Skip to main content
Biology LibreTexts

4.E: Mutation and Variation (Exercises)

  • Page ID
    4026
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    These are homework exercises to accompany Nickle and Barrette-Ng's "Online Open Genetics" TextMap. Genetics is the scientific study of heredity and the variation of inherited characteristics. It includes the study of genes, themselves, how they function, interact, and produce the visible and measurable characteristics we see in individuals and populations of species as they change from one generation to the next, over time, and in different environments.

    Q4.1

    How are polymorphisms and mutations alike? How are they different?

    Q4.2

    What are some of the ways a substitution can occur in a DNA sequence?

    Q4.3

    What are some of the ways a deletion can occur in a DNA sequence?

    Q4.4

    What are all of the ways an insertion can occur in a DNA sequence?

    Q4.5

    In the context of this chapter, explain the health hazards of smoking tobacco.

    Q4.6

    You have a female fruit fly, whose father was exposed to a mutagen (she, herself, wasn’t). Mating this female fly with another non-mutagenized, wild type male produces offspring that all appear to be completely normal, except there are twice as many daughters as sons in the F1 progeny of this cross.

    1. Propose a hypothesis to explain these observations.
    2. How could you test your hypothesis?

    Q4.7

    You decide to use genetics to investigate how your favourite plant makes its flowers smell good.

    1. What steps will you take to identify some genes that are required for production of the sweet floral scent? Assume that this plant is a self-pollinating diploid.
    2. One of the recessive mutants you identified has fishy-smelling flowers, so you name the mutant (and the mutated gene) fishy. What do you hypothesize about the normal function of the wild-type fishy gene?
    3. Another recessive mutant lacks floral scent altogether, so you call it nosmell. What could you hypothesize about the normal function of this gene?

    Q4.8

    Suppose you are only interested in finding dominant mutations that affect floral scent.

    1. What do you expect to be the relative frequency of dominant mutations, as compared to recessive mutations, and why?
    2. How will you design your screen differently than in the previous question, in order to detect dominant mutations specifically?
    3. Which kind of mutagen is most likely to produce dominant mutations, a mutagen that produces point mutations, or a mutagen that produces large deletions?

    Q4.9

    hich types of transposable elements are transcribed?

    Q4.10

    You are interested in finding genes involved in synthesis of proline (Pro), an amino acid that is normally synthesizes by a particular model organism.

    1. How would you design a mutant screen to identify genes required for Pro synthesis?
    2. Imagine that your screen identified ten mutants (#1 through #10) that grew poorly unless supplemented with Pro. How could you determine the number of different genes represented by these mutants?
    3. If each of the four mutants represents a different gene, what will be the phenotype of the F1 progeny if any pair of the four mutants are crossed?
    4. If each of the four mutants represents the same gene, what will be the phenotype of the F1 progeny if any pair of the four mutants are crossed?

    Chapter 4 - Answers

    4.1 Polymorphisms and mutations are both variations in DNA sequence and can arise through the same mechanisms. We use the term polymorphism to refer to DNA variants that are relatively common in populations. Mutations affect the phenotype.

    4.2 Misreading of bases during replication can lead to substitution and can be caused by things like tautomerism, DNA alkylating agents, and irradiation.

    4.3 Looping out of DNA on the template strand during replication; strand breakage, due to radiation and other mutagens; and (discussed in earlier chapters) chromosomal aberrations such as deletions and translocations.

    4.4 Looping out of DNA on the growing strand during replication; transposition; and (discussed in earlier chapters) chromosomal aberrations such as duplications, insertions, and translocation.

    4.5 Benzopyrene is one of many hazardous compounds present in smoke. Benzopyrene is an intercalating agent, which slides between the bases of the DNA molecule, distorting the shape of the double helix, which disrupts transcription and replication and can lead to mutation.

    4.6 a) One possible explanation is that original mutagenesis resulted in a loss-of-function mutation in a gene that is essential for early embryonic development, and that this mutation is X-linked recessive in the female. Because half of the sons will inherit the X chromosome that bears this mutation, half of the sons will fail to develop beyond very early development and will not be detected among the F1 progeny. The proportion of male flies that were affected depends on what fraction of the female parent’s gametes carried the mutation. In this case, it appears that half of the female’s gametes carried the mutation.

    b) To test whether a gene is X-linked, you can usually do a reciprocal cross. However, in this case it would be impossible to obtain adult male flies that carry the mutation; they are dead. If the hypothesis proposed in a) above is correct, then half of the females, and none of the living males in the F1 should carry the mutant allele. You could therefore cross F1 females to wild type males, and see whether the expected ratios were observed among the offspring (e.g. half of the F1 females should have a fewer male offspring than expected, while the other half of the F1 females and all of the males should have a roughly equal numbers of male and female offspring).

    4.7 a) Treat a population of seeds with a mutagen such as EMS. Allow these seeds to self-pollinate, and then allow the F1 generation to also self-pollinate. In the F2 generation, smell each flower to find individuals with abnormal scent.

    b) The fishy gene appears to be required to make the normal floral scent. Because the flowers smell fishy in the absence of this gene, one possibility explanation of this is that fishy makes an enzyme that converts a fishy-smelling intermediate into a chemical that gives flowers their normal, sweet smell.

    Note that although we show this biochemical pathway as leading from the fishy-smelling chemical to the sweet-smelling chemical in one step, it is likely that there are many other enzymes that act after the fishy enzyme to make the final, sweet-smelling product. In either case, blocking the pathway at the step catalyzed by the fishy enzyme would explain the fishy smell.

    c) In nosmell plants, the normal sweet smell disappears. Unlike fishy, the sweet smell is not replaced by any intermediate chemical that we can easily detect. Thus, we cannot conclude where in the biochemical pathway the nosmell mutant is blocked; nosmell may normally therefore act either before or after fishy normally acts in the pathway:

    Alternatively, nosmell may not be part of the biosynthetic pathway for the sweet smelling chemical at all. It is possible that the normal function of this gene is to transport the sweet-smelling chemical into the cells from which it is released into the air, or maybe it is required for the development of those cells in the first place. It could even be something as general as keeping the plants healthy enough that they have enough energy to do things like produce floral scent.

    4.8 a) Dominant mutations are generally much rarer than recessive mutations. This is because mutation of a gene tends to cause a loss of the normal function of this gene. In most cases, having just one normal (wt) allele is sufficient for normal biological function, so the mutant allele is recessive to the wt allele. Very rarely, rather than destroying normal gene function, the random act of mutation will cause a gene to gain a new function (e.g. to catalyze a new enzymatic reaction), which can be dominant (since it performs this new function whether the wt allele is present or not). This type of gain-of-function dominant mutation is very rare because there are many more ways to randomly destroy something than by random action to give it a new function (think of the example given in class of stomping on an iPod).

    b) Dominant mutations should be detectable in the F1 generation, so the F1 generation, rather than the F2 generation can be screened for the phenotype of interest.

    c) Large deletions, such as those caused by some types of radiation, are generally less likely than point mutations to introduce a new function into a protein: it is hard for a protein to gain a new function if the entire gene has been removed from the genome by deletion.

    4.9 Class I. see Figure 4.5 on Transposable Elements.

    4.10 a) Mutagenize a wild type (auxotrophic) strain and screen for mutations that fail to grow on minimal media, but grow well on minimal media supplemented with proline.
    b) Take mutants #1-#10) and characterize them, based on (1) genetic mapping of the mutants (different locations indicate different genes); (2) different response to proline precursors (a different response suggests different genes); (3) complementation tests among the mutations (if they complement then they are mutations in different genes).
    c) If the mutations are in different genes then the F1 progeny would be wild type (able to grow on minimal medium without proline).
    d) If the mutations are in the same gene then the F1 progeny would NOT be wild type (unable to grow on minimal medium without proline).


    This page titled 4.E: Mutation and Variation (Exercises) is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by Todd Nickle and Isabelle Barrette-Ng via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.