gen unit 2 hw (5-8)
major differences between DNA and RNA
(i) In RNA the sugar is ribose instead of deoxyribose. (ii) RNA contains the base U instead of T. (iii) Most DNA molecules found in nature are double stranded while most RNA molecules are single stranded, but there are exceptions to both cases. (iv) DNA strands can be very long -more than 100,000,000 nucleotides in a human chromosome, for example. The longest naturally occurring RNA molecules are much shorter —about 20,000 nucleotides at most.
biosynthetic pathways
----- to + one at a time
what are the building blocks of DNA and how are they subdivided with what kind of chemical bonds linking them together
-Each of the four building blocks is a nucleotide. -Each nucleotide is made of the sugar deoxyribose and one of the four nitrogenous bases (A, C, G or T) and a phosphate group. -Deoxyribose plus a base makes a nucleoside. -When phosphate groups are added, these become nucleotides. - In a strand of DNA, the adjacent nucleotides are connected by phosphodiester bonds that link a phosphate group to both the 3′ carbon atom of the deoxyribose of one nucleotide and the 5′carbon atom of the deoxyribose of the next nucleotide in the chain.
Why are A-T rich regions the first to denature when a double stranded DNA molecule is exposed to high temperature?
-The A-T base pairs have only two hydrogen bonds, so it takes less heat energy to denature these base pairs. -G-C base pairs have three hydrogen bonds holding them together. It thus takes more energy to break the bonds between Cs and Gs. Remember that the DNA of different species can vary a great deal in the proportions of A-T base pairs relative to G-C base pairs. -In most organisms, the regions between genes have a higher proportion of A-T base pairs (they are A-T rich) than the genes themselves.
Explain the structure of renatured DNA molecules and why they have the bubble like formation
-The denatured single-stranded DNA must contain stretches of nucleotides that are complementary to a nearby sequence but in an inverted orientation. -These stem-loop structures are regions where the single strand of DNA formed a double-stranded region. -The loops as well as the strings holding the stems together are made from the same single molecule that is a single strand of DNA.
How is it possible for DNA to carry complex genetic info if its structure is so simple
-The information in the DNA is contained in the order of its building blocks. -There are 3 billion nucleotides in the complete haploid set of 23 human chromosomes. This amount of sequence can potentially provide a huge amount of information. -Although there are only 4 different building blocks ,they can be combined in a huge number of combinations. (For instance, when considering a short 10-nucleotide long piece of DNA,410 or 1,048,576 different possible sequence permutations exist. The information may be recognized by proteins that bind directly to DNA or it can be copied into RNA to direct the synthesis of proteins.)
In human DNA, 70% of cytosine residue followed by guanine are methylated to form 5-methylcytosine, if this undergoes spontaneous deamination, it becomes thymine. Why are methylated CpG hotspots for point mutations in human DNA?
.Deamination of 5-methylcytosines produces thymine (T); after replication this change could be fixed in the genome, producing C-G to T-A transitions. Deamination of 5-methylcytosines would produce T-G base pairs, and you would be correct in thinking that cells ought to have molecular mechanisms to correct these mismatches. However, the repair is inefficient, so errant Ts formed by deamination of 5-methylcytosines are repaired only rarely.
Results after conservative replication after 2 rounds of DNA Synthesis on 14N
1/4 of the DNA is in an H/H (15N/15N) band near the bottom of the gradient; 3/4 of the DNA is in an L/L (14N/14N) band near the top.
Results after semiconservative replication after 3 rounds of DNA synthesis on 14N
1/4 of the DNA is in an L/H (14N/15N) band in the middle of the gradient; 3/4 of the DNA is in an L/L (14N/14N) band near the top.
Results after conservative replication after 3 rounds of DNA synthesis on 14N
1/8 of the DNA is in an H/H (15N/15N) band near the bottom of the gradient; 7/8 of the DNA is in an L/L (14N/14N) band near the top.
RNA poly moves along in the
3 to 5 along the template
What way can we use forward mutation to test a compound for mutagenicity
A wild-type (His+) strain could be grown in the presence of a potentially mutagenic compound (plus rat liver enzymes) and then plated for single colonies on minimal medium + histidine. Replica-plating on minimal medium (without a histidine supplement) would identify colonies that are His−; they would fail to grow
Plant breeders studying genes influencing leaf shape in the plant Arabidopsis thaliana identified six independent recessive mutations that resulted in plants that had unusual leaves with serrated rather than smooth edges. The investigators started to perform complementation tests with these mutants, but some of the tests could not be completed because of an accident in the greenhouse. Exactly what experiment was done to fill in individual boxes in the table with a plus (+) or a minus (−) ? What does + represent? What does − represent? Why are some boxes in the table filled in green?
A −indicates a lack of complementation and the resulting plants have serrated leaves.A +means that the two mutations complement each other, and so the resulting plant has wild-type leaves with smooth edges. The colored boxes in the table represent crosses that were not performed because the information is redundant
Results after Dispersive replication after 3 rounds of DNA synthesis on 14N
After one round, a single L/H band near the middle of the gradient would be expected. In subsequent rounds this band would be expected to spread gradually toward the top of the gradient as the heavy DNA is diluted across all the new molecules in a random fashion. As a result, you would not expect discrete bands. Instead you would see a broad "smear".
In human DNA, 70% of cytosine residue followed by guanine are methylated to form 5-methylcytosine, if this undergoes spontaneous deamination, it becomes thymine. The frequency of CpG is much lower than the predicted calculation because?
Any CpG dinucleotides in the genome that are not important functionally would mutate to TpG over time.Another interesting idea is that evolution may select against some CpG dinucleotides whose methylation could have deleterious consequences.
why does the AMES test use the reversion rate to test for mutagency
Identification of forward mutations is through a screen, not a selection.In the screen technique to find forward mutations after exposure to the compound, both His+ and His−colonies were allowed to grow, and they were screened by replica-plating for the His−mutant phenotype. Even though the forward mutation rate is higher than the reversion rate, this screening process is much more labor-intensive than the selection that is employed when testing for revertants.In the selection for His+ revertants, millions of bacteria can be spread on a single petri plate containing minimal medium and only the rare revertants will form colonies, making them simple to identify and quantify even though the reversion rate is low.
where are the + and - mutations in mutant DNA
Look at where deletions(-)and additions(+) were made`
Design an experiment to show that RNA, rather than protein, acts as the hereditary material in TMV
MixRNA from virus type 1 with protein from virus type 2 to reconstitute a hybrid virus. In a parallel experiment, mix RNA from virus type 2 with protein from virus type 1. Infect cells with each of these reconstituted hybrid viruses separately, and analyze the protein in the progeny viruses that result from each infection. You will find that the progeny viruses in each case have the protein that corresponds to the type of RNA in the parent hybrid virus. The protein in the progeny did not correspond to the protein in the parent hybrid virus. Thus, RNA is the hereditary material in TMV virus while protein is not.
where would you find the BP encoding the 5' cap
No base pairs encode the 5′cap because the cap is added post-transcriptionally.
how can you find alterations in amino acids
The AA that are different from wild type
In human DNA, 70% of cytosine residue followed by guanine are methylated to form 5-methylcytosine, if this undergoes spontaneous deamination, it becomes thymine. How frequently in the human genome would you expect to find CpG?
The frequency of each base is 1/4, and thus the expected frequency of CpG at a given location in the genome is the same as that ofany dinucleotide: 1/4 ×1/4 = 1/16 ≈ 6.25%.
Albinism in animals is caused by recessive mutations in an autosomal gene required for synthesis of melanin, a precursor for skin and eye pigment. Albino animals are sometimes confused with leucistic animals that are white due to recessive mutations in a gene required for a different pathway What happens if one leucistic hummingbird mates with one albino hummingbird
The progeny of all matings between an albino bird and a leucistic bird will be colorful because the parents are necessarily homozygous for recessive alleles of different genes.
Albinism in animals is caused by recessive mutations in an autosomal gene required for synthesis of melanin, a precursor for skin and eye pigment. Albino animals are sometimes confused with leucistic animals that are white due to recessive mutations in a gene required for a different pathway What happens if two albino hummingbirds mate
The progeny of two albinos will be either bewhite (albino)due to noncomplementation(if the albino parents are both homozygous for recessive alleles of the same gene needed for melanin formation/deposition) or colorful due to complementation(if the albino birds are homozygous for recessive alleles of different genes needed for melanin formation/deposition).
Albinism in animals is caused by recessive mutations in an autosomal gene required for synthesis of melanin, a precursor for skin and eye pigment. Albino animals are sometimes confused with leucistic animals that are white due to recessive mutations in a gene required for a different pathway What happens if two leucistic hummingbirds mate
The progeny of two leucistic birds can be either white due to non complementation(if the leucistic birds are homozygous for recessive alleles of the same gene affecting the formation of all pigments)or colorful due to complementation(if the leucistic birds are homozygous for recessive alleles of different genes affecting the formation of all pigments).
where would you find the BP encoding the promoter
The promoter is upstream of the 5′UTR and so it is not included in any of the DNA sequences listed in the problem.
What DNA single base change could result in the termination of the 3rd AA
The protein would terminate after the third AA due to a nonsense mutation making the AA after the 3rd a stop
where would you find the sequences encoding the poly A tail
The sequence encoding the poly-A tail does not exist in the gene because poly-A is added post transcriptionally by poly-A polymerase.
why are the AA different in the double mutant compared to wild type but the mutant has the same phenotype
Those four amino acids must not be in a crucial functional domain of the protein (for example, they do not carry out catalysis), nor do they alter the overall structure of the protein significantly enough to affect its function.
is it possible that a known mutagen would be able to revert a particular His - mutant used in the AMES test
Yes; specific mutagens can revert only particular types of mutations. Proflavin, for example, can revert only single-base insertions or deletions; it cannot revert nucleotide substitutions caused by mutagens such as base analogs. The Ames test deals with this issue by testing potentially mutagenic compounds for their ability to revert His−strains that have different types of mutations at the molecular level.
how large is the mature mRNA in bases
add up all the exons alone and add bases of poly-Atail
how large is a gene in BP
add up all the exons and introns and ignore the 5′UTR and 3′UTR
where would you find the BP encoding the stop codon
compare the size of the 3′ UTR to that of the last exon. Part or all the first exon must be the 3′UTR, if it is longer than the last exon then it must be in the previous one
where would you find the BP encoding the initiation codon
compare the size of the 5′ UTR to that of the first exon. Part or all the first exon must be the 5′UTR, if it is longer than the first exon then it must be in the next one
effects of nonconservative missense mutations affecting the non active site of the protein
could be severe if it affects the protein structure enough to hinder the action of the protein, or could be mild if the protein′s function can tolerate the substitution
how many AA are in the protein
divid BP by 3
how large is the primary transcript for genes in bases
exons plus introns, +poly A tail, + 1 for 5' cap
where would you find the sequences encoding the C terminus
find the exon containing the stop codon, and the codon preceding it encodes the final (C-terminal) amino acid.
how do you find the sequence of nucleotides in the mRNA
find the specific bases that code in a codon for the AA were looking for, they may be gaps in between and indicate that with a "I"
Replication bubble
look at image
how to tell if its RNA or DNA being coded
look at the AA on the 5' to 3' and find the AA from the codons that resemble what AA we are looking for
effects of conservative missense mutations
mild to no effect as the replacement is with an amino acid with similar chemical properties
effects of silent mutations
no effect because by definition, a silent mutation maintains the same amino acid
effects of frameshift mutations occurring in the sequences encoding AA near C terminus
probably mild effect as only the last few amino acids will be affected
effects of nonsense mutations occurring in sequences encoding AA near C terminus
probably mild effect if none of the last few amino acids are important for function
effects of nonconservative missense mutations affecting the active site of the protein
severe effect as the mutation is likely to destroy the protein′s activity
how large is the coding region of the gene in BP
subtract the 5'UTR and 3'UTR from the exons on corresponding sides, add the remaining and rest of exons in the middle
which intron interrupts the 3'UTR?
the exon that the 3'UTR ends on
which mutant effects the synthesis pathway
the mutant that acts first in the synthesis pathway dictates which compound accumulates.
effects of frameshift mutations occurring in the sequences encoding AA near N terminus
very severe effect as most of the amino acids in the protein will be incorrect
effects of nonsense mutations occurring in sequences encoding AA near N terminus
very severe effect as there will be no functional protein
