High school · ScienceAbout 150 minutes
ScienceHigh schoolAbout 150 minutes

Analyze original and mutated DNA segments to identify gene and chromosomal mutations, predict their effects on the protein and phenotype, and connect those effects to natural selection

Several sessions: instruction with the worked model, a laboratory or data-analysis investigation, independent practice and a written explanation.

Ohio learning standards this draft targets

science-2018:B.H.4

Read the standard text

science-2018:B.H.4 Mutations

Provisional targets. Clause-by-clause educational review is still pending.

Section 1 · LearnLearn the idea
Starting from the coding strand ATG AAA GGC TTT TAA, which encodes methionine-lysine-glycine-phenylalanine-stop, four gene mutations are analyzed: changing the third base of the second codon gives AAG, still lysine (silent); changing its first base to G gives GAA, glutamic acid (missense); changing it to T gives UAA in the mRNA, a stop codon (nonsense); deleting one A shifts the reading frame of every later codon, turning glycine into alanine and destroying the stop codon (frameshift). Two of the new codons happen to code for the same amino acids as before.Starting from the coding strand ATG AAA GGC TTT TAA, which encodes methionine-lysine-glycine-phenylalanine-stop, four gene mutations are analyzed: changing the third base of the second codon gives AAG, still lysine (silent); changing its first base to G gives GAA, glutamic acid (missense); changing it to T gives UAA in the mRNA, a stop codon (nonsense); deleting one A shifts the reading frame of every later codon, turning glycine into alanine and destroying the stop codon (frameshift). Two of the new codons happen to code for the same amino acids as before.One original coding sequence and fourmutationsChangeCoding DNAProteinTypeNoneATG AAA GGC TTTTAAMet Lys Gly PhestoporiginalAAA→AAGATG AAG GGC TTTTAAMet Lys Gly PhestopsilentAAA→GAAATG GAA GGC TTTTAAMet Glu Gly PhestopmissenseAAA→TAAATG TAA GGC TTTTAAMet stopnonsensedelete one AATG AAG GCT TTTAAMet Lys Ala Phe,no stopframeshift
Starting from the coding strand ATG AAA GGC TTT TAA, which encodes methionine-lysine-glycine-phenylalanine-stop, four gene mutations are analyzed: changing the third base of the second codon gives AAG, still lysine (silent); changing its first base to G gives GAA, glutamic acid (missense); changing it to T gives UAA in the mRNA, a stop codon (nonsense); deleting one A shifts the reading frame of every later codon, turning glycine into alanine and destroying the stop codon (frameshift). Two of the new codons happen to code for the same amino acids as before.

Learning goal: Recall the types of mutations and describe the effects they may have on a protein; classify mutations as gene mutations (insertion, deletion, substitution) or chromosomal mutations (such as trisomy and monosomy); given an original and a mutated DNA segment, analyze the mutation and identify its impact on the phenotype; and explain how natural selection might favor, select against or be neutral toward the resulting change.

What a mutation is. A mutation is a change in the sequence of DNA. Genes can be altered by the insertion, deletion or substitution of a segment of DNA, and an altered gene is copied by replication into every cell that develops from the cell in which it occurred. In animals such as humans, whose gamete-forming cells are set apart from the rest of the body, a mutation in a body (somatic) cell affects only that cell's descendants within the individual, as in most cancers, and is not passed to offspring (organisms that reproduce from body tissue, such as plants grown from cuttings, can pass a somatic mutation on). A mutation in a gamete, or in the cell line that produces gametes, can be passed to offspring and then to every cell of the offspring's body; these germline mutations are the raw material of heredity and evolution. Mutations arise from copying errors during replication, which repair enzymes correct but not perfectly, and from damage by ultraviolet light, ionizing radiation and certain chemicals. They are not directed at any need: a mutation that turns out useful arises no more often because it is useful.

Gene mutations and their effects on the protein. Because the ribosome reads mRNA in three-base codons, the effect of a gene mutation depends on what it does to the reading frame and to the codons. A substitution replaces one base with another. If the new codon specifies the same amino acid, the mutation is silent and the protein is unchanged; the redundancy of the code makes this common. If the new codon specifies a different amino acid, the mutation is missense, and its effect ranges from none, if the substitute amino acid is similar and sits in an unimportant part of the chain, to severe, if it alters the active site or the folding. If the new codon is a stop codon, the mutation is nonsense: translation ends early, and a truncated protein usually cannot work. An insertion or deletion, within the translated coding sequence, of a number of bases that is not a multiple of three shifts the reading frame, so every codon downstream is misread; such a frameshift typically scrambles the rest of the protein and often introduces an early stop. Insertions or deletions of exactly three bases (or a multiple) add or remove whole amino acids without shifting the frame, which is often less damaging, though not always: cystic fibrosis most often results from a three-base deletion that removes a single, critical amino acid. An insertion or deletion outside the coding sequence, in an intron or an untranslated region, does not shift the reading frame, although it can still matter if it disturbs RNA processing or regulation. Mutations in the regulatory DNA that controls when a gene is transcribed change how much protein is made rather than what it is; the adult ability to digest milk sugar in many human populations arose this way.

A fully worked analysis. Original coding strand: ATG AAA GGC TTT TAA. Its mRNA is AUG AAA GGC UUU UAA, translated methionine-lysine-glycine-phenylalanine-stop. Mutation 1, substitution of the last A of AAA by G: AAG still codes for lysine, so the protein is unchanged; silent. Mutation 2, substitution of the first A of AAA by G: GAA codes for glutamic acid, so lysine becomes glutamic acid; missense, and because lysine carries a positive charge and glutamic acid a negative one, the change could well alter the protein's folding or binding. Mutation 3, substitution of the first A of AAA by T: TAA in the coding strand gives UAA in the mRNA, a stop codon, so the protein is just methionine; nonsense, and the protein is lost. Mutation 4, deletion of the first A of AAA: the strand becomes ATG AAG GCT TTT AA, read as methionine-lysine-alanine-phenylalanine and then an incomplete codon. Every codon after the deletion is read in a new frame (AAG, GCT, TTT, AA instead of AAA, GGC, TTT, TAA); by coincidence two of the new codons still give lysine and phenylalanine, glycine becomes alanine, and the stop codon is destroyed, so the ribosome would read on past the end of the gene into whatever follows. That is a frameshift: in a longer gene the new frame almost always changes most of the downstream amino acids.

Chromosomal mutations. Some mutations change chromosomes rather than single genes. A segment of a chromosome may be deleted, duplicated, inverted (reversed end to end) or translocated (moved to another chromosome). Chromosome number can change when chromosomes fail to separate during a cell division, an error called nondisjunction. It can happen in mitosis, leaving one body with a mixture of cells that have different chromosome numbers, but the inherited cases worked here arise in meiosis, when homologs or sister chromatids fail to separate: one gamete receives an extra copy and another receives none. An offspring with three copies of a chromosome has a trisomy; three copies of chromosome 21 causes Down syndrome. An offspring with a single copy has a monosomy; a single X chromosome with no second sex chromosome causes Turner syndrome, and monosomy of most other chromosomes is not survivable. Because a whole chromosome carries hundreds or thousands of genes, changes in chromosome number alter the amount of many proteins at once, which is why their effects are broad.

Mutation, phenotype and natural selection. A mutation's effect on the organism's success in its environment may be harmful, helpful or negligible, and which of these it is depends on the environment. Natural selection then acts on the resulting phenotype. A mutation that removes the function of an essential enzyme is selected against because carriers survive or reproduce less. A silent mutation, or one in DNA that is never used, is usually neutral (occasionally a silent change still affects how much protein is made), and its frequency drifts by chance. Some mutations are favored: a change in a bacterial gene that lets the cell survive an antibiotic spreads rapidly wherever the antibiotic is used, and is nearly neutral where it is not. The sickle cell allele of the beta-globin gene shows all of this in one case. A single substitution changes a GAG codon to GTG in the coding strand, so the sixth amino acid of the protein is valine instead of glutamic acid; the altered hemoglobin sticks together in low oxygen and deforms red cells into a sickle shape. Two copies cause sickle cell disease, a serious condition, so selection acts against the allele; yet one copy gives substantial protection against severe malaria, so in regions where malaria is common the allele is maintained at high frequency by the advantage to heterozygotes. The same allele is favored in one environment and disfavored in another.

Chromosome maps and linkage groups. Mutations are also the tools by which geneticists map chromosomes. A gene can only be followed if it has more than one allele, and mutant alleles are what make that possible. Recombination frequencies between mutant markers place genes along a chromosome, and genes that recombine with one another at less than 50 percent belong to one linkage group, while genes that recombine at 50 percent behave as unlinked, usually because they are on different chromosomes (genes very far apart on one chromosome also reach 50 percent). Evaluating a chromosome map means reading such distances and deciding which genes travel together.

Misconceptions. "Mutations are always harmful" is wrong; most are neutral, some are harmful and a few are beneficial, and the judgment depends on the environment. "Organisms mutate in response to need" is wrong; mutations arise at random with respect to their usefulness, and selection afterward sorts them. "A substitution always changes the protein" ignores silent mutations and the redundancy of the code. "A mutation in my skin cell will be passed to my children" confuses somatic with germline mutations.

Investigation you can do. Model the reading frame with a sentence of three-letter words: THE BIG CAT ATE THE RAT. Write it on a strip, then make one strip for each mutation type: substitute one letter (THE BIG CAT ATE THE HAT: one word changed, missense), substitute to make a meaningless word (THE BIG XAT), delete one letter (THE BIG ATA TET HER AT: frameshift), delete three letters (THE CAT ATE THE RAT: one word lost, frame intact). Then take real practice: for a hypothetical original coding sequence of 15 bases supplied by your teacher and four mutated versions, write the mRNA and amino acids for each and classify the mutation. A result in which single-letter deletions scramble every following word while three-letter deletions remove one word supports the claim that the reading frame explains why frameshifts are so damaging. What the model cannot show: words do not fold, so the model cannot show how a changed amino acid alters protein shape, and it cannot show that some substitutions are silent because letters, unlike codons, are not redundant.

Section 2 · ModelWork through the examples

Worked analysis of an original and mutated segment: Original coding strand ATG CCC GAA TGG TAA: mRNA AUG CCC GAA UGG UAA, protein methionine-proline-glutamic acid-tryptophan-stop. Mutated: ATG CCC TAA TGG TAA. The change GAA to TAA gives the mRNA codon UAA, a stop; the protein is methionine-proline and ends. Classification: gene mutation, substitution, nonsense. Impact on phenotype: the protein is truncated and almost certainly nonfunctional, so if the gene were for an essential enzyme, natural selection would act against the allele; if the enzyme's job could be done by another, the effect might be small.

Worked frameshift: Original ATG GCT AAA GGC TAG (methionine-alanine-lysine-glycine-stop). An extra T is inserted after ATG: ATG TGC TAA AGG CTA G. The mRNA AUG UGC UAA ... gives methionine-cysteine-stop. The insertion of one base shifted the frame, changed the second amino acid and created an early stop at the third codon. Classification: gene mutation, insertion, frameshift. Compare an insertion of three bases, ATG GCT GCT AAA GGC TAG, which simply adds a second alanine and leaves the rest intact.

Worked chromosomal classification: (a) A child has 47 chromosomes with three copies of chromosome 21: chromosomal mutation, trisomy, most often arising from nondisjunction in a parent's meiosis (the chromosome count alone does not show when the error happened). (b) A person has 45 chromosomes with a single X: chromosomal mutation, monosomy. (c) A stretch of chromosome 9 is found attached to chromosome 22: chromosomal mutation, translocation; a particular translocation of this kind produces a fused gene that drives a form of leukemia, which shows that chromosomal changes can create abnormal proteins as well as change gene dosage. (d) One base changed in one gene: gene mutation, substitution, to be classified further by its effect on the codon.

Worked selection reasoning: In a hypothetical insect population, a substitution makes an enzyme that breaks down a pesticide. Where the pesticide is sprayed, carriers survive and reproduce more, so the allele is favored and spreads. Where no pesticide is used, the extra enzyme costs a little energy and the allele is slightly selected against or is nearly neutral. The same mutation is therefore beneficial in one environment and useless or mildly harmful in another; the mutation did not arise because of the spraying, but spraying decided its fate.

Section 3 · DoPractice, demonstrate, and explain

Supported record: Build a table with columns Original codons, Mutated codons, mRNA, Amino acids, Type of mutation, Likely effect on protein, and fill five rows for the coding strand ATG AAA GGC TTT TAA: the original and one silent, one missense, one nonsense and one frameshift version of your own design. Check every codon against a codon table. Below the table, sort these terms into two lists headed Gene mutations and Chromosomal mutations: substitution, trisomy, deletion of one base, monosomy, insertion, translocation, inversion.

Independent performance: Write an analysis of the sickle cell mutation for a reader who knows what a codon is. Success criteria: (a) it states the original and mutated codons and the original and substituted amino acids; (b) it classifies the mutation by type and explains the effect on the hemoglobin protein and on red blood cells; (c) it explains, with the words favor, select against and neutral, how natural selection treats the allele in a region with malaria and in a region without, and why the answers differ; (d) it explains why a person with one copy and a person with two copies have different phenotypes; (e) it contrasts this gene mutation with a chromosomal mutation such as trisomy 21 in terms of how many genes are affected; (f) it states one thing the paper-strip model cannot show about mutations.

Section 4 · Check7-question lesson practice test

An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.

Problem 1

Classify each of the following as a gene mutation or a chromosomal mutation and name the specific type: (a) one base replaced by another; (b) an extra copy of chromosome 21; (c) two bases removed from the middle of a gene; (d) a chromosome segment reversed end to end.

Show worked solution

(a) Gene mutation, substitution. (b) Chromosomal mutation, trisomy, usually caused by nondisjunction during a parent's meiosis. (c) Gene mutation, deletion; if the two bases are inside the coding sequence, it also shifts the reading frame, because two is not a multiple of three, so it is a frameshift. (d) Chromosomal mutation, inversion. Gene mutations change the sequence within one gene; chromosomal mutations change the structure or number of whole chromosomes and therefore affect many genes at once.

Problem 2

The coding strand ATG TGG AAA TAA is mutated to ATG TGA AAA TAA. Write both proteins and classify the mutation (UGG is tryptophan, AAA is lysine, UGA and UAA are stops).

Show worked solution

Original mRNA AUG UGG AAA UAA gives methionine-tryptophan-lysine-stop. Mutated mRNA AUG UGA AAA UAA: UGA is a stop codon, so the protein is methionine only. This is a gene mutation, a substitution of G by A, and because it creates a premature stop it is a nonsense mutation. The truncated product lacks nearly the whole chain and would almost certainly be nonfunctional.

Problem 3

Explain why deleting one base from the coding sequence of a gene is usually far more damaging than substituting one base there, and describe the exception in which a deletion is comparatively mild.

Show worked solution

A substitution changes at most one codon, and because the genetic code is redundant it may not change the amino acid at all; only if it creates a stop or alters a critical amino acid is it severe. Deleting one base from the coding sequence shifts the reading frame, so every codon after the deletion is read differently, scrambling the remainder of the protein and often producing an early stop. The exception is a deletion of exactly three bases, or a multiple of three, within the coding sequence, which removes whole amino acids while keeping the frame, so the rest of the protein is unchanged; it is often milder, though removing a critical amino acid can still be severe.

Problem 4

A mutation occurs in a skin cell of an adult man, and a different mutation occurs in one copy of an autosomal gene in a diploid sperm-forming cell in his testis, before that cell copies its DNA for meiosis. Compare the consequences for the individual and for the next generation.

Show worked solution

The skin cell mutation is somatic: it is copied into every cell that develops from that skin cell, so it could affect a patch of skin, and if it disrupts growth control it could contribute to a tumor, but it cannot be inherited and ends with the individual. The mutation in the sperm-forming cell is germline: because the sperm-forming cell is diploid and the mutation is on one of its two copies, it will be in about half of the sperm descended from that cell (had it arisen later, in a cell that was already haploid, it would be in that one sperm only); if such a sperm fertilizes an egg the mutation will be present in every cell of the child, and the child can in turn pass it to about half of his or her own children, so it can continue down the generations.

Problem 5

Using the sickle cell allele as the example, explain how natural selection can favor and select against the same mutation.

Show worked solution

The allele arises from one substitution that changes glutamic acid to valine at the sixth position of beta-globin. People with two copies have sickle cell disease, so selection acts against the allele wherever it occurs. People with one copy are largely healthy and are substantially protected against severe malaria. In regions where malaria is common, heterozygotes survive and reproduce better than either homozygote, so the allele is maintained at a high frequency; where malaria is absent, the advantage disappears and only the cost remains, so the allele declines. The direction of selection depends on the environment.

Problem 6

Explain what nondisjunction is, in which cell divisions it can occur, and how nondisjunction in meiosis leads to trisomy in one offspring and monosomy in another.

Show worked solution

Nondisjunction is the failure of chromosomes to separate during a cell division. It can occur in mitosis, which leaves a mixture of cell lines in one body, or in meiosis, when homologous chromosomes or sister chromatids fail to separate. In meiosis one resulting gamete receives two copies of that chromosome and another receives none. If the gamete with two copies is fertilized by a normal gamete, the zygote has three copies, a trisomy; if the gamete with none is fertilized, the zygote has one copy, a monosomy. Because a chromosome carries many genes, both conditions change the dosage of hundreds of proteins at once, which is why most are severe or not survivable.

Problem 7

Three genes show recombination frequencies of 8 percent between K and L, 15 percent between L and M and 23 percent between K and M; a fourth gene N recombines at 50 percent with all three. Draw the chromosome map in words and state which genes form a linkage group.

Show worked solution

Because 8 + 15 = 23, L lies between K and M: the map is K at 0, L at 8 and M at 23 map units (estimates from the recombination percentages). These three genes recombine with one another at less than 50 percent, so they lie on the same chromosome and form one linkage group. N recombines with each at 50 percent, the rate independent assortment produces, so N shows no linkage to them; most likely it is on a different chromosome, though a gene very far beyond K or M on the same chromosome, where multiple crossovers make observed recombination level off near 50 percent, would give the same numbers. Chromosome maps are read this way: distances below 50 place genes together in one linkage group, and 50 percent means no detectable linkage.