged with amino acids by aminoacyl-tRNA synthetases using ATP. Charged tRNAs then deliver amino acids to the ribosome according to base pairing between mRNA codons and tRNA anticodons.

How do gene work1

Jannat Iftikhar

This document discusses gene function and the relationship between genes and proteins. It begins by explaining the one-gene-one-enzyme hypothesis proposed by Beadle and Tatum, which demonstrated that mutations in specific genes resulted in defects in single enzymes. The document then discusses protein structure and how amino acid sequences encode higher-level structures. It explores how gene mutations can result in single amino acid changes in proteins. Finally, it discusses how genes control cellular metabolism and provides examples of genetic diseases caused by enzyme deficiencies.

Structure-Function Analysis of POR Mutants

AYang999

This document summarizes protein structure and function, with a focus on mutations in the protein cytochrome P450 oxidoreductase (POR). It describes how mutations can alter a protein's amino acid sequence and structure. Specifically, it investigates the S102P and R550Q mutations in POR, which were found in humans. Experiments expressed and purified these mutant POR proteins, and will test their electron transfer activity compared to the wild type protein using a cytochrome c reduction assay. This will help determine if these mutations impair POR function.

Gene expression in bacteria and bacteriophages

Vinod Pawar

1. The document discusses gene regulation in bacteria, focusing on the lac operon in E. coli. which controls lactose metabolism. 2. The lac operon contains 3 genes - lacZ, lacY, lacA - that are regulated together. When lactose is present, it is broken down into allolactose, which binds the lac repressor and prevents it from blocking transcription of the operon. 3. Experiments by Jacob and Monod identified mutations affecting regulation, including mutations in the operator or repressor gene that result in constitutive expression of the lac genes.

Pairwise ortholog matrix of cg.ezbiocloud.net

Jon Jongsik Chun

IB Proteins & Enzymes Review (2.4-2.5-7.3)

Jacob Cedarbaum

This document contains review questions for a biology exam on proteins and enzymes. It includes questions about protein structure, functions of different protein types, enzyme structure and function, and the effects of temperature, pH and substrate concentration on enzyme activity. Specific review questions cover topics like protein denaturation, enzyme specificity, and the industrial production of lactose-free milk.

Discuss how descriptive epidemiology by person- place- and time may as.pdf

Discovery of the operon- Jacob and Monod In study of induction of lact.pdf

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