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Chapter 3, Gene function

Specific Reading Assignment. The entire chapter will be covered, with the following exceptions: skip Mechanism of exon splicing in pp 65-66, Colinearity of gene and protein structure in pp 78, and Division of labor in p84. Crucial Figures are 3, 4, 5, 6, 7, 8, 9a&c, 11, 12 (bases in bold only), 16, 17a, 20 (although you do not need to memorize it!), 21, 24!!!, Table 4 (some examples), Important Figures are 7, 9b, 17b, 18, 19, 22, 23, 25, 26, 27, 28, 29, 30, Table 1,  Skip Figures 10, 13, 14, 15,  31, Tables 2, 3, and 5.

RNA (pp. 55-60)

Objectives. To learn: 1) How RNA is organized. 2) What the main classes of RNA, and their functions, are. 3) What the properties of nucleic acids are that allow them to perform their very diverse functions.

1a) Draw comparative diagrams of the structure of uracil and thymine, and ribose and deoxyribose.

1b) Draw a four nucleotide long segment of RNA in the style of Fig. 2-5a.

2) Classes of RNA.

2a) The terms Informational and Functional RNAs used in the book are not of widespread use, but the other designations for RNAs are. What are the main types of RNA and what are their functions?

2b) What are the terms used to designate the transfer of information from DNA to RNA and from RNA to polypeptides?

2c) What are the two properties of polynucleotides that are at the root of most nucleic acid functions?

MAKING AND PROCESSING TRANSCRIPTS (pp. 60-65 and p. 66, message and review)

Objectives. To learn 1) The relationship (complementarity, identity, and relative polarities) between DNA sequence and RNA sequence. 2) The concepts of template and non-template strands. 3) The RNA polymerases. 4) The concept of promoter. 5) The important elements of prokaryotic promoters. 6) The concept of intron and exon. 7) The signals that control intron splicing. 8) The steps in eukaryotic RNA processing to produce mRNA.

3) DNA as transcription template.

3a) Is the RNA copied off a segment of DNA (the transcript), identical in sequence to the template or non-template strand (allowing for the equivalence between T and U)? Using Fig. 3-5a as a model, give an example (invent a sequence) indicating polarity (5' and 3' ends). Show, in your example, the first and last nucleotides laid down in the RNA molecule.

3b) When dealing with sequences, it is laborious and confusing to represent both strands of DNA. Since the sequence on one strand dictates the sequence of the other, it is a convention to represent the double-stranded DNA molecule with a single strand. Represent with a single sequence the DNA molecule of your example from 3a).

3c) Make a schematic diagram of one of the rRNA genes shown in Fig. 3-6. Show the DNA molecule as a single line, 8-10 RNA molecules, and RNA polymerases. Indicate the 5' and 3' ends of the RNA molecules.

4a) List all the polymerases of prokaryotes and eukaryotes, and their functions.

5a) Using the symbolism of question 4a in Chapter 2, diagram a prokaryotic gene identifying promoter, initiation and termination sequences. Label all components.

5b) Diagram the reaction represented by the chemical equation in p. 63 (Addition of a nucleotide to a growing ribonucleotide chain).

5c) Diagram transcription of a gene in the style of Fig. 3-5a. Make two drawings depicting the process at two points in time. Use a specific (invented) sequence. If you want to, you could bring in the idea of simultaneous transcription of the same gene by two molecules of polymerase (Fig. 3-6).

6a) Using the symbolism of question 4a in Chapter 2, diagram a eukaryotic gene identifying promoter, initiation and termination sequences, polyadenylation and splicing sites. Label all components.

6b) What is missing in the processing of primary transcript shown in Fig. 3-11?

6c) Diagram the splicing of a primary transcript.

PROTEIN  (pp. 66-74)

Objectives. To learn: 1) The general structure of amino acids. 2) The peptide bond. 3) The classes of amino acids and the names of 10 of them. 4) The structure of some simple amino acids. 5) The levels of protein organization. 6) The concept of α helix. 7) The conceptual function of the elements in translation: mRNA, tRNA, aminoacyl-tRNA synthetase 8) The concept of the genetic code, initiation and termination codons. 9) The parts and signals of the mRNA 10) The structure of tRNA molecules. 11) The process of protein synthesis: initiation, elongation and termination.

7) Protein structure

7a) Draw the general formula for an amino acid and a dipeptide.

7b) Draw the structure of the amino acids glycine and alanine (They are not shown in the book; you will need to do a little search).

7c) Using the Table of amino acids in the web site, learn the names, the groups to which they belong and the 3-letter abbreviation for 10 amino acids.

7d) Would you say that polypeptide chains are polarized? Why? What do they remind you of?

7e) What are the levels of organization of proteins. Explain, briefly, what are the primary, secondary, tertiary and quaternary structure of a protein. Which level(s) are determined directly by the nucleotide sequence (codons), and which levels are determined indirectly by the codons?

7f) By analogy to protein organization, what level of organization would you assign to the DNA double helix?

7g) Do all proteins have quaternary structure? What is the quaternary structure of hemoglobin?

7h) What are the main differences between globular and fibrous proteins.

7i) Review Figs. 18, 19, 26 and Chapter 2-27. They are representations of various proteins in different ways.

8) Translation, Codon translation by tRNA (pp. 68-71)

8a) The following elements can be identified in all tRNAs: a 3' end, a 5' end, four stems and three loops. How are the four stems formed? What are the three loops? What happens in the middle loop? What happens at the 3' end?

8b) What is the most important feature that differentiates one tRNA from another? How many different amino acids can associate to each tRNA? How many tRNA can associate with each amino acid?

8c) How does a tRNA recognize its amino acid and binds to it?

8d) What is the Awobble@ and what does it allow?

9) Ribosomes are Protein Factories.

9a) List the components of prokaryotic ribosomes. Note that S (as in 18S) is a unit of velocity of sedimentation and an indirect measure of molecular size.

9b) Make two diagrams (in the style of Fig. 3-24, but omitting the details of the amino acid atoms shown in that figure) showing two successive steps in the elongation of a polypeptide chain. Indicate 5' and 3' polarity of mRNA and amino- and carboxy-terminus polarity of the polypeptide chain.

9c) Make a diagram similar to the ones in 9b), but showing the first two amino acids in the chain. Indicate polarities and position of the Shine-Delgarno sequence.

9d) How often do you expect to find termination codons in a random stretch of DNA in which all four nucleotides are equally frequent?

9e) The figure below is the sequence of the mRNA for a small human peptide hormone, gastrin. There are three possible Aframes@ in which this sequence can be read. In the first frame the triplets start with the first base (ACA CUC AUC A.. etc.). In the second frame they start with the second base (..A CAC UCA UCA etc.). In the third frame they start with the third base (.AC ACU CAU CA.). If we advanced the Areading frame@ one more position, we would be back in the first frame. Obviously the three frames have different meaning in the amino acid sequence, and only one of the three is the one used by the ribosome.

Make copies of this figure so you will have three clean sequences. In one of them , Aread@ the triplets in the first frame by drawing an arch above each triplet. When you arrive at a start signal (AUG) in that frame, circle it in color and proceed. When you arrive at a stop signal (UAG, UGA, UAA), circle it in a different color and proceed to the end of the message. Repeat this exercise with the second frame (and if you wish the third). Which one seems to be the most likely frame to be utilized to synthesize this protein? Where does translation start and where does it end?

9f) If you had the sequence of the gene that corresponds to this mRNA, would you expect to find a long trail of Ts to encode for the many As at the end of the mRNA?

9g) Polyadenylation occurs at position 307 (???) **** and there is a 130 nucleotide intron inserted between positions 202 and 203. Make a diagram of the gastrin gene as in Orientation Question 6a in this chapter. Include the length, in nucleotides, of the various elements. Identify leader, coding sequence, trailer, and any other functional genetic element in the sequence.

PROTEIN FUNCTION AND MALFUNCTION IN CELLS (pp. 74-81)

Objectives. To learn: 1) The basic mode of action of enzymes (concept of active site). 2) How enzymes take part in metabolic pathways. 3) The concepts of allele, phenotype and genotype at the molecular level (We will revisit these topics when we study inheritance and Mendel=s experiments). 4) How mutations cause enzymes to lose activity and thus alter the phenotype. 5) The observations of Garrod, and Beadle and Tatum. 6) The concepts of prototroph and auxotroph. 7) A few examples of human diseases caused by mutations.

10) How proteins function (Concentrating on enzymes only)

10a) What is the active site of an enzyme? Are all the codons that encode for the amino acids at the active site clustered together?

10b) What is a metabolic pathway?

10c) Most enzymes are proteins of what type?

11a) What are alleles (In the context of genes encoding proteins)?

11b) In that same context, what are Awild type@ and Amutant@ alleles? What is meant by Aphenotype@?

11c) How would you define Agenotype@?

11d) Does the statement >This strain of Neurospora can grow without adding arginine to the medium= describe a phenotype or a genotype?

11e) Does the statement >This strain of Neurospora is arg+= describe a phenotype or a genotype?

11f)Which is usually the wild-type allele, the one that leads to a prototrophic phenotype or the one that leads to an auxotrophic phenotype?

11g) Can mutations in two different genes produce the same (or apparently similar) phenotype? Explain.

11h) How can single-base mutations be classified according to their effect on translation and protein product?

11i) How can mutations be classified according to their effect on a specific cellular function (assume the gene codes for an enzyme)?

11j) The mutations in question 11h) refer to changes in the coding region of the gene. What other base substitutions or deletions can have an effect on gene expression?

11k) Describe briefly three human hereditary diseases and the molecular explanation for cystic fibrosis.

DEFECTIVE PROTEINS AND DOMINANCE AND RECESSIVENESS (pp. 81-83)

Objectives. To learn: 1) The concept of dosage effect. 2) The concepts of haplo-sufficient and haplo-insufficient genes. 3) The molecular explanation for dominance or recessiveness of mutations in haplo-sufficient and haplo-insufficient genes.

12a) How can you explain the observation that for many enzyme-coding genes (Table 3-4), the wild type allele is dominant over mutant null alleles?

12b) Suppose you are dealing with a haplo-sufficient gene from which each normal allele produces 50 units of product.. You find a leaky mutation that produces 50% of normal activity (that is to say, 25 units of product), would you expect that an organism that carries two such alleles will have normal or mutant phenotype? What about an individual that carries one leaky allele and one null allele?

12c) On the basis of the material covered in this section, what would be an operational definition of whether a mutant allele is dominant or recessive?

Answers to GENES AND RNA

1a) See pages 27 and 59

1b) Note that the bases do not appear H-bonded because RNA is single stranded. If one segment of the RNA sequence is complementary to another segment, however, the molecule can fold back on itself (like a hair-pin) and form a limited amount of double stranded helix, similar to DNA (As in Fig. 3-21 and 3-22).

2a) messenger RNA, the nucleotide sequence of an mRNA contains the information that determines the amino acid sequence of a polypeptide (protein) chain. transfer RNAs (tRNAs) are short (less than 100 nucleotides long) molecules that can associate with specific amino acids and with specific mRNA sequences, thus bringing the two together. ribosomal RNAs (rRNAs)form part of the ribosome, a complex particle that also includes many proteins and on which protein synthesis occurs. small nuclear RNAs (snRNA) are involved in the processing of mRNAs. small cytoplasmic RNAs (scRNAs) are involved in protein transport across the various cellular compartments.

2b) Transcription and translation.

2c) 1) The ability to form helical double-stranded complexes (even over short distances) when they find the complementary sequence and 2) the ability of proteins to recognize base sequences, very short sequences, as little as four nucleotides can be recognized with great specificity.

Answers to MAKING FUNCTIONAL TRANSCRIPTS

3a) The RNA sequence is identical to the nontemplate strand, complementary to the template strand. Notice that the molecule is drawn with polarity that is rotated 180^o relative to Fig. 3-5a. This was done so that the 5’to 3’direction of the RNA would be from left to right, which is the convention. Note that the polarity of the RNA is opposite that of the template strand and the same as that of the non-template strand. Here we represent a segment of DNA and RNA in the middle of a gene; imagine that the RNA extends further to the left; the arrowhead is the position where the RNA is growing.

3b) Diagram. The DNA molecule is always represented by the non-template strand:

3c) Note that the spacing of the RNA polymerase molecules in Fig. 3-6 is very close; that means that they are starting to transcribe one after the other indicating a very high level of transcription (many RNAs are being made at the same time). In genes that are transcribed at much lower levels, we may find only one or two polymerases on at a time.

4a) Only one RNA polymerase in prokaryotes for all RNAs. In eukaryotes, Pol I transcribes rRNA genes, Pol II transcribes protein coding genes, and Pol III small non-mRNA genes such as tRNA and snRNA.

5a) Same as the answer to Chapter 2, Question 4a.

5b)

A molecule of the nucleotide triphosphate (in this case ATP) is paired to the T on the template strand, and it is about to make an ester bond between the phosphate and the 3’ OH group of the previous nucleotide in the growing RNA chain. When the ester bond is formed, the diphosphate (P~P) is released as shown below. At this point the next nucleotide triphosphate comes to take its place.

5c) The top diagram of the figure below represents the same sequence as in Question 3a. The yellow circle represents the RNA polymerase in the process of elongating near the left end of the figure (again, this represents the middle, not the beginning of a gene), and the three red dots represent RNA that is already synthesized and is hanging off the DNA. In the middle diagram the polymerase has moved further downstream; in this diagram (unlike the one in Question 3a) we show the RNA detaching itself from the DNA template as synthesis proceeds. In the bottom diagram we illustrate simultaneous synthesis of two RNA chains a small distance apart. Verify the consistent polarity, sequence and complementarity of the sequences in the diagrams for Questions 3a), 5b) and 5c); they all represent the same segment of DNA.

6a) Fig.3-16a

6b) The splicing of introns.

6c) Fig. 3-16b

Answers to PROTEIN STRUCTURE

7a) See Fig. 3-17a.

7b) These are the two simplest amino acids. In Glycine R is H, in Alanine, it is a methyl group.

7c) See Table of amino acids in the web site

7d) Yes, because at one end there is a free NH2 group (The N- or amino terminus) and at the other there is a free COOH (The C- or carboxy terminus). By Afree@ here we mean Anot associated in a peptide bond@. The situation is similar to the 5' and 3' ends of polynucleotides.

7e) The paragraph on p 67 explains this very well.

7f) The double helix in DNA is determined by the angles between neighboring nucleotides, and is not dependent in base composition; in this sense it is similar to the secondary structure of proteins. On the other hand, it requires two polynucleotide strands to form a double helix, and in this sense it would be similar to the quaternary structure of proteins. (Different people may come to different answers. This is a question to encourage you to think about these issues.)

7g) No, some proteins are made up of a single polypeptide chain. The quaternary structure of hemoglobin is α₂β₂: two chains of alpha globin and two chains of beta globin (Fig. 3-18c&d).

7h) Globular proteins are usually rounded or egg-shaped; they are usually found in suspension inside or outside the cell; if they are made up of several polypeptides (multimeric), there is a well-defined number of subunits (dimer, tetramer, hexamer, etc.). The compact, Aglobular@ shape is maintained by hydrophobic side-chains that project toward the interior of the molecule forming a minute lipid droplet. Globular proteins are maintained in suspension by the presence of charged side-chains pointing toward the outside of the molecule. This charged surface turns the protein into a giant ion that interacts with the water molecules and is thereby kept in solution. Examples are hemoglobin and many enzymes.

Fibrous proteins are very large aggregates of subunits in indeterminate numbers. The subunits themselves are often elongated in shape, and they associate end-to-end and side-by-side to create fibrillar aggregates. Examples are fibrillar collagen and muscle fibers. In collagen three polypeptides of two different kinds associate very tightly into elongated subunits called tropocollagen. These subunits associate by cross-linking to form collagen fibers and sheets. In muscle, molecules of actin interact with myosin and other proteins to form fiber bundles that fill the striated muscle cells.

There are also other kinds of proteins that do not fall under either category, for example, transmembrane proteins. Microtubule proteins can be considered under both categories, because they exist both as globular subunits in the cytoplasm, or aggregated into microtubules in the cytoskeleton and mitotic spindle.

The immediate causes of the properties of proteins are their tertiary and quaternary structures: their shape, position of reactive or charged amino acids, etc. Tertiary and quaternary structures, however, are ultimately determined by the primary structure: the amino acid sequence.

Answers to TRANSLATION

8a) The four stems are formed by complementarity of bases in different regions of the molecule, so that Watson-and-Crick style double helices can be formed. The loops are segments in between the regions of complementarity, where the bases cannot pair. The middle loop contains the anticodon triplet. The amino acid is attached to position 3' of the ribose at the 3' end.

8b) The most important feature that differentiates one tRNA from another is the anticodon loops. Only one amino acid can associate with each tRNA. Several tRNAs can associate with most amino acids

8c) Trick question. A tRNA does not recognize its amino acid; rather an enzyme, the aminoacyl-tRNA synthetase recognizes a tRNA and its corresponding amino acid and binds them together.

8d) The Awobble@ is the ability of the 5'-most base in the anticodon to pair with more than one base in the 3' position of the codon, thus allowing one tRNA to recognize more than one synonymous codon.

9a) Fig. 3-23 (upper half)

9b)

9c)

9d) There are 64 possible triplets, all equally likely if the nucleotides are in equal amounts. Since three out of 64 triplets are termination codons, these occur with a frequency of approximately one every 21 triplets, or one every 64 nucleotides.

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9e) The most likely frame seems to be the first, because in that frame occurs a long open reading frame after an AUG, which could serve as initiation. Notice how many termination codons interrupt the other reading frames. Also in the first reading frame there are several termination codons in the leader and trailer regions (i.e., outside of the coding region). Translation would start at 112 and terminate at the UAA in position 265. This was found to be the correct reading frame.

9f) No, because this is the poly(A) tail added during mRNA processing.

9g) Sequence of the human gastrin gene. Numbering is from the transcription initiation site (TIS). Compare the gene sequence to that of the mRNA. This sequence includes nucleotides upstream of the TIS, downstream of the poly(A) site, and in the intron, that are not present in the mRNA. The TATA box and poly(A) signal are underlined and the TIS is marked with an arrow. Verify the splicing signals.

Answers to PROTEIN FUNCTION AND MALFUNCTION IN CELLS

10a) The active site is a region of the enzyme where the substrate(s) bind and chemical reactions take place. Usually a group of neighboring amino acids make up most of the active site, although more distant amino acids can be brought to form part of it. The three dimensional specificity of the active site is part of the tertiary structure of the protein.

10b) A metabolic pathway is a series of chemical reactions linked in such a way that the product(s) of one reaction are the substrate(s) of the next. Each reaction is in turn catalyzed by an enzyme.

10c) Most enzymes are globular proteins.

11a) Alternative forms of a gene that differ in their sequence, and, therefore, in the sequence of the protein they encode. In some alleles the sequence leads to a severely altered form of the protein product, or to no protein at all; these constitute null alleles.

11b) Wild type alleles are those found in the normal population; they produce fully functional products in appropriate amounts. Mutant alleles are rare variants in which insufficient product is made, or the product is deficient (for example, an enzyme without enzymatic activity). Phenotype is the outward appearance of an organism; the consequence of interaction between the genes and the environment. Phenotype can be studied at different levels, for example, we can describe a phenotype as albinism, or, once we understand the biochemical basis, as a deficiency of tyrosinase. In another example, we can say two strains of mold have similar phenotypes because both require arginine to grow; however one strain may be able to use either arginine or citrulline to grow, while the other strain can only grow when arginine is provided. So the phenotypes are not identical, and the differences can only be detected by further studies.

11c) Genotype is the genetic endowment of an individual; it is specified by the alleles present at each gene. As a practical matter we always concentrate on one or a few genes of interest and ignore the rest.

11d) Phenotype.

11e) Genotype. Note that Neurospora is haploid, i.e., it has only one allele for each gene.

11f) Prototrophic

11g) Yes, see the answer to 11b)

11h) Single-base (or single-nucleotide) mutations can be a) base substitutions, b) base additions, or c) base deletions.

Single base substitutions can lead to: a1) no amino acid changes (silent substitutions) (for example, UAU ÷ UAC, see table of codons), a2) an amino acid substitution (missense mutation)(for example, UAU ÷ UCU), or a3) premature termination because of change to a stop or nonsense codon (nonsense mutation) (for example UAU ÷ UAA).

Single base addition or deletion always leads to a frame shift mutation in which all amino acids downstream(*) of the mutation are changed until a termination codon is reached. (*) By making an analogy between the ribosome running down the mRNA as a canoe floating down river, the terms >upstream= and >downstream= are used instead of >before= and >after= to indicate relative positions. The same applies for RNA synthesis and the transcription process.

11i) If the mutation deforms the active site it may lead to total loss of enzymatic activity and function (null mutation), if the mutation has a milder effect on the structure of the protein it may lead to a partial reduction of function (leaky mutations), finally an amino acid substitution may have no effect on protein function at all (silent or neutral mutation).

11j) a) mutations in intron splicing signals, b) mutations in the promoter, which affect transcription, c) mutations in the trailer region, which may affect mRNA stability. Notice that these changes can also constitute null, leaky or neutral mutations depending on their effect on function.

Answers to DEFECTIVE PROTEINS AND DOMINANCE AND RECESSIVENESS

12a) Let us say that a normal cell has a certain amount of a particular enzyme (E) and we will call that amount 100%. 50% of that amount is the result of transcription and translation from each of the two alleles present in the cell, and if a cell carries one null mutant allele and one normal allele, the amount of product is 50%. [This proportionality between the number of functional copies of a gene and amount of gene product is sometimes called dosage effect.] It turns out that for most enzymes, 50% is sufficient to fulfill all the functions of that enzyme, thus, only one wild type allele is sufficient to produce a wild type phenotype (haplo-sufficient genes). In other words, in a heterozygote between a mutant and wild type alleles, the phenotype is wild type, which indicates that the wild type allele is dominant.

12b) A homozygote for the leaky mutation will produce 50% of the normal level of enzymatic activity (25% from each allele), and will be of normal phenotype, since it is haplo-sufficient. A null mutation contributes 0% of enzymatic function; thus, a heterozygote between the leaky and the null alleles will have a total of 25% enzyme function and will probably have a mutant phenotype.

12c) In terms of phenotype,

if a⁺/a⁺ = a⁺/a … a/a then a⁺ (the wild type allele) is dominant over a

if B⁺/B⁺ … B⁺/B = B/B then B⁺ (the wild type allele) is recessive to B

Notice that often genes are named after their mutations, and we chose a lower case symbol if the mutant allele is recessive and an upper case symbol if the mutant allele is dominant.

Problems 1, 2, 3, 4, 7, 8, 9, 10, 12, 18, 19,  24.