In: Biology
Activity 2: Transcription
Imagine that you’re having a house built. The architect draws up a set of blueprints for the builder. The builder, in turn, has to hire several subcontractors for the electrical, plumbing, and heating systems. The builder doesn’t wish to give up the original set of blueprints, so copies are made for each of the subcontractors.
This same principle applies to a cell. The nucleus contains the master set of blueprints, the DNA. DNA molecules are needed to direct the activities that occur in the cytoplasm of each cell.
When other organelles need to manufacture materials for the cell, copies of the master blueprint are made, so that the originals are not damaged or lost. This process is referred to as transcription. The copies made by transcription will be in the form of another molecule called messenger RNA (mRNA)
There are several differences between DNA and mRNA:
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In this activity you’ll transcribe a DNA code into mRNA. The DNA sequence in this activit represents a gene located on one of the chromosomes in the nucleus of a cell that codes for a specific protein.
Table 9 – 2 mRNA Base pairing
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Fill in the appropriate mRNA sequence in the second column.
ATG is the first triplet on the DNA sequence. What is the corresponding codon?
FIGURE 9-2
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2 DNA |
3 Amino Acids |
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Activity 3: Translation and protein synthesis
Having copied the DNA into mRNA leaves the nucleus of the cell and travels to a ribosome.
Reminder: Ribosomes are the sites of protein synthesis within a cell
Proteins are composed of chains of amino acids. A chain of amino acids is also referred to as a polypeptide chain because the amino acids are held together with a type of covalent bond called a peptide bond.
Amino acids are brought to the ribosome by another type of RNA called transfer RNA (tRNA). The codons in the mRNA determine the sequence and number of amino acids being bonded into the corresponding protein.
To do this, just match each codon in your mRNA sequence to the corresponding amino acid as listed in the table.
Fill in column three of Figure 9-2 with the correct amino acids.
In: Biology
Animals evolved certain traits in order to adapt to their environments. These traits inform their life histories such as where the species lives, what it eats, and how it behaves. Explain at least two anatomical differences between amphibians and reptiles which explain their very different relationships to water. Explain two anatomical features that would shed light on the life history of a bird species, such as what it eats, where it lives, or any other feature. Explain the anatomical feature that would most help a scientist trying to figure out what kind of food a mammal species eats.
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1. The alpine sky pilot, Polemonium viscosum is a common flowering plant in the Rocky Mountains. At high elevations, P. viscosum is pollinated mainly by B. kirbyellus bumblebees; At lower elevations, it is pollinated mainly by flies. A biologist observed that alpine sky pilot populations growing at higher elevations had significantly larger flower corollas than those growing at lower elevations.
a) Explain why flower size would be likely to affect the fitness of sky pilots at different elevations.
b) Explain why, based on fitness differences, average flower size has become different at high and low elevations.
2. A biologist claims that although the two sky pilot populations have different flower sizes, they are the same biological species.
a) How would the biologist test this hypothesis?
b) What experimental outcome would support the hypothesis that the two populations are the same biological species?
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What is the significance that primates generally give birth to one offspring at a time and the elongations of gestation and infant dependency? What does this mean about the degree of learning found among prosimians, monkeys and apes.
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A number of other related drugs with a similar indication to tropisetron are not being considered for use in Alzheimer’s disease. What distinguishes tropisetron from other drugs? Consider the paper and its specific findings.
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Biology review Sheet - NEED ANSWERS ASAP - the last person did this very poorly
14. Compare and contrast the life cycles of bryophytes, pteridophytes, gymnosperms, and angiosperms (some of this included in the list for objective 6 above).
part 2
part 3
16. Briefly explain why antibiotics are effective at killing bacterial cells, but do not seem to impact human sells. → Hint: This is in your assignment.
part 4
part 5
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The general structure of the replication fork and the enzymes and steps involved in resolving Okazaki fragments.
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2.) Consider the following RNA sequence, which represents the beginning of an open reading frame: 5’-AUGGGACUAGAUACC-3’. Examine each of the sequences below, describe the change to the sequence, and state whether it represents a silent, missense, nonsense, or frameshift mutation. If frameshift, be sure to include the consequences of the frameshift (nonsense or massive missense). Use the codon table. a) 5’-AUGGGUCUAGAUACC-3’ b) 5’-AUGCGACUAGAUACC-3’ c) 5’-AUGGGACUAGUUACC-3’ d) 5’AUGGGACUAAGAUACC-3’ 3. Using a couple of sentences each, describe the function of each of the following in regulating eukaryotic gene expression: a) promoter b) enhancer c) histone acetyltransferase d) 5-methylcytosine
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On the planet Proboscide, there are elephants that are the size
of flies. They genomic makeup is also very similar to the genomic
make up of earth drosophila. There are two genes of interest for
the elephants. They are PNT, and WGT.
The PNT gene produces two phenotypes, those elephants that like to
eat peanuts and wild types that do not.
The WGT gene produces a wild-type phenotype of elephants near
mean-weight, and a mutant phenotype where the weights are extreme
(very heavy). Both genes have two alleles at their loci.
The mode of inheritance for the WGT gene is autosomal dominant,
homozygous lethal. The mode of inheritance for the PNT gene is
autosomal recessive. The two genes are unlinked.
In a cross between a male elephant with both extreme weight and
that likes peanuts, and female elephant with wild-type phenotypes
for both loci (true breeding at each locus)offspring are produced.
Among these offspring, two elephants, both with extreme weight and
wild-type PNT phenotypes (don't like peanuts) are crossed. In an
F2 offspring pool of approximately 1000 elephants, the
most likely pairs of phenotype counts are:
A. (Phenotype 01, Phenotype 02) Count
(WGT, PNT) 500
(WGT, +) 500
(+, PNT) 0
(+, +) 0
Here, + is the wild-type phenotype at for the repsective gene.
B. (Phenotype 01, Phenotype 02) Count
(WGT, PNT) 570
(WGT, +) 180
(+, PNT) 187
(+, +) 62
Here, + is the wild-type phenotype at for the respective gene.
C. (Phenotype 01, Phenotype 02) Count
(WGT, PNT) 333
(WGT, +) 334
(+, PNT) 167
(+, +) 162
Here, + is the wild-type phenotype at for the repsective gene.
D. (Phenotype 01, Phenotype 02) Count
(WGT, PNT) 0
(WGT, +) 681
(+, PNT) 0
(+, +) 319
Here, + is the wild-type phenotype at for the repsective gene.
E. (Phenotype 01, Phenotype 02) Count
(WGT, PNT) 252
(WGT, +) 252
(+, PNT) 240
(+, +) 261
Here, + is the wild-type phenotype at for the repsective gene.
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Please answer in detail.(2 pages)
At the core of a fluorescence microscope is the dichroic mirror.
What function does the dichroic mirror perform?
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