Questions and Answers for Chapter 15
Question1. In the chapter-opening photograph of kernels on an ear of corn , what is the genetic basis of the following :
a. the fully pigmented kernel?
b. the un-pigmented kernels? Note that they can arise in two different ways.
Answer:
a. The fully pigmented kernel resulted from the wild-type copy of the C gene expressed in all cells.
b. The un-pigmented kernels can result from a loss of the C gene by Ac- activated chromosome breakage at a Ds element, resulting in appearance of the recessive, colorless phenotype. Additionally, insertion of the Ds element into the C gene in the absence of an Ac element to mediate its excision will result in colorless kernels.
Question3. For Figure 15-7, draw out a series of steps that could explain the origin of this large plasmid containing many transposable elements.
Answer:One series of events leading to the generation of the R plasmid in figure 15-7 is shown. For example, the insertion of Tn3 into Tn4 could have happened in a previous host genome or plasmid, and both elements could transfer together to this plasmid.
Question 8. Describe the generation of multiple-drug resistant plasmids.
Answer: R plasmids are the main carriers of drug resistance. These plasmids are self-replicating and contain any number of genes for drug resistance, as well as the genes necessary for the transfer by conjugation (called the RTF region).it is the R plasmid’s ability to transfer rapidly to other cells, even those related species that allows drug resistance to spread so rapidly. R plasmids acquire drug resistance genes through transposition. Drug resistance genes are found flanked by IR (inverted repeat) sequences and as a unit are known as transposons. Many transposons have been identified, and as a set they encode a wide range of drug resistances. Because transposons can jump between DNA molecules (e.g., from one plasmid to another or from a plasmid to the bacterial chromosome and vice versa), plasmids can continue to gain new drug-resistance genes as they mix and spread through different strains of cells. It is a classical example of evolution through natural selection. Those cells harboring R plasmids with multiple drug resistances survive to reproduce in the new environment of antibiotic use.
Question 11. Although class 2 elements are abundant in the genomes of multicellular eukaryotes, class I elements usually make up the largest fraction of very large genomes such as those from humans (~2500 Mb), maize (~2500 Mb) and barley (~5000 Mb). Given what you know about class I and class 2 elements what is it about their distinct mechanisms of transposition that would account for this consistent difference in abundance?
Most eukaryotic class 2 elements transpose by a cut and paste mechanism that involves their excision from one site ans reinsertion into a new site elsewhere in the genome. Class 2 element can increase their copy number in a few ways; one is by excising from a site that has been replicated into another site that has not yet been replicated. On the other hand, for class 1 elements RNA transcript is the transposition intermediate (it is copied into double-stranded DNA and reinserted into the genome). As such, a single class 1 element can give rise to many transcripts , and each can theoretically be copied into double stranded DNA and reinserted. However, because the reinsertion of so many DNA copies into the genome would almost certainly harm the host, mechanisms have evolved to prevent this from happening. For example, the transcription of class 1 elements is usually repressed. In addition, the host controls the conversion of the RNA transcript into –stranded double DNA.
Question 12. As you saw in Figure 15-22, the gens of multicellular eukaryotes often contain many transposable elements. Why do most of these elements not affect the expression of the gene?
Answer12: The transposable elements shown in Figure 15-22, like many in the human genome reside in non –coding sequences, such as introns.
Question 13. What are safe heavens? Are there any places in the much more complicated bacterial genomes that might be a safe haven for insertion elements?
Answer: Some transposable elements have evolved strategies to insert into safe havens-regions of the genome where they will do minimal harm. Safe havens include duplicate genes (such as t RNA or rRNA genes) and other transpsosable elements. Safe havens in bacterial genomes might be very specific sequences between genes or the repeated rRNA genes.
Question 15
The insertion of transposable elements into the genes can alter the normal pattern of expression. In the following situations, describe the possible consequences on gene expression.
a. A LINE inserts into an enhancer of a human gene.
b. A transposable element contains a binding site for a transcriptional repressor and inserts adjacent to a promoter.
c. An Alu element inserts into the 3’ splice (AG) site of an intron.
d. A Ds element that was inserted into the exon os a gene excises imperfectly and leaves 3 base pairs behind in the exon.
e. Another excision by the same Ds element leaves 2 base pairs behind in the exon.
f. A Ds element that was inserted into the middle of an intron excises imperfectly and leaves 5 base pairs behind in the intron.
Answer:
a. The consequences will be different for different genes and different insertions. in the simplest scenario, the insertion prevents binding of transcriptional activators that are required for the ultimate binding of RNA polymerase to the promoter. In this case, the gene will not be expressed ( no mRNA will be synthesized).
In a more complex scenarios, the gene may be regulated by many enhancers ( as is the case for most human genes). For example, one enhancer might be required for transcription in the liver, one required for transcription in muscle cells, etc. In this instance, insertion of the LINE into the liver enhancer may prevent transcription of the gene in the liver but not interfere with its transcription in muscle.
b. Again, the consequences will differ depending on the gene that has sustained the insertion. In the simplest scenario, the presence of transposable element will provide a binding site for the transcriptional repressor to bind near the promoter and prevent the binding of RNA polymerase II.
c. The Alu element will be transcribed into RNA with rest of the gene sequences and will prevent the splicing of the intron that is has inserted into. The insertion will almost certainly result in a null allele as the Alu sequence and the intron will now be translated. The intron, Alu, or both probably will contain stop codons.
d. the insertion and excision will result in a 3-bp indel in the exon and slightly alter the amino acis sequence of the protein, but it will not produce a frameshift mutation. the minor change in the amino acid sequence may or may not affect the function of the encoded protein.
e. This insertion and excision will produce a frameshift mutation and is more likely than (d) to impair protein function.
f. Chances are that this mutation will not affect gene expression as the intron will probably still be spliced correctly.
Question 16
Before the integration of a transposon, its transposase makes a staggered cut in the host target DNA. If the staggered cut is at the sites of the arrows below, draw what the sequence of the host DNA will be after the transposon has been inserted. Represent the transposon as a rectangle.
AATTTGGCC TAGTACTAATTGGTTGG
TTAAACCGGATCATGATT AACCAACC
AATTTGGCCTAGTACTAA TAGTACTAATTGGTTGG
TTAAACCGGATCATGATT ATCATGATTAACCAACC
Question 18.
Consider two maize plants:
a Genotype C/cm; Ac/Ac+, where cm is an unstable allele caused by Ds insertion.
b. Genotype C/ cm, where cm is an unstable allele caused by Ac insertion.
What phenotypes would be produced and in what proportions when (1) each plant is crossed with a base-pair-substitution mutant t c/c and (2) the plant part a is crossed with the plant in par b?? Assume that Ac and c are unlinked, that the chromosome- breakage frequency is negligible , and that mutant c/ C is Ac+.
Answer:
In the Ac/ Ds system Ac can produce an unstable allele that is autonomous. Ds can revert only in the presence of Ac and is nonautonomous. In the following, Ac+ indicates the absence of the Ac regulator gene.
cross 1 P C/cDS; Ac/Ac+ X c/c; Ac/Ac+
F1 ¼ C/c; Ac/Ac+ ( solid pigment)
¼ C/c; Ac+/Ac+ ( solid pigment)
¼ cDS/c; Ac/Ac+( unstable colorless or spotted)
¼ cDS/c; Ac+/Ac+ ( colorless)
Overall: 2 solid : 1 spotted : 1 colorless
Cross 2
P C/cAC X c/c
F1 ½ C/c (solid pigment)
½ C/cAc ( unstable colorless or spotted)
Overall: 1 solid: 1 spotted
Cross 3
P C/cDS ; Ac/Ac+ X C/cAc ; Ac+/Ac+
F1 1/8 C/C ; Ac/Ac+ (solid pigment)
1/8 C/CAc ; Ac/Ac+ ( solid pigment)
1/8 C/C ; Ac+/Ac+ (solid pigment)
1/8 C/cAc; Ac+/Ac+ (solid pigment)
1/8 C/cDs; Ac+/Ac+ (solid pigment)
1/8 C/cDs; Ac+/Ac (solid pigment)
1/8 cDs/cAc ; Ac+/Ac+ (unstable colorless or spotted)
1/8 cDs/cAc ; Ac+/Ac (unstable colorless or spotted)
Overall: 3 solid : 1 spotted
Question 19.
You meet your friend, a scientist, at the gym and she begins telling you about a mouse gene that she is studying in the lab. The product of this gene is an enzyme required to make the fur brown. The gene is called FB and the enzyme is called FB enzyme. When FB is mutant and cannot produce the FB enzyme, the fur is white. The scientist tells you that she has isolated the gene from two mice with brown fur and that, surprisingly, she found that the two genes differ by the presence of a 250-bp SINE (like the human Alu element) in the FB gene of one mouse but not in the gene of the other. She does not understand how this difference is possible, especially given that she determined that both mice make the FB enzyme. Can you help her formulate a hypothesis that explains why the mouse can still produce FB enzyme with a transposable element in its FB gene?
Answer:
It would not be surprising to find a SINE element in an intron of a gene, rather than an exon. Processing of the pre-mRNA would remove the transposable element as part of the intron, and translation of the FB enzyme would not be effected.
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