Biology 1001A Lecture 13: bio 1001a lecture 13

58 views1 pages

emeraldcattle11

7 May 2018

School

Western University

Department

Biology

Course

Biology 1001A

Professor

Beth Mac Dougall- Shackleton

For unlimited access to Class Notes, a Class+ subscription is required.

Could’t you hae oegee to get

the same AA/nucleotide sequence

Could’t you hae oegee to get the sae AA/uleotide seuee?

 Poteis a oege to siila futios, ut they do’t eessaily

have the same nucleotide/AA sequence

 This is because natural selection acts on phenotype (protein function)

not genotype (AA/nucleotide sequence). There is no selective pressure to

change nucleotide sequence so it would look similar to another sequence

such that they would be considered homologous (convergent evolution

does’t esult i to sequences looking homologous!)

 Convergence onto highly similar nucleotide sequence is rare.

Therefore, similarity between sequences are considered to be evidence of

homology because there is no force to drive convergence of nucleotide/AA

sequences

There is no reason for 2 sequences to proteins to share high identity

across their entire sequence

There would be localized regions of high similarity, but entire nucleotide

sequence would not be similar

- The functionality of a protein is not represented by its entire

sequence, but by localized regions:

o Location of cysteine amino acids (give rise disulfide (S-S)

bonding )

o Amino acids necessary for catalysis

o DNA binding domains, transcription factors, receptor binding

sites

find more resources at oneclass.com

Unlock document

This preview shows half of the first page of the document.
Unlock all 1 pages and 3 million more documents.

Already have an account? Log in

Document Summary

Could(cid:374)"t you ha(cid:448)e (cid:272)o(cid:374)(cid:448)e(cid:396)ge(cid:374)(cid:272)e to get the same aa/nucleotide sequence. P(cid:396)otei(cid:374)s (cid:272)a(cid:374) (cid:272)o(cid:374)(cid:448)e(cid:396)ge to si(cid:373)ila(cid:396) fu(cid:374)(cid:272)tio(cid:374)s, (cid:271)ut they do(cid:374)"t (cid:374)e(cid:272)essa(cid:396)ily have the same nucleotide/aa sequence. This is because natural selection acts on phenotype (protein function) not genotype (aa/nucleotide sequence). There is no selective pressure to change nucleotide sequence so it would look similar to another sequence such that they would be considered homologous (convergent evolution does(cid:374)"t (cid:396)esult i(cid:374) t(cid:449)o sequences looking homologous!) Convergence onto highly similar nucleotide sequence is rare. Therefore, similarity between sequences are considered to be evidence of homology because there is no force to drive convergence of nucleotide/aa sequences. There is no reason for 2 sequences to proteins to share high identity across their entire sequence. There would be localized regions of high similarity, but entire nucleotide sequence would not be similar.

Related Questions

Part B

Think about the DNA coding sequence of a gene. If an A were swapped for a T, what kind of mutation could it cause and why?

a) It could cause a frameshift nonsense or frameshift missense mutation because it would change the reading frame of the codon triplet.

b) It could cause a silent, missense, or nonsense mutation because those are the types that can be caused by a nucleotide-pair substitution like this one.

c) It could cause a nonsense mutation because the sequence would no longer be the same, so the protein would be shorter and non-functional.

d) It could cause a silent mutation because A and T are complementary to each other so it is not really a substitution mutation.

Part C

Why is a frameshift missense mutation more likely to have a severe effect on phenotype than a nucleotide-pair substitution missense mutation in the same protein?

a) A substitution missense mutation causes the protein to be shorter and thus non-functional.

b) A frameshift missense will cause the codons to be out of order, but a substitution missense does not change the order of the codons.

c) A frameshift missense mutation will cause an early Stop codon, but a substitution missense might be silent.

d) A substitution missense affects only one codon, but a frameshift missense affects all codons downstream of the frameshift.

Part D

Which of the following sequences shows a frameshift mutation compared to the wild-type mRNA sequence?

a) wild-type	5â-AUGCAUACAUUGGAGUGA-3â
mutant	5â-AUGCAUACAGAGUGA-3â

b) wild-type	5â-AUGCAUACAUUGGAGUGA-3â
mutant	5â-AUGCAUACGUUGGAGUGA-3â

c) wild-type	5â-AUGCAUACAUUGGAGUGA-3â
mutant	5â-AUGCAUACAUCUGGAGUGA-3â

d) wild-type	5â-AUGCAUACAUUGGAGUGA-3â
mutant	5â-AUGCAUGUGACAUUGGAGUGA-3â

Part F

Suppose that the triplet of nucleotides indicated in bold (AGC) spans two codons, that is, CTA and GCC. If the triplet AGC were deleted from this DNA coding sequence, what effect would it have on the resulting protein?

5â-ATGCTAGCCTATCGTAAC-3â

a) The two flanking codons would be altered, but the rest of the amino acid sequence would be the same because there would be no frameshift.

b) The entire amino acid sequence would be altered due to the frameshift.

c) All of the amino acids after the deletion would be altered due to the frameshift.

d) All of the amino acids up to the deletion would be altered due to the frameshift.

mauvetrout554

1). Thyroid hormone (TH) has been identified as one of the factors promoting phenotype development in the Mexican axolotl salamander. Which of the following statements regarding the impact of TH is FALSE?

A). Axolotls maturing in water with high concentration of TH (thyroxine; T4) metamorphose into a terrestrial form.

B). A genome-wide scan of both the axolotl and its sister species found a large reduction in the number of mRNA transcripts that regulate the production of TH.

C). Adding TH (thyroxine; T4) to the water in which axolotl individuals are reared causes them to remain in a larval stage.

D). Most salamander species produce a burst of TH when they move from water to land.

2).What would be an appropriate evolutionary explanation of the fact that the ordering of some homeotic genes (such as Hox genes) on vertebrate chromosomes parallels the ordering of homeotic genes on fruit fly chromosomes?

A). The similarity is a coincidence, not documented in other invertebrates and largely a result of random mutations.

B). Homeotic genes show that insects and vertebrates are closer relatives than the traditional classification system indicates.

C). Homeotic genes of different animal lineages might be ordered in a similar fashion, yet their sequences are completely different.

D). The similar ordering of homeotic genes in animals, both vertebrate and insect, is due to ancient homology.

3). Which of the following correctly defines "homeotic genes"?

A). Homeotic genes encode proteins that control the cellular environment in a differentiated cell tissue.

B). Homeotic genes are DNA sequences that turn off, but not on, the expression of other genes and are controlled by "master-switch" genes.

C). Homeotic genes are "master-switch" genes that control other genes in a set sequence and thereby affect cell size, shape, division, and the positioning of the cells within the organism.

D). Homeotic genes are DNA sequences that turn on, but not off, the expression of other genes and are controlled by "master-switch" genes.

4). Which of the following is NOT considered a type of heterochrony (change in the rate and timing of development)?

A). hypermorphosis

B). acceleration

C). neoteny

D). ontogeny

5). How does the dominance theory explain why the heterogametic sex is at a disadvantage in Haldane's rule?

A). Beneficial dominant alleles will never be fully expressed in hybrids as they only carry one copy of the sex chromosome that carries the allele.

B). If a dominant allele has a negative effect on the fitness of hybrids, it will always be expressed in the heterogametic hybrids that possess that allele because they only need one copy of the respective sex chromosome.

C). If a recessive allele linked to a sex chromosome has a negative effect on the fitness of hybrids, that allele will always be expressed in the heterogametic hybrids that possess that allele because they only have one copy of the sex chromosome that carries the allele.

D). If a recessive allele linked to a sex chromosome has a positive effect on the fitness of hybrids, that allele will never be expressed in the heterogametic hybrids that possess that allele because they only have one copy of the sex chromosome that carries the allele.

6). In a hypothetical flowering plant species, one population evolves a different response to environmental stimuli and begins to bloom significantly later in the season than nearby populations. What type of reproductive isolating mechanism would this be?

A). hybrid sterility

B). postzygotic

C). prezygotic

D). gametic incompatibility

7). In a study of red spruce trees and black spruce trees, scientists found that the red spruce was living in a smaller geographic area and had much less genetic variation than the black spruce. They proposed that the red spruce might have arisen from a southern population of black spruce, which became geographically isolated from other black spruce populations at some point during the Pleistocene glaciations. What is this an example of?

A). allopatric speciation with the vicariance model

B). sympatric speciation

C). allopatric speciation with the peripheral isolate model

D). sympatric speciation with the ring species model

8). Which of the following defines Haldane's rule and its predictions?

A). If among hybrid offspring one sex is absent, rare, or sterile, that sex is always male.

B). If among hybrid offspring one sex is absent, rare, or sterile, that sex is always female.

C). If among hybrid offspring one sex is absent, rare, or sterile, that sex is heterogametic (has two different sex chromosomes).

D). If among hybrid offspring one sex is absent, rare, or sterile, that sex is homogametic (has two of the same sex chromosomes).

8.) Which of the following statements best explains secondary reinforcement?

A). If the reproductive isolating mechanisms that developed during the geographic isolation are strong but the hybrids have higher fitness than the parent species, then the speciation process will cease.

B). If the reproductive isolating mechanisms that developed during the geographic isolation are somewhat weak but the hybrids between two populations have lower fitness, then the speciation process may continue.

C). If the reproductive isolating mechanisms that developed during the geographic isolation are sufficiently weak as to allow free interbreeding across reunited populations, then the speciation process will halt and a single species will remain.

D). When geographic isolation comes to an end, two populations always complete the divergence process and become separate species.

ceruleanzebra430

You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?

	Using an RNA probe complementary to the region not removed by the truncation.
	Using an RNA probe complementary to the region removed by the truncation.
	Using an DNA probe complementary to the region not removed by the truncation.
	Using an DNA probe complementary to the region removed by the truncation.

To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?

	As both are composed of nucleic acids, using either would result in identical results.
	An RNA probe because RNA has uracil bases.
	An RNA probe because it could also be used in a translation experiment.
	A DNA probe because it is more stable than RNA.
	A DNA probe because RNA cannot bind to DNA.

Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?

	No, because only DNA can be labeled with radioactivity.
	No, because the RNA genome would not enter the bacteria upon infection.
	No, because while DNA and RNA nucleotides are similar, they are not identical.
	Yes, because DNA and RNA nucleotides are similar.
	Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly.

The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?

	Random DNA mutations generally won't affect RNA and protein function.
	It is faster to duplicate the genome when these are present.
	The existence of introns can lead to multiple variations of proteins encoded by a single gene.
	It is unlikely transposons would exist in the genome if there was too much protein coding DNA.

You accidentally add a mutant dNTP (which has an H instead of an OH connected to the 3â carbon) to a reaction where DNA is being replicated. Which of the following is true of this mutant dNTP?

	It can be incorporated into DNA strand but cannot form a phosphodiester bond with an incoming wild type dNTP
	It can be incorporated into a DNA strand but cannot base pair with a complementary nucleotide
	It can be incorporated into a DNA strand and can form a phosphodiester bond with an incoming dNTP, but only if it is another mutant dNTP
	It cannot be incorporated into a DNA strand.

Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (âade) and histidine (âhis). The plasmid she is currently working with consists of an origin of replication and the Ade gene.

Following her transformation of the plasmid into her yeast, what media will the cells be plated on to select for cells that have picked up the plasmid?

	Media containing histidine
	Media containing adenine
	Media lacking adenine
	Media lacking histidine

She starts by attempting to add the centromere DNA into a plasmid containing the origin of replication. Unfortunately, when adding the centromere sequence, the origin of replication is removed, thus leaving a plasmid with only a centromere and selection marker. Following plasmid transformation, what growth result will she see on her plates?

	No colonies
	Little colonies
	Big colonies

She fixes the mistakes from the previous experiment and now has a complete plasmid (selection marker, origin of replication, centromere). She then inserts telomere sequences into the plasmid. How will this impact her transformation?

	It will not impact her transformation
	Her transformation will no longer work because plasmids donât require telomeres
	She will now see much larger colonies
	She will now see fewer but larger colonies
	She will now see smaller colonies

In the Meselson/Stahl experiment, E. coli were first grown in media containing heavy nitrogen, 15N, and then transferred to light nitrogen, 14N, at the beginning of the experiment.

Imagine that their data showed that replication occurs in a conservative manner instead of semi-conservative. What fraction of the DNA helices will consist of mixed DNA after 4 rounds of replication in this case?

	None
	More than 75%
	25-50%
	51-75%
	Less than 25%

peachmarten804

Biology 1001A Lecture 13: bio 1001a lecture 13

Document Summary

Get access

Related Documents

Biology 1001A Study Guide - Quiz Guide: Hoatzin, Foregut, Photoisomerization

Biology 1001A Lecture Notes - Lecture 19: Neofunctionalization, Lux, Polyadenylation

Biology 1001A Lecture Notes - Lecture 14: National Center For Biotechnology Information

Related Questions