BIOL4200, Solution to Homework 1

Written or Computer Assignment

  1. In this exercise, please use Mathematica, which you can download Mathematica from here. NOTE: To download software for which BC has a site license, if you are not physically on campus (hence recognized as being a member of Boston College), then you MUST use the Boston College virtual private network (VPN) -- i.e. use Cisco connect to enable your computer to be recognized as being associated with Boston College.

    You'll find the following demo very useful: biostatsBoxplotSummary.nb (PDF file is here). Please download the data Bear Weights, consisting of weights of a collection of bears.

    1. Using Mathematica, please create a histogram for the bear weights.
    2. Please create a relative frequency histogram for the bear weights. (To do this, use the function
      Histogram[data,bspec,hspec]
      where "bspec" stands for "bin specification" and "hspec" for "height specification". For "bspec", you can write (for instance) "{min,max,step}", where "min" is the minimum, "max" is the maximum value, and "step" is the distance between one bin and the following bin. Thus "{1,200,5}" gives bins containing values in each of the successive intervals: [1-5], [5-10], etc. To obtain a relative frequency histogram, the argument "hspec" should be either "Probability" or "PDF" (which latter stands for probability density function).
    3. Please create a box wiskers plot for the bear weights.
    4. Using Mathematics, please compute the number of bears, the arithmetic mean, 10% trimmed mean, geometric mean, harmonic mean, median, sample variance, sample standard deviation, population variance, population standard deviation, first, second and third quartile (the second quartile is the median), range (range is the maximum minus the minimum), and midrange (midrange is the average of the maximum and minimum) values for the data.

    Solution: mathematicaSol1.pdf

  2. Explain the two ways in which DNA differs from RNA, and draw the structure of both. Your drawing should give all atoms in the sugar and phosphate group, but the base can be indicated by an attached circle containing the letter A,C,G,T or U.
    Solution:
    1. DNA has bases A,C,G,T while RNA has bases A,C,G,U. It is believed that in a primordial "RNA world", RNA played both an information carrying role (now taken over by DNA) as well as a catalytic role (in many, but not all cases taken over by proteins). DNA is a more reliable information carrier because of DNA repair mechanisms, including SOS. In contrast, RNA suffers from the problem that methylated cytosine spontaniously deaminates and forms uracil. Since uracil is not a valid DNA base, this error can be corrected in DNA though not in RNA. To the best of my knowledge, there is no known error correction mechanism for RNA.
    2. RNA has a hydroxyl group on the 2' carbon of the pentose sugar, in contrast to DNA which has hydrogen in place of the OH group. This difference allows RNA to form many more hydrogen bonds than DNA; in particular, RNA is generally a single-stranded structure, which forms H-bonds with itself in stable secondary structures.

  3. Suppose that 5'-GGATCGTA-3' is a genomic DNA, which is transcribed into RNA. What is the RNA transcript (i.e. reverse complement)? Be sure to indicate 5' and 3' orientations.
    Solution: 5'-UACGAUCC-3'.

  4. What is a TATA-box? If TATAAA is the consensus TATA-box hexamer, then what is the expected number of TATA-box occurrences in the human genome?
    Solution:
    1. A TATA box is a DNA sequence located 25-35 base pairs upstream of the transcription start site in eukaryotes. Transcription factors (TFs) bind to the TATA box, which then recruit RNA polymerase, which then transcribes mRNA from the gene. The consensus sequence for TATA boxes is TATAAA. Note that only eukaryotes and archaea contain a TATA box, while most prokaryotes contain a Pribnow box, thought to be function as a TATA box, which usually consists of the nucleotides TATAAT.
    2. If genes were transcribed from only one strand of DNA, then the answer would be 3000000000/(4**6) = 732421.875; however, taking into account both the + and - strands, the answer is be 6000000000/(4**6) = 1464843.75. Being meticulous, we cannot include the last 5 nucleotides, so one can argue that the "true" answer is 600000000.-10)/(4**6) = 146484.372559. But even this doesn't account for the fact that the 3 billion base pairs of double-stranded DNA lie on 46 linear chromosomes (23 pairs of double-stranded chromosomes). So to be meticulous, one would need make separate computations for each chromosome, where the plus and minus strands are taken into account of, and the last 5 nucleotides of each strand of each chromosome must be ignored. Moreover, so far, our computation only concerned the haploid genome, so if we account for the diploid genome, we must double the final answer. For the answer to this question, I'm looking for your reasoning process, not the exact answer, which as we see involves multiple considerations. However, the numerical differences between the values when accounting for the end effect or not are "teeninsy", so one should not account for end effects in this case. Moreover, we have assumed that the proportion of each nucleotide A,C,G,T is 0.25 in the human genome. This is a technically incorrect assumption. If we search for H. sapiens chromosome 22, complete sequence (since chromosome 22 is one of the smallest of the 23 pairs of chromosomes) at the European Nucleotide Archive of the EBI, we find files in the primary assembly, one of which has the EMBL accession code AC005301.22. Immediately before the nucleotide sequence appears, we find that
      Sequence 113688 BP; 37158 A; 22385 C; 21889 G; 32256 T; 0 other
      from which we determine that
      a/bp = 0.32684188304834283
      c/bp = 0.19253571177257053
      g/bp = 0.19253571177257053
      t/bp = 0.28372387587080433
      Following convention, the values should be rounded, for instance to the second decimal place yielding
      a/bp = 0.33
      c/bp = 0.19
      g/bp = 0.19
      t/bp = 0.28
      If you now do a more correct computation for the probability of TATAAA, you find a probability of (0.33)4 * (0.28)2 = 0.09025921, which is considerably different than 0.256 = 0.000244140625! (This of course assumes that the compositional frequency in the entire genome is the same as this fragment of chromosome 22.) Note that Chargoff's second "law" (not really a law) approximatly holds.

      In any case, this is how you go about performing a "Milchmädchenrechnung" (milk maid's computation, the German expression for "back of the envelope computation").

  5. Find human proteins containing the pentapeptide Glu-Leu-Val-Ile-Ser, by using the Protein Information Resource (PIR). Enter the IUPAC single-letter amino acid codes into the search field.
    Solution: Search/Analysis -> Peptide Match. Then enter "Homo sapiens" for organism, and use the default settings to get the following:

    Notice that many of the proteins that contain ELVIS are transcription factors and zinc fingers, both classes of protein that bind DNA. This raises the question of whether ELVIS is a motif for binding DNA -- you would have to do additional research to determine whether this is the case.

    Now, if you hadn't first selected "Search analysis -> peptide match", and just entered ELVIS in the search box at the top right, then you get files that simply contain "ELVIS" somewhere in the file, as the examples below indicate.

    where the only occurrences of ELVIS are in the non-sequence data, for instance, in the first hit, one can look at the GenPept entry for

    cytochrome oxidase subunit 1, partial (mitochondrion) [Glossosoma elvisso]
    where ELVIS is a portion of the organism name. This appears in several similar places in the file, such as in "mitochondrion Glossosoma elvisso", but NOT in the amino acid sequence of the protein.

    If instead you use NCBI BLASTp (protein blast), then you find the following:

    The above search was not restricted in BLAST to H. sapiens. By restricting to H. sapiens, I found different output, which included, for instance a hit with link to the GenPept file XP_024308862.1 for titin isoform X2 [Homo sapiens]:

    14161 lsayaelvis psersdkgiy tlklenrvkt isgeidvnvi arpsapkelk fgditkdsvh
    Titin (also called connectin) has primary function to give elastic stabilization in myosin and actin filaments.

  6. Name an amino acid that has physicochemical properties similar to (a) leucine, (b) aspartic acid, (c) threonine. We expect that such substitutions would in most cases have relatively little effect on the structure and function of a protein. Name an amino acid that has physicochemical properties very different from (d) leucine, (e) aspartic acid, (f) threonine. Such substitutions might have severe effects on the structure and function of a protein, especially if they occur in the interior of a protein structure. Using the BLOSUM62 matrix, for each of your answers (a)-(f), give the BLOSUM62 similarity score between the amino acid in this question and that which you give as your answer. For instance, if you state that serine has physicochemical properties similar to leucine in (a), then give the BLOSUM62 similarity score between leucine and serine.
    Solution:
    1. Isoleucine (IUPAC code I) is similar to leucine (IUPAC code L), as both are hydrophobic and have BLOSUM62 similarity 2.
    2. Glutamic acid (IUPAC code E) is similar to aspartic acid (IUPAC code D), as both are negatively charged and have BLOSUM62 similarity of 2.
    3. Serine (IUPAC code S) is similar to threonine (IUPAC code T), as both are small polar amino acids and have BLOSUM62 similarity of 1.
    4. Histidine (IUPAC code H) is different from leucine, since histidine is polar or hydrophilic, while leucine is hydrophobic. Their BLOSUM62 similarity is -3.
    5. Cysteine (IUPAC code C) is different from glutamic acid (IUPAC code E), since cysteine is neutral, while glutamic acid is negatively charged, and their BLOSUM62 similarity is -4.
    6. Tryptophan (IUPAC code W) is different from threonine, since tryptophan is hydrophobic while threonine is polar, and their BLOSUM62 similarity is -2.

  7. In designing an antisense sequence, estimate the minimum length required to avoid exact (reverse) complementarity to many random regions of the human genome?
    Solution: The expected number of random hybridizations (Watson-Crick reverse complements) of an n-length antisense sequence in the 6 billion nucleotides of the human genome is 0.25n * 6 * 109. For this expression to be less than 1, by taking natural logarithms, we need that n*ln(1/4)+ln(6)+9*ln(10) < 0. Algebra yields that n must be greater than (ln(6)+9*ln(10))/ln(4) = 16.24. So choose n to be 17.

  8. Search for the complete sequence (whole genome) of HIV-1 in the "nucleotide" database of NCBI server. Find the GenBank file with accession number AF224507.1. What is a gag protein? What is the amino acid sequence of the HIV-1 gag protein? What is the genome sequence length? What is the number of each nucleotide A,C,G,T.

    Hint: The latter information is not in the GenBank file; however, it IS in the EMBL file. Go to the EMBL (EBI) server, and search for the GenBank accession number, click the link "Text" to see the EMBL formatted file, and look at the line that begins with "SQ" (sequence information) immediately before the nucleotide sequence of the HIV-1 genome. Later we will learn how to write a Python program to compute the number of each nucleotide A,C,G,T.
    Solution: Information can be found at EBI. The acronym `gag', which stands for `group-specific antigen', refers to a coding region in retroviruses. In HIV-1, gag is the gene in the 9238 nt viral genome that codes for the core structural proteins. The translated gag protein has amino acid sequence given by: 

    MGARASILSGGELDQWEKIRLRPGGKKKYRLKHLVWASRELERFA
VNPGLLETSEGCRQILGQLQPSLQTGSEELKSLFNAVAVLYCVHQRIEIKDTKEALEKI
EEEQSKSKKKAQQATADTGSSSQVSQNYPIVQNLQGQMVHQPISPRTLNARVKVIEEKA
FSPEVIPMFSALSEGATPQDLNTMLNTVGGHQAAMQMLKETINEEAADWDRLHPVHAGP
IAPGQMREPRGSDIAGTTSTLQEQIGWMTNNPPIPVGEIYKRWIILGLNKIVRMYSPAS
ILDIRQGPKEPFRDYVDRFYKTLRAEQASQEVKNWMTETLLVQNANPDCKTILKALGPG
ATLEEMMTACQGVGGPSHKARVLAEAMSQATNSATIMMQRGNFRNQRRTVKCFNCGKEG
HIAKNCRAPRKKGCWKCGKEGHQMKDCTERQANFLGKIWPSHKGRPGNFLQSRPEPSAP
PEESFRFGEETTTPSQKQEPIDKELYPLASLRSLFGNDPSSQ
    The genome size and compositional frequency is given by
    Sequence 9238 BP; 3301 A; 1640 C; 2241 G; 2056 T; 0 other;

  9. Consider the datasets of ages of a female cohort FAGE.txt and of ages of a male cohort of unrelated males MAGE.txt. Using EXCEL, determine the means of each set (use formula average), the sample standard deviation (use formula stdev.s), and with 95% confidence whether the two means are equal. Hint: you reject the null hypothesis H0 (that ages of men and women are equal on average) if the p-value obtained by the Excel formula T.TEST with 2 tails is less than 5%.
    Solution: 
    Fmean	33.225		
Fstdev	12.45399225		
Mmean	35.475				
Mstdev	13.92652422				

null hypothesis			H0: Fmean=Mmean
alternative hypothesis		H1: Fmean≠Mmean
					
2-tailed p-value is 0.44858032, as computed by
T.TEST(Fage,Mage,2,3)
p-value is NOT < 0.05 (or 0.01) so we can NOT					
reject the null hypothesis that women's age equal

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