53 terms

Genetics Exam 3 - Study Questions

What three general characteristics must the genetic material possess?
1) Must contain complex info that encodes the phenotype

2) Must replicate or be replicated faithfully

3) Must be able to mutate to generate diversity
How does an RNA nucleotide differ from a DNA nucleotide?
DNA nucleotides = Lacks O on 2' Carbon of Sugar, Thymine

RNA nucleotides = Has O on 2' Carbon of Sugar, Uracil
How does a purine differ from a pyrimidine? What purines and pyrimidines are found in DNA and RNA?
Purine - six sided ring attached to five sided ring

Pyrimidine - one six sided ring

DNA - Purines = AG Pyrimidines = TC

RNA - Purines = AG Pyrimidines = UC
Which bases are capable of forming hydrogen bonds with each other.
A - T 2

C - G 3
For each organism, compute the ratio of (A + G)/(T + C) and the ratio of (A + T)/(C + G).

Organism and tissue A-G-C-T
Sheep thymus 29.3-21.4-21.0-28.3
Pig liver 29.4-20.5-20.5-29.7
Human thymus 30.9-19.9-19.8-29.4
Human sperm 30.9-19.1-18.4-31.6
Salmon sperm 29.7-20.8-20.4-29.1
Herring sperm 27.8-22.1-20.7-27.5
Sheep Thymus 1.03 and 1.36

Pig Liver 0.99 and 1.44

Human Thymus 1.03 and 1.36

Human Sperm 1.00 and 1.67

Salmon Sperm 1.02 and 1.43

Herring Sperm 1.04 and 1.29

b. Are these ratios constant or do they vary among the organisms? Explain why.
The ratios for the (A + G)/(T + C) are constant at approximately 1.0 for the different organisms. Each of these organisms contains a double-stranded genome. The percentages of guanine and cytosine are almost equal to each other and the percentages of adenine and thymine are almost equal to each other as well. In other words, the percentage of purines should be equal to the percentage of pyrimidines for double-stranded DNA. This means that (A + G) = (C + T). The (A + T)/(C + G) ratios are not constant. The relative numbers of AT base pairs and GC base pairs are unique to each organism and can vary between the different species.
c. Is the (A + G)/(T + C) ratio different for the sperm samples? Would you expect it to be? Why or why not?
The ratios for the two sperm samples are essentially the same. The equal ratio should be expected. As stated in the answer to part b. of this question, the percentage of purines should equal the percentage of pyrimidines.
Which of the following relations will be found in the percentages of bases of a double-stranded DNA molecule?
A double-stranded DNA molecule will contain equal percentages of A and T nucleotides and equal percentages of G and C nucleotides. The combined percentage of A and T bases added to the combined percentage of the G and C bases should equal 100.

ie. A + T = G + C NO
A + G = T + C YES
If a double-stranded DNA molecule has 15% thymine, what are the percentages of all the other bases?
The percentage of thymine (15%) should be approximately equal to the percentage of adenine (15%). The remaining percentage of DNA bases will consist of cytosine and guanine bases (100% - 15% - 15% = 70%); these should be in equal amounts (70%/2 = 35%). Therefore, the percentages of each of the other bases if the thymine content is 15% are adenine = 15%; guanine = 35%; and cytosine = 35%.
Describe the composition and structure of the nucleosome. How do core particles differ from chromatosomes?
Nucleosome core particle = 2 moles of each histone H2A, H2b, H3, and H4 which form a protein core with 145-147 bp of DNA wound around the core.

Chromatosomes contain the nucleosome core with a molecule of histone H1.
Describe in steps how the double helix of DNA, which is 2 nm in width, gives rise to a chromosome that is 700 nm in width.
DNA - > Nucleosomes

Nucleosomes - > 30nm fiber

30nm fiber - > 250nm fiber

250nm fiber - > 700nm chromatid
What is the difference between euchromatin and heterochromatin?
Euchromatin undergoes regular cycles of condensation during mitosis and decondensation during interphase.

Heterochromatin remains highly condensed throughout the cell cycle, except transiently during replication.

Nearly all transcription takes place in euchromatic regions, with little or no transcription of heterochromatin.
How did Meselson and Stahl demonstrate that replication in E. coli takes place in a semiconservative manner?
15N was incorporated in the DNA of E. coli cells
These cells were switched into a 14N medium
Samples removed at each generation
Centrifuging separate them by weight over generations
Eventually created only 2 bands
What similarities and differences exist in the enzymatic activities of DNA polymerases I, II, and III? What is the function of each type of DNA polymerase in bacterial cells?
All 3 have 5' to 3' polymerase activity.

Only DNA Poly I has 3' to 5' exonuclease activity as well.

DNA Poly I - Proofreading. Removes and replaces RNA primers

DNA Poly II - DNA repair polymerase. Restarts replication after DNA dmg has halted replication. Also proofreading.

DNA Poly III - Primary replication enzyme and also has a proofreading function in replication.
Why is primase required for replication?
Primase is a DNA-dependent RNA polymerase.

Primase synthesizes the short RNA molecules, or primers, that have a free 3'-OH to which DNA polymerase can attach deoxyribonucleotides in replication initiation.

The DNA polymerases require a free 3'-OH to which they add nucleotides, and therefore they cannot initiate replication.

Primase does not have this requirement.
How does replication licensing ensure that DNA is replicated only once at each origin per cell cycle?
Only replication origins to which replication licensing factor (RPF) has bound can undergo initiation. Shortly after the completion of mitosis, RPF binds the origin during G1 and is removed by the replication machinery during S phase.
In what ways is eukaryotic replication similar to bacterial replication, and in what ways is it different?
Origin serves as starting point
Primers provide 3-OH' for polymerases to begin synth
Synth is 5'-3'
Template strand is read in 3' to 5' direction
Deoxyribonucleoside triphosphates are substrates
Continuous on leading and discontinuous on lagging

Euk has multiple origins of replication per chromosome
Euk has several diff DNA polymerases with diff functions
Euk has assembly of nucleosomes take place immediately after DNA replication
What is the end-of-chromosome problem for replication? Why, in the absence of telomerase, do the ends of chromosomes get progressively shorter each time the DNA is replicated?
For DNA polymerases to work, they need the presence of a 3' OH group to which to add a nucleotide. At the ends of the chromosomes when the RNA primer is removed, there is no adjacent 3' OH group to which to add a nucleotide, thus no nucleotides are added leaving a gap at the end of the chromosome. Telomerase can extend the single stranded protruding end by pairing with the overhanging 3' end of the DNA and adding a repeated sequence of nucleotides. In the absence of telomerase, DNA polymerase will be unable to add nucleotides to the end of the strand. After multiple rounds of replication without a functional telomerase the chromosome will become progressively shorter.
Outline in words and pictures how telomeres at the end of eukaryotic chromosomes are replicated.
Telomeres are replicated by the enzyme telomerase. Telomerase, a ribonucleoprotein, consists of protein and an RNA molecule that is complementary to the 3' end of the DNA of a eukaryotic chromosome. The RNA molecule also serves as a template for the addition of nucleotides to the 3' end. After the 3' end has been extended, the 5' end of the DNA can be extended as well, possibly by lagging strand synthesis of a DNA polymerase using the extended 3' end as a template.

1) DNA replication of the linear eukaryotic chromosomes generates a 3' overhang. Part of the RNA sequence within telomerase is complementary to the overhang.

2) Telomerase RNA sequence pairs with the 3' overhang and serves as a template for the addition of DNA nucleotides to the 3' end of the DNA molecule, which serves to extend the 3' end of the chromosome.

3) Additional nucleotides to the 5' end are added by DNA synthesis using a DNA polymerase with priming by primase.
Phosphorous is required to synthesize the deoxyribonucleoside triphosphates used in DNA replication. A geneticist grows some E. coli in a medium containing nonradioactive phosphorous for many generations. A sample of the bacteria is then transferred to a medium that contains a radioactive isotope of phosphorus (32P). Samples of the bacteria are removed immediately after the transfer and after one and two rounds of replication. What will be the distribution of radioactivity in the DNA of the bacteria in each sample? Will radioactivity be detected in neither, one, or both strands of the DNA?
In the initial sample removed immediately after transfer, no 32P should be incorporated into the DNA because replication in the medium containing 32P has not yet occurred. After one round of replication in the 32P containing medium, one strand of each newly synthesized DNA molecule will contain 32P, while the other strand will contain only nonradioactive phosphorous. After two rounds of replication in the 32P containing medium, 50% of the DNA molecules will have 32P in both strands, while the remaining 50% will contain 32P in one strand and nonradioactive phosphorous in the other strand.
A circular molecule of DNA contains 1 million base pairs. If DNA synthesis at a replication fork occurs at a rate of 100,000 nucleotides per minute, how long will theta replication require to completely replicate the molecule, assuming that theta replication is bidirectional? How long will replication of this circular chromosome take by rolling-circle replication? Ignore replication of the displaced strand in rolling-circle replication.
In bidirectional replication there are two replication forks, each proceeding at a rate of 100,000 nucleotides per minute. Therefore, it would require 5 minutes for the circular DNA molecule to be replicated by bidirectional replication because each fork could synthesize 500,000 nucleotides (5 minutes × 100,000 nucleotides per minute) within the time period. Because rolling-circle replication is unidirectional and thus has only one replication fork, 10 minutes will be required to replicate the entire circular molecule.
A bacterium synthesizes DNA at each replication fork at a rate of 1000 nucleotides per second. If this bacterium completely replicates its circular chromosome by theta replication in 30 minutes, how many base pairs of DNA will its chromosome contain?
Each replication complex is synthesizing DNA at each fork at a rate of 1000 nucleotides per second. So for each second, 2000 nucleotides are being synthesized by both forks (1000 nucleotides / second × 2 forks = 2000 nucleotides / second) or 120,000 nucleotides per minute. If the bacterium requires 30 minutes to replicate its chromosome, then the size of the chromosome is 3,600,00 nucleotides (120,000 nucleotides / minute × 30 minutes = 3,600,000).
How would DNA replication be affected in a cell that is lacking topoisomerase?
Topoisomerase II or gyrase reduces the positive supercoiling or torsional strain that develops ahead of the replication fork due to the unwinding of the double helix. If the topoisomerase activity was lacking, then the torsional strain would continue to increase, making it more difficult to unwind the double helix. Ultimately, the increasing strain would lead to an inhibition of the replication fork movement.
If the gene for primase were mutated so that no functional primase was produced, what would be effect on theta replication? On rolling-circle replication?
Primase is required for replication initiation in theta form replication. If primase is nonfunctional then replication initiation would not take place resulting in no replication occurring. Rolling-circle replication does not require primase. A single-stranded break within one strand provides a 3 OH group to which nucleotides can be added so rolling circle replication could occur without a functional primase.
How is the structure of RNA similar to that of DNA? How is it different?
RNA - ribose sugar, uracil, single stranded

DNA- deoxyribose sugar, thymine, double stranded
Why is DNA more stable than RNA?
2' OH in ribose of RNA is more susceptible to degradation under alkaline conditions.
What parts of DNA make up a transcription unit?
Promoter | Transcription start site | Terminator

RNA coding region includes Transcription start site and terminator
What is the substrate for RNA synthesis? How is this substrate modified and joined together to produce an RNA molecule?
4 ribonucleoside triphosphates:
adenosine triphosphate
guanosine triphosphate
cytosine triphosphate
uridine monophosphate

Enzyme RNA Polymerase uses DNA polynucleotide strand as a template to synth a complementary RNA strand.

Nucleotides added to RNA one at a time, at 3-OH of RNA molecule

As each nucleoside triphosphate is added, two phosphates are removed from the 5' end of the nucleotide.

The remaining phosphate is linked to the 3-OH of the RNA forming a phosphodiester bond.
Give the names of the three RNA polymerases found in eukaryotic cells and the types of RNA they transcribe.
RNA Pol I - transcribes rRNA

RNA Pol II - transcribes pre-mRNA, snoRNAs, and some miRNAs and snRNAs

RNA Pol III - transcribes small RNA molecules such as 5S rRNA, tRNAs, and some snRNAs and miRNAs

RNA Pol IV - transcribes siRNAs in plants
What are the three basic stages of transcription? Describe what happens at each stage.
Initiation: Transcription proteins assemble at the promoter to form the basal transcription apparatus and begin synthesis of RNA

Elongation: RNA polymerase moves along the DNA template in a 3' to 5' direction, unwinding the DNA and synthing the RNA in a 5' to 3' direction

Termination: Synth of RNA is terminated and the RNA molecule separates from the DNA template.
How is the process of transcription in eukaryotic cells different from that in bacterial cells?
Eukaryotic transcription requires the action of three RNA polymerases. Each type of polymerase recognizes and transcribes from different types of promoters. Binding to the promoter and initiation from the promoter requires the action of many protein transcription factors; different promoters require different sets of protein factors. The RNA molecule produced by transcription in eukaryotic cells usually requires extensive posttranscriptional processing, such as the addition of a 5' cap, a 3' poly(A) tail, and the removal of introns prior to becoming functional. Bacterial promoters tend to be more uniform in composition, and only one RNA polymerase does transcription. Bacterial RNAs are typically functional once transcription has taken place.
Compare and contrast transcription and replication. How are these processes similar and how are they different?
Utilize DNA template
Synth molecules 5' to 3'
Synth molecules antiparallel and complementary to template
Use nucleotide triphosphates as substrates
Involve complexes of proteins and enzymes necessary for catalysis

Bidirectional synth of two strands of nucleic acid
Initiates from replication origins

Unidirectional synth of only a single strand of nucleic acid
Initiation does not require a primer
Subject to numerous regulatory mechs
Each gene is transcribed separately
How are the processes of transcription in archaeans and eukaryotes different? How are they similar?
A comparison of transcription between eukaryotes and archaea shows that transcription in eukaryotes shares more similarities with archaeal transcription than they do with transcription in bacteria.

1) Organisms in domain archaea use a single RNA pol for transcription, while euk have three RNA poly.
2) Archaea lack a nucleus so transcription occurs in cytoplasm.
3) RNA Poly from archaea is similar to RNA poly of euk
4) Archaea possess a TATA binding protein, also found in euk
5) Archaea TATA binding protein binds to TATA box with help of TFIIB, which is also found in euk.
An RNA molecule has the following percentages of bases: A = 23%, U = 42%,
C = 21%, G = 14%.
a. Is this RNA single-stranded or double-stranded? How can you tell?

b. What would be the percentages of bases in the template strand of the DNA that contains the gene for this RNA?
a) The RNA molecule is likely to be single-stranded. If the molecule was double-stranded, we would expect nearly equal percentages of adenine and uracil, as well as equal percentages of guanine and cytosine. In this RNA molecule, the percentages of these potential base pairs are not equal, so the molecule is single-stranded.

b) Because the DNA template strand is complementary to the RNA molecule, we would expect equal percentages for bases in the DNA complementary to the RNA bases. Therefore, in the DNA we would expect A = 42%, T = 23%, C = 14%, and G = 21%.
The following diagram represents DNA that is part of the RNA-coding sequence of a transcription unit. The bottom strand is the template strand. Give the sequence found on the RNA molecule transcribed from this DNA and label the 5′ and 3′ ends of the RNA.

5′-A T A G G C G A T G C C A-3′
3′-T A T C C G C T A C G G T-5′ <= template strand
The RNA molecule would be complementary to the template strand, contain uracil, and be synthesized in an antiparallel fashion. The sequence would be:
5′-A U A G G C G A U G C C A-3′.
The RNA strand contains the same sequence as the nontemplate DNA strand except that the RNA strand contains uracil in place of thymine.
The following sequence of nucleotides is found in a single-stranded DNA template:
Assume that RNA polymerase proceeds along this template from left to right.
a. Which end of the DNA template is 5′ and which end is 3′?

b. Give the sequence and label the 5' and 3' ends of the RNA copied from this template.
a) RNA is synthesized in a 5′ to 3′ direction by RNA polymerase, which reads the DNA template in a 3′ to 5′ direction. So, if the polymerase is moving from left to right on the template then the 3′ end must be on the left and the 5′ end on the right.
3′-A T T G C C A G A T C A T C C C A A T A G A T-5′

b) 5′-U A A C G G U C U A G U A G G G U U A U C U A-3′
RNA polymerases carry out transcription at a much slower rate than DNA polymerases carry out replication. Why is speed more important in replication than in transcription?
DNA polymerases are required to replicate much larger regions of DNA, such as entire chromosomes. Speed is essential to complete the replication process in a timely manner. RNA polymerases typically transcribe only small areas of the chromosomes. The speed required for replication by DNA polymerases is not needed by the RNA polymerases to transcribe these smaller regions.
Most RNA molecules have three phosphates at their 5′ end, but DNA molecules never do. Explain this difference.
During initiation of DNA replication, DNA nucleoside triphosphates must be attached to a 3′-OH of a RNA molecule by DNA polymerase. This process removes the terminal two phosphates of the nucleotides. If the RNA molecule is subsequently removed, then a single phosphate would remain at the 5′ end of the DNA molecule. RNA polymerase does not require the 3′-OH to initiate synthesis of RNA molecules. Therefore, the 5′ end of a RNA molecule will retain all three of the phosphates from the original nucleotide triphosphate substrate.
The following diagram represents a transcription unit in a hypothetical DNA molecule:
a. On the basis of the information given, is this DNA from a bacterium or from a eukaryotic organism?

b. If this DNA molecule is transcribed, which strand will be the template strand and which will be the nontemplate strand?

c. Where, approximately, will the start site of transcription be?
a) The DNA is from a bacterium as evidenced by the TATAAT sequence that corresponds to the -10 consensus sequence of a bacterial promoter and the TTGACA sequence that is identical to the -35 consensus sequence of a bacterial promoter.

b) 5′...TTGACA...TATAAT...3′ Nontemplate
3′...AACTGT...ATATTA...5' Template

c) The start site should be located approximately 9 nucleotides downstream from the last nucleotide of the -10 consensus sequence (TATAAT) or the +1 nucleotide.
What is the function of the Shine-Dalgarno consensus sequence?
The Shine-Dalgarno consensus sequence functions as the ribosome-binding site on the mRNA molecule.
a. What is the 5′ cap?
b. How is the 5′ cap added to eukaryotic pre-mRNA?
c. What is the function of the 5′ cap?
a) The 5′ end of eukaryotic mRNA is modified by the addition of the 5′ cap. The cap consists of an extra guanine nucleotide linked 5′ to 5′ to the mRNA molecule. This nucleotide is methylated at position 7 of the base. The ribose sugars of adjacent bases may be methylated at the 2′ -OH.

b) How is the 5′ cap added to eukaryotic pre-mRNA?
Initially, the terminal phosphate of the three 5′ phosphates linked to the end of the mRNA molecule is removed. Subsequently, a guanine nucleotide is attached to the 5′ end of the mRNA using a 5′ to 5′ phosphate linkage. Next, a methyl group is attached to position 7 of the guanine base. Ribose sugars of adjacent nucleotides may also be methylated, but at the 2′-OH.

c)CAP binding proteins recognize the 5′ cap and stimulate binding of the ribosome to the 5′ cap and to the mRNA molecule. The 5′ cap may also increase mRNA stability in the cytoplasm. Finally, the 5′ cap is needed for efficient splicing of the intron that is nearest the 5′ end of the pre-mRNA molecule.
How is the poly(A) tail added to pre-mRNA? What is the purpose of the poly(A) tail?
Initially, a complex consisting of several proteins forms on the 3′ UTR of the pre-mRNA molecule. Cleavage and polyadenylation specificity factor (CPSF) bind to the AAUAAA consensus sequence, which is located upstream of the 3′ cleavage site. Another protein, cleavage stimulation factor (CsTF), binds downstream of the cleavage site. Two cleavage factors (CFI and CFII) and polyadenylate polymerase (PAP) also become part of the complex. Once the complex has formed, the pre-mRNA is cleaved. CsTF and the two cleavage factors leave the complex. PAP adds approximately 10 adenine nucleotides to the 3′ end of the pre-mRNA molecule. The addition of the short poly(A) tail allows for the binding of the poly(A) binding protein (PABII) to the tail. PABII increases the rate of polyadenylation, which subsequently allows for more PABII protein to bind the tail.
The presence of the poly(A) tail increases the stability of the mRNA molecule through the interaction of proteins at the poly(A) tail. The length of the poly(A) tail influences the time in which the transcript remains intact and available for translation. The poly(A) tail also assists with the binding of the ribosome to the mRNA.
What makes up the spliceosome? What is the function of the spliceosome?
The spliceosome consists of five small ribonucleoproteins (snRNPs). Each snRNP is composed of multiple proteins and a single small nuclear RNA molecule or snRNA. The snRNPs are identified by which snRNA (U1, U2, U3, U4, U5, or U6) each contains. Splicing of pre-mRNA nuclear introns takes place within the spliceosome.
Explain the process of pre-mRNA splicing in nuclear genes.
Removal of an intron from the pre-mRNA requires the assembly of the spliceosome complex on the pre-mRNA, cleavage at both the 5′ and 3′ splice sites of the intron, and two transesterification reactions ultimately leading to the joining of the two exons. Initially, snRNP U1 binds to the 5′ splice site through complementary base pairing of the U1 snRNA. Next, snRNP U2 binds to the branch point within the intron. The U5 and U4-U6 complex joins the spliceosome, resulting in the looping of the intron so that the branch point and 5′ splice site of the intron are now adjacent to each other. U1 and U4 now disassociate from the spliceosome and the spliceosome is activated. The pre-mRNA is then cleaved at the 5′ splice site, producing an exon with a 3′ -OH. The 5′ end of the intron folds back and forms 5′-2′ phosphodiester linkage through the first transesterification reaction with the adenine nucleotide at the branch point of the intron. This looped structure is called the lariat. Next, the 3' splice site is cleaved and then immediately ligated to the 3′-OH of the first exon through the second transesterification reaction. Thus, the exons are now joined and the intron has been excised.
Summarize the different types of processing that can take place in pre-mRNA.
Addition of the 5′ 7-mG cap to the 5′ end of the pre-mRNA
Cleavage of the 3′ end of a site downstream of the AAUAAA consensus sequence of the last exon
Addition of the poly(A) tail to the 3′ end of the mRNA immediately following cleavage
Removal of the introns (splicing)
How do the mRNAs of bacterial cells and the pre-mRNAs of eukaryotic cells differ? How do the mature mRNAs of bacterial and eukaryotic cells differ?
Bacterial mRNA is translated immediately upon being transcribed. Eukaryotic pre-mRNA must be processed. Bacterial mRNA and eukaryotic pre-mRNA have similarities in structure. Each has a 5′ untranslated region as well as a 3′ untranslated region. Both also have protein-coding regions. However, the protein-coding region of the pre-mRNA is disrupted by introns. The eukaryotic pre-mRNA must be processed to produce the mature mRNA. Eukaryotic mRNA has a 5′ cap and a poly(A) tail, unlike bacterial mRNAs. Bacterial mRNA also contains the Shine-Dalgarno consensus sequence. Eukaryotic mRNA does not have the equivalent.
Identify the following items and, for each item, give a brief description of its function.

a. 5 untranslated region
b. promoter
c. AAUAAA consensus sequence
d. Transcription start site
e. 3 untranslated region
f. introns
g. exons
h. Poly(A)tail
i. 5' cap
a. 5′ untranslated region
The 5′ untranslated region lies upstream of the translation start site. In bacteria, the ribosome binding site or Shine-Dalgarno sequence is found within the 5′ untranslated region. However, eukaryotic mRNA does not have the equivalent sequence, and a eukaryotic ribosome binds at the 5′ cap of the mRNA molecule.
b. Promoter
The promoter is the DNA sequence that the transcription apparatus recognizes and binds to initiate transcription.
c. AAUAAA consensus sequence
The AAUAAA consensus sequence lies downstream of the coding region of the gene. It determines the location of the 3′ cleavage site in the pre-mRNA molecule.
d. Transcription start site
The transcription start site begins the coding region of the gene and is located 25 to 30 nucleotides downstream of the TATA box.
e. 3′ untranslated region
The 3′ untranslated region is a sequence of nucleotides at the 3′ end of the mRNA that is not translated into proteins. However, it does affect the translation of the mRNA molecule as well as the stability of the mRNA.
f. Introns
Introns are noncoding sequences of DNA that intervene within coding regions of a gene.
g. Exons
Exons are transcribed regions that are not removed in intron processing. They include the 5′UTR, coding regions that are translated into amino acid sequences, and the 3′UTR.
h. Poly(A) tail
A poly(A) tail is added to the 3' end of the pre-mRNA. It affects mRNA stability.
i. 5′ cap
The 5′ cap functions in the initiation of translation and mRNA stability.
Suppose that a mutation occurs in an intron of a gene encoding a protein. What will the most likely effect of the mutation be on the amino acid sequence of that protein? Explain your answer.
Because introns are removed prior to translation, an intron mutation would have little effect on a protein's amino acid sequence unless the mutation occurred within the 5′ splice site, the 3′ splice site, or the branch point. If mutations within these sequences altered splicing, then the mature mRNA would be altered, thus altering the amino acid sequence of the protein. The result could be a protein with additional amino acid sequence. Or, possibly, the altered splicing could introduce a stop codon that stops translation prematurely. If a mutation in the intron induced a frameshift, the reading frame and the amino acid sequence would be altered.
How is biological evolution defined?
Biological evolution is change in the genetic composition of a population over time.
What are the two steps in the process of evolution?
First, mutation causes genetic variation in a population and recombination creates new combinations of genetic variants. Second, the frequencies of the variants change over time (from generation to generation), as a result of random processes (e.g., drift) or selection.
What is the biological species concept? What are some of the problems with it?
The biological species concept defines species as a group of individuals that can potentially interbreed with each other but are reproductively isolated from members of other species. One obvious problem is applying this concept to asexually reproducing organisms. Another is that, in practice, most species are defined by phenotypic or anatomical differences, and it may be difficult or impossible to determine whether individuals have the potential to interbreed if they existed at different times or their full life cycles have not been observed and studied.
What is the difference between prezygotic and postzygotic reproductive isolating mechanisms. List the different types of each.
Prezygotic mechanisms operate before fertilization of the egg by sperm (or fusion of gametes), and postzygotic mechanisms operate after fertilization. Prezygotic mechanisms include ecological isolation, behavioral isolation, temporal isolation, mechanical isolation, and gametic isolation. Postzygotic mechanisms include hybrid inviability, hybrid sterility, and hybrid breakdown.
What is the basic difference between allopatric and sympatric modes of speciation?
Allopatric speciation involves populations separated by a geographic barrier that precludes gene flow between the populations. Sympatric speciation takes place between populations occupying the same geographical area.
Briefly outline the process of allopatric speciation.
First, a population is split by a geographical barrier that prevents gene flow between the two groups on either side of the barrier. The two groups then evolve independently; they accumulate genetic differences through various evolutionary processes such as mutation, selection, and random drift. If these genetic differences lead to reproductive isolation, meaning these individuals cannot or will not interbreed, then speciation has occurred.
What is the molecular clock?
The molecular clock is the concept (hypothesis) that the rate at which nucleotide changes take place in a DNA sequence is relatively constant over long periods of time, and therefore the number of nucleotide substitutions that have taken place between two organisms can be used to estimate the time since they last shared a common ancestor