BIOC2201 Lecture 13: Theme 3 Notes

Its one problem with 7 questions

A) Most eukaryotic cells transcribe a relatively high portion of their genome, but most of those RNA transcripts are likely not encoding for proteins. Elucidating the function(s) of that "junk" RNA is one of the next big challenges in molecular cell biology. You isolate one of those transcripts and determine that its sequence consists of 50% C, 30% U, and 20% A. What is the percentage of Gs in this transcript? (don't write the unit)

B) Which stop codon(s) can be encoded by the RNA in the previous question ("How many times do we need to tell you to stop? 1").
List the codon(s) alphabetically. If you need to enter fewer codons than the alloted number of "blanks", enter N/A .

C) From what you can tell, the sequence of the transcript you isolated in the previous question* appears quite random. Assuming that it is random, you can expect to see the various combinations of any three nucleotides at a probability that depends of the frequency of each of the nucleotides. For example, you naturally should expect to see the triplet CCC more times than AAA. What is the expected percentage of triplets corresponding to stop codon(s) in this transcript (round to one decimal place and enter just the number without "%")? *(How many times do we need to tell you to stop? 1)

D) If the length of the transcript in the previous two questions is 2,500 nucleotides, how many stop codons do you expect to find in it, based on the frequency you calculated above?

E) On average, how many bases separate each of the occurrence of the stop codon(s) (round to the nearest integer).

F) The sequence between two potential stop codons is an "open reading frame" (ORF*). Upon closer examination of the RNA sequence from the previous questions** you determine the presence of an ORF that may encode for a putative protein which is 600 amino acid residues-long. How long is the ORF?

*Open reading frame: a stretch of DNA or RNA which is uninterrupted by a stop codon and may therefore encode for a protein.

**Questions "How many times do we need to tell you to stop" 1 through 5.

G) What may be the ramifications of your analyses of the RNA you isolated*? It is clearly containing an ORF which is longer than what you expected to find based on the sequence analysis you have conducted. What would be a logical conclusions based on the comparison of the lengths of the actual and predicted ORFs? Questions "How many times do we need to tell you to stop" 1 through 6. Pick 1, 2 or 3 below

Question 9

Codominance is a form of inheritance in which two different allelesfor a gene are both expressed, and neither allele is dominant overthe other. One example of codominance is fur color in cattle; thealleles for red fur and white fur are codominant. If a homozygousred cow and a homozygous white bull mate, what would you expect tofind in their offspring?.912.l.16.2>

They will all have white fur.

Their fur will be a mixture of red and white hairs.

Their fur will be a lighter shade of red.

They will all have red fur.

Question 10

If a tRNA molecule has an anticodon which reads AUG, what will itmatch up with and what amino acid is it carrying?
.912.l.16.5>

It is carrying arginine and will match with a CGU codon on themRNA.

It is carrying methionine and will match with a TAC codon on theDNA.

It is carrying methionine and will match with a UAC codon on themRNA.

It is carrying tyrosine and will match with a UAC codon on themRNA.

Question 11

While the DNA in a human skin cell was being replicated, a singlebase was miscopied. What will be the most likely result of this forthe cell in which it happened?.912.l.16.3>

All the proteins the cell creates from the miscopied strand willdo different jobs than the old ones.

If the new sequence codes for the same amino acid as theoriginal cell, it will function normally.

Both new DNA strands will end up together in a new cell, and theinaccurate one will be discarded.

Any miscopied DNA will be replaced with an accurate DNA copyonce the cell divides.

Question 12

A ferret's haploid number of chromosomes is 20. How would thenumber of chromosomes in the ferret's body cells compare to thenumber of chromosomes in its gametes?.912.l.16.17>

Its body cells would have 20 chromosomes, and its gametes wouldhave 40 chromosomes.

Its body cells and gametes would both have 40 chromosomes.

Its body cells and gametes would both have 20 chromosomes.

Its body cells would have 40 chromosomes, and its gametes wouldhave 20 chromosomes.

Question 13

A body cell is in the longest stage of its life cycle. The cellgrows, synthesizing proteins and increasing in size. Eventually,the cell will grow too large to carry out normal activities. Whichof the following is the best conclusion you can make about the lifecycle of this cell?.912.l.16.14>

The cell is close to the end of its life cycle, and a chemicalsignal will initiate cell death.

The cell is ready to undergo mitosis, and a chemical signal willsend the cell to prophase.

The cell is in the G1 phase of interphase, and a chemical signalwill send the cell to the S phase.

The cell is in the S phase of interphase, and a chemical signalwill move the cell to the G2 phase.

Question 14

While mRNA strands are being created a sequence is sometimesmiscopied. What is the best possible outcome for the cell shouldthis take place?.912.l.16.5>

The miscopied sequence codes for the same amino acids as theoriginal sequence.

The new sequence creates a protein that serves a differentfunction from the original.

The ribosomes will correct the mistake before the tRNA matchesan amino acid to it.

The mRNA will only be used to create non-critical proteins forthe cell.

Question 15

Over the last several decades, the scientific community hasgathered a large amount of information regarding genetics andgenetic variation. What are two main sources that lead toincreased genetic variation?.912.l.15.15>

Selective breeding

Gamete mutations

Recombination

Genetic drift

Cystic Fibrosis (CF) is an autosomal recessive genetic disease that is characterized by the secretion of thick, abnormal mucous in the lungs. The mutated gene that leads these symptoms is the CF transmembrane conductance regulator (CFTR gene). The encoded protein is a chloride channel that moves chloride ions out of epithelial cells. In a homozygous individual, the lack of this chloride channel function in a mutant slows the flow of water into extracellular spaces. This lack of water export leads to the heavy mucus in the lungs of CF patients. Although children with CF would die in childhood, current treatments provide life expectancies into the 30’s.

CF is the most frequently inherited genetic disorder among American children. It affects 1 of every 2,500 children; 30,000 currently have the disease. Although its frequency is higher in European populations, it affects all human populations. Heterozygous carrier frequencies in the US are estimated to be 1 out of 29 Caucasians, 1 out of 46 Hispanics, 1 out of 61 African Americans and 1 out of 90 Asians. There are genetic screens for carriers of CF but they are not routinely performed despite this high frequency of carriers. At least all states in the US proscribe testing of newborns to identify children with CF early so that treatment can be started. Newborn tests for CF: http://www.cff.org/AboutCF/Testing/NewbornScreening/ScreeningforCF/ . If CF is such a common genetic disease, and such a simple Mendelian trait, why is genetic screening of parents not done more frequently? Is it just economic, a pain to do, or is it more complex such that the results of genetic testing are not just positive or negative?

Answer the following questions:
1. If two parents are heterozygous carriers of CF, show a pedigree with four children showing the incidence of carriers and afflicted children. If each unafflicted child (carrier or not) mates with another person, who has a 1 out of 30 chance of being a carrier of CF, what is the expected incidence CF disease in the third generation (frequency of CF in any one of grandchildren of the original couple)?

2. Pleiotropy of Cystic Fibrosis. Watch the two videos on the left at:

http://www.cftrscience.com/MOD-animation
or, if you can not view these, read the short NIH Genetics Home Reference for Cystic Fibrosis http://ghr.nlm.nih.gov/condition/cystic-fibrosis

Connect the description with one of the following videos, or related ones (there are many) you see:
video (10 minutes) of daily life of a child with Cystic Fibrosis http://www.youtube.com/watch?v=WUsuGKzdDg8
Laura’s (20 year old) video diary of life with CF http://www.youtube.com/watch?v=Pgrf8qJXhpc&feature=related

In your own words, describe the range of CF phenotypes. List the other organs besides lungs that are affected and how does it affects those organs.

3.The DNA sequence of the CFTR gene is 170,000 base pairs in length and encodes the CFTR protein that is 1480 amino acids long. Open the accompanying text file (in Canvas/Files/Discussion folder) “CFTR gene DNA sequence”. You will see an alignment of the NIH human genome sequence of the CFTR gene – called “Reference” rather than “wild-type”- against a gene sequence containing a common mutation in CFTR. Look at the alignment and determine how the DNA sequence differs between them. What is the position number where they differ? How many and what nucleotides do they differ in? A “-“ mark in one sequence in the alignment means that that base sequence is missing the corresponding bases in the other sequence. The sequence is a long one but the differences are in bolded bright red text. Matching sequence is marked with a | linking the two sequences. Look near the sequence AAATATCATC if you use FIND tool in Word.

In another file, you will an alignment of the reference translated into a protein sequence, compared to the mutant allele translated into protein sequence. The differences are again in red whereas the areas where the proteins match are in black. Look at the alignment. How do these two proteins differ? Again, find the position number and describe how they differ at that location. A “-“ mark in the alignment means that amino acid is missing when compared to the other sequence.

4. Given that the CFTR gene can be analyzed for a mutation that causes CF, couples interested in being tested as carriers face several choices given the results.
What could be a consequence of a false positive result (couple is identified as carriers but they would not produce any children with symptoms)?

What could be a consequence of a false negative result (a test indicates the couple is not both carriers but they would produce some children with CF)?

5. The sensitivity of the test can vary with ethnic group. The following is a chart from the NCBI Genetic testing site. A sensitivity less than 100% would be likely to produce false negative results from an inability to determine that someone is carrier.

Why might different ethnic populations differ in the ability of the genetic test to identify carriers of CF?