ECEN 689  Statistical Computation in Genomic Signal Processing
Fall 2009
Homework 1 – Solutions
1. (a) Starting at index 133 (initiation codon), we have
5
′

ATG CAG AGG TCG CCT CTG .
..

3
′
Hence, by the “table of life,” the aminoacid sequence is MetGlnArgSerProLeu.
..
or, in oneletter notation, MQRSPL.
..
(b) First case: If “A” and “a” denote a normal and defective CFTR allele, respectively,
then the parents are both Aa. The possible equallylikely combination of alleles in
the children are AA, Aa, aA, and aa. Therefore, there exists a 25% probability of
a healthy noncarrier, 50% probability of a healthy carrier, and 25% probability of
a diseased individual. If the couple has 4 children, the expected numbers are 1, 2,
and 1. Second case: One parent is AA and the other is Aa. The possible equally
likely combination of alleles in the children are AA, Aa, AA, and Aa. Therefore,
there exists a 50% probability of a healthy noncarrier, 50% probability of a healthy
carrier, and 0% probability of a diseased individual. If the couple has 4 children, the
expected numbers are 2, 2, and 0.
2. One can tell this is a prokaryote mRNA due to absence of a polyA tail. In addition, the
ShineDalgarno AGGAGGU sequence existent in E. Coli can be seen in the 5’ UTR region.
The 5’UTR, ORF, and 3’UTR extend respectively from positions 1–17, 18–47, and 48–57.
The ShineDalgarno sequence appears at positions 6–12. The ORF contains