MBB 197.
You have the choice of ASIAN PARLIAMENTARY or MOCK TRIAL.
ASIAN PARLIAMENTARY
1. Government: Prime Minister, Deputy Prime Minister, Government Whip
2. Opposition: Leader of the Opposition, Deputy Leader of the Opposition, Opposition Whip
3. Government opens, then alternates with Opposition. Each of the three speakers gets 7 min max. They may be interrupted by the opposite side AND by the audience.
4. When all 6 have spoken, there will be an open forum with the audience.
5. Then Opposition Reply makes a 4 min speech, followed by Government Reply. The Reply speakers are chosen from one of the first two (e.g., Prime Minister or Deputy Prime Minister).
MOCK TRIAL
1. Plaintiff: One Lawyer, 2 witnesses
2. Defense: One Lawyer, 2 witnesses
3. Audience is the "Jury".
4. Plaintiff Lawyer opens with 5 min speech, followed by Defense Lawyer
5. Plaintiff presents first witnesses and does direct exam (5 min). If Defense wishes, it may cross examine (3 min). If Plaintiff wishes, it may re-direct; Defense may re-cross. Time: 15 min max. Then go to the next Plaintiff witness. Then the first Defense witness, in which case Defense does direct and Plaintiff does cross; then the second Defense witness, and so on.
6. Then Plaintiff does a closing speech (4 min), followed by Defense (4 min). Plaintiff may rebut, but strictly based on the Defense closing speech (2 min or whatever remains).
7. Then everybody leaves the room, including judge. Jury (audience) deliberates, then hands the verdict. Since these are civil cases, the jury will award or not award a monetary amount to the plaintiff.
CONVERSION
Government equivalent to Plaintiff.
Opposition equivalent to Defense.
Sunday, December 20, 2009
Debate topics and debaters, in sequence
Castillo-Cayabyab-Cruz versus Valbuena-Villamil-Yao
In 2015 government scientists illegally produce a chimp/human chimera with a human IQ of 85, a TOEFL score of 510/677, and a human age equivalent of 21 years. It kills a human guard while escaping, is captured, and is held in a government facility. Government wants chimera "executed", but human and animal rights groups, who got to know of the case through the extensive media coverage, don't, and they go to court to stop the "execution". Will they succeed in saving "its" life?
Laqui-Lim-Lim versus Pamintuan-Pandy-Panlilio
In 1990 a Florida dentist was accused of malpractice by 8 of his patients who claimed he infected them with HIV. Evidence based on gp120 sequence data were presented to "prove" he was the source of the infection.
Reference: http://www.righto.com/theories/dentist.html
Recto-Roa-Salud Bautista versus Aquino-Babaran-Bejarin
During a paternity test, an additional screen was made on the child's DNA (child is 1 month old) without the knowledge of either parent (the father, is of course, in doubt). Doctor reveals to Mom, verbally, that child is at risk for a serious disease, and shows her the results on paper. Mom gets mad and accuses Doctor of breach of privacy. She sues.
Emralino-Espigar-Faustino versus Fortuno-Jamerlan-Juntila
Mom and Dad go for IVF and produce 6 viable embryos. One is implanted, and 9 months later a baby girl is born; the others are frozen. Three years late Mom and Dad file for annullment. Dad wants the frozen embryos discarded; Mom doesn't. They go to court.
Lorenzo-Marayag-Orpilla versus Zaulda-Bejarin-Recto
In a celebrated and oft-cited case, Moore goes to court against the Regents of the University of California, accusing them of patenting and getting royalties out of a cultured cell line that they established without his knowledge.
Reference: http://www.lawnix.com/cases/moore-regents-california.html
In 2015 government scientists illegally produce a chimp/human chimera with a human IQ of 85, a TOEFL score of 510/677, and a human age equivalent of 21 years. It kills a human guard while escaping, is captured, and is held in a government facility. Government wants chimera "executed", but human and animal rights groups, who got to know of the case through the extensive media coverage, don't, and they go to court to stop the "execution". Will they succeed in saving "its" life?
Laqui-Lim-Lim versus Pamintuan-Pandy-Panlilio
In 1990 a Florida dentist was accused of malpractice by 8 of his patients who claimed he infected them with HIV. Evidence based on gp120 sequence data were presented to "prove" he was the source of the infection.
Reference: http://www.righto.com/theories/dentist.html
Recto-Roa-Salud Bautista versus Aquino-Babaran-Bejarin
During a paternity test, an additional screen was made on the child's DNA (child is 1 month old) without the knowledge of either parent (the father, is of course, in doubt). Doctor reveals to Mom, verbally, that child is at risk for a serious disease, and shows her the results on paper. Mom gets mad and accuses Doctor of breach of privacy. She sues.
Emralino-Espigar-Faustino versus Fortuno-Jamerlan-Juntila
Mom and Dad go for IVF and produce 6 viable embryos. One is implanted, and 9 months later a baby girl is born; the others are frozen. Three years late Mom and Dad file for annullment. Dad wants the frozen embryos discarded; Mom doesn't. They go to court.
Lorenzo-Marayag-Orpilla versus Zaulda-Bejarin-Recto
In a celebrated and oft-cited case, Moore goes to court against the Regents of the University of California, accusing them of patenting and getting royalties out of a cultured cell line that they established without his knowledge.
Reference: http://www.lawnix.com/cases/moore-regents-california.html
Saturday, February 7, 2009
MBB 141 Exercises 090207
1. Genes met and thi are linked in Neurospora crassa; we want to locate arg with respect to met and thi. From the cross arg + + x + thi met (+ means wild-type, while italicized means mutant), the following random ascospores were obtained
arg thi met, 26
arg thi +, 17
arg + met, 3
+ thi met, 56
arg + +, 51
+ thi +, 4
+ + met, 14
+ + +, 29
Map the three genes.
2. Genes a and b are linked, with 10 percent recombination. What would be the phenotypes, and the probability of each, among progeny of the following cross?
(a b+/a+ b) x (a b/a b)
(a+ means mutant while a means wild-type; the organisms are diploid, indicated by the slash)
3. Use the following two-point recombination data to map the genes concerned, and show the order and the length of the shortest intervals:
Gene loci, % recombination
a,b, 50
a,c, 15
a,d, 38
a,e, 8
b,c, 50
b,d, 13
b,e, 50
c,d, 50
c,e, 7
d,e, 45
4. One parent heterozygous for three linked alleles (C,c,Sh,sh,Bz,bz) is test-crossed with a parent homozygous for the recessive version of all three alleles (c,c,sh,sh,bz,bz)
Breeding data are shown in this table. Position the three genes on a genetic map using this information.
arg thi met, 26
arg thi +, 17
arg + met, 3
+ thi met, 56
arg + +, 51
+ thi +, 4
+ + met, 14
+ + +, 29
Map the three genes.
2. Genes a and b are linked, with 10 percent recombination. What would be the phenotypes, and the probability of each, among progeny of the following cross?
(a b+/a+ b) x (a b/a b)
(a+ means mutant while a means wild-type; the organisms are diploid, indicated by the slash)
3. Use the following two-point recombination data to map the genes concerned, and show the order and the length of the shortest intervals:
Gene loci, % recombination
a,b, 50
a,c, 15
a,d, 38
a,e, 8
b,c, 50
b,d, 13
b,e, 50
c,d, 50
c,e, 7
d,e, 45
4. One parent heterozygous for three linked alleles (C,c,Sh,sh,Bz,bz) is test-crossed with a parent homozygous for the recessive version of all three alleles (c,c,sh,sh,bz,bz)
Breeding data are shown in this table. Position the three genes on a genetic map using this information.
| Group | Genotype | Crossovers | Number | Totals |
| 1 | CShBz | None; the parentals | 479 | 952 |
| 2 | cshbz | 473 | ||
| 3 | C|shbz | Single; between C and others | 15 | 28 |
| 4 | c|ShBz | 13 | ||
| 5 | CSh|bz | Single, between Bz and others | 9 | 18 |
| 6 | csh|Bz | 9 | ||
| 7 | C|sh|Bz | Double recombinants | 1 | 2 |
| 8 | c|Sh|bz | 1 | ||
| Totals | 1000 |
Wednesday, January 7, 2009
MBB 141 Question 36
36. Comparative genomics offers insights into the relationship between homologous genes and the organization of genomes. When the genome of C. elegans was sequenced, it was striking that some types of sequences were distributed nonrandomly. Consider the data obtained for chromosome V and the X chromosome shown here. The following figure shows the distribution of genes, the distribution of inverted and tandem repeat sequences, and the location of ESTs in C. elegans that are highly similar to yeast genes.
a. How do the distributions of genes, inverted and tandem repeat sequences, and conserved genes compare?
b. Based on your analysis in (a), what might you hypothesize about the different rates of DNA evolution (change) on the arms and central regions of autosomes in C. elegans?
c. Curiously, meiotic recombination (crossing-over) is higher on the arms of autosomes, with demarcations between regions of high and low crossing-over at the boundaries between conserved and nonconserved genes seen in the physical map. Does this information support your hypothesis in (b)?
MBB 141 Questions 27-35
27. BACs I-IV have been aligned in the contig shown in the figure above after screening a BAC library with a series of numbered STS and EST markers.
a) During the construction of BAC libraries, genomic DNA segments from different chromosomal regions sometimes are cloned together into a single BAC clone. This generates a chimeric BAC. How would you verify that none of the BACs in this contig was chimeric?
b) How would you identify the chromosomal region from which the contig derived?
c) How would you identify the orientation of the contig on a chromosome (i.e., which end of the contig is closer to the telomere or centromere)?
d) How would you identify the distance (in bp) of the markers in each BAC from each other and from the end of the insert?
28. When Celera Genomics sequenced the human genome, they obtained 13,543,099 reads of plasmids having an average insert size of 1,951 bp, and 10,894,467 reads of plasmids having an average insert size of 10,800 bp.
a) Dideoxy-chain termination sequencing provides only about 500-550 nucleotides of sequence. About how many nucleotides of sequence did they obtain from sequencing these two plasmid libraries?
b) Why did they sequence plasmids from two libraries with different-sized inserts?
c) They only sequencedf the ends of each insert. How did they determine the sequence lying between the sequenced ends?
29. Eukaryotic genomes differ in their repetitive DNA content. For example, consider the typical euchromatic 50-kb segment of human DNA that contains the human beta T-cell receptor. About 40 percent of it is composed of various genome-wide repeats, about 10 percent encodes three genes (with introns), and about 8 percent is taken up by a pseudogene. Compare this to the typical 50-kb segment of yeast DNA containing the HIS4 gene. There, only about 12 percent is composed of a genome-wide repeat, and about 70 percent encodes genes (without introns). The remaining sequences in each case are untranscribed and either contain regulatory signals or have no discernible information. Whereas some repetitive sequences can be interspersed throughout gene-containing euchromatic regions, others are abundant near centromeres. What problems do these repetitive sequences pose for sequencing eukaryotic genes? When can these problems be overcome, and how?
30. How has genomic analysis provided evidence that Archaea is a branch of life distinct from Bacteria and Eukarya?
31. Once a genomic region is sequenced, computerized algorithms can be used to scan the sequence to identify potential ORFs.
a) Devise a strategy to identify potential prokaryotic ORFs by listing features assessable by an algorithm checking for ORFs.
b) Why does the presence of introns within transcribed eukaryotic sequences preclude direct application of this strategy to eukaryotic sequences?
c) The average length of exons in humans is about 100-200 base pairs while the length of introns can range from about 100 to many thousands of base pairs. What challenges do these findings pose for identifying exons in uncharacterized regions of the human genome?
d) How might you modify your strategy to overcome some of the problems posed by the presence of introns in transcribed eukaryotic sequences?
32. One powerful approach to annotate genes is to compare the structures of cDNA copies of mRNAs to the genomic sequences that encode them. However, during the synthesis of cDNA, reverse transcriptase may not always copy the entire length of the mRNA and so a cDNA that is not full-length can be generated. This approach to gene annotation often uses ESTs that are not full-length cDNA copies of mRNA. Recently, a large collaboration involving 68 research teams analyzed 41,118 full-length cDNAs to annotate the structure of 21,037 human genes (see http://www.h-invitational.jp/).
a) What types of information can be obtained by comparing the structures of cDNAs with genomic DNA?
b) Why is it desirable, when possible, to use full-length cDNAs in these analyses?
c) The research teams characterized the number of loci per Mb of DNA for each chromosome. Among the autosomes, chromosome 19 had the highest ratio of 19 loci per Mb while chromosome 13 had the lowest ratio of 3.5 loci per Mb. Among the sex chromosomes, the X had 4.2 loci per Mb while the Y had only 0.6 loci per Mb. What does this tell you about the distribution of genes within the human genome? How can these data be reconciled with the idea that chromosomes have gene-rich regions as well as gene deserts?
d) The research teams were able to map 40,140 cDNAs to the current human genome sequence. Of the 978 cDNAs that could not be mapped, 907 could be roughly mapped to the mouse genome. Why might some (human) cDNAs be unable to be mapped to the current human genome sequence while they could be mapped to the mouse genome sequence? (Hint: consider where errors and limited information might exist.)
33. A central theme in genetics is that an organism’s phenotype results from an interaction between its genotype and the environment. Because some diseases have strong environmental components, researchers have begun to assess how disease phenotypes arise from the interaction of genes with their environments, including the genetic background in which the genes are expressed. (See http://pga.tigr.org/desc.shtml for additional discussion.) How might DNA microarrays be useful in a functional genomic approach to understanding human diseases that have environmental components, such as some cancers?
34. Mutations in the dystrophin gene can lead to Duchenne muscular dystrophy. The dystrophin gene is among the largest known: it has a primary transcript that spans 2.5 Mb, and produces a mature mRNA that is about 14 kb. Many different mutations in the dystrophin gene have been identified. What steps would you take if you wante to use a DNA microarray to identify the specific dystrophin gene mutation present in a patient with Duchenne muscular dystrophy?
35. Distinguish between structural, functional, and comparative genomics by completing the following exercise. The following list describes specific activities and goals associated with genome analysis. Indicate the area associated with the activity or goal by placing a letter (S, F, C) next to each item. Some items will have more than one letter associated with them.
a) Aligning DNA sequences within databases to determine the degree of matching.
b) Annotation of sequences within a sequenced genome.
c) Characterizing the transcriptome and proteome present in a cell at a specific developmental stage or in a particular disease state.
d) Comparing the overall arrangements of genes and nongene sequences in different organisms to understand how genomes evolve.
e) Describing the function of all genes in a genome.
f) Determining the functions of human genes by studying their homologs in nonhuman organisms.
g) Developing a comprehensive two-dimensional polyacrylamide gel electrophoresis map of all proteins in a cell.
h) Developing a physical map of a genome.
i) Developing DNA microarrays (DNA chips).
j) Identifying homologs to human disease genes in organisms suitable for experimentation.
k) Identifying a large collection of simple tandem repeat or microsatellite sequences to use as DNA markers within one organism.
l) Identifying expressed sequence tags.
m) Making gene knockouts and observing the phenotypic changes associated with them.
n) Mapping a gene in one organism using the lod score method.
o) Sequencing individual BAC clones aligned in a contig using a shotgun approach.
p) Using oligonucleotide hybridization analysis to type an SNP.
MBB 141 Questions 24-26
24. YAC clone contigs can be assembled following STS mapping. First, the locations of STSs on YAC clones are mapped using PCR. Then YAC clones that share STSs are aligned. The first step usually involves repeated screening of a library of YAC clones with different STS markers. In practice, some of the STSs may be well characterized and even localized to specific chromosomal regions or genes, whereas others may be less well characterized. In a typical screen of a YAC library constructed from the genome of a higher eukaryote, most STSs identify only a few YACs. However, a few STSs identify dozens of YACs. What might be the basis of this difference? How does the difference influence the assembly of a contig?
25. The average size of fragments, in base pairs, observed after genomic DNA from 8 different species was individually cleaved with each of 6 different restriction enzymes, is shown below:
continuing the table...
a) Under the assumption that each genome has equal amounts of A, T, G, and C, and that on average these bases are evenly distributed, what average fragment size is expected following digestion with each enzyme?
b) How might you explain each of the following?
i. There is a large variation in the average fragment sizes when different genomes are cut with the same enzyme.
ii. There is a large variation in the average fragment sizes when the same genome is cut with different enzymes that recognize sites having the same length (e.g., ApaI, HindIII, SacI, and SspI).
iii. Both SrfI and NotI, which each recognize an 8-bp site, cut the Mycobacterium genome more frequently than SspI and HindIII, which each recognize a 6-bp site.
c) Based on this data, which enzymes would be good choices for constructing a restriction map of a chromosome (or a large segment of a chromosome) in each organism? Explain your choices.
26. STS mapping has been useful to generate clone contig maps. Perform the following exercise to consider the logistics of locating STSs using PCR.
a) A plasmid library contains 500-bp inserts generated from randomly sheared mouse DNA. How would you identify clones harboring STRs with the dinucleotide repeat (AT)N or the trinucleotide repeat (CAG)N?
b) How would you use these STRs as STSs in a mapping experiment?
c) Nusbaum and colleagues generated a YAC-based physical map of the mouse genome by localizing 8,203 STSs onto 960 YAC clones. If each of the STSs were assayed in each of the 960 YAC clones, how many different PCRs would need to be analyzed?
d) Although PCRs can be performed robotically, each reaction consumes time and material resources, and with each there is a certain chance of a false positive result or other error. It is therefore advantageous to reduce the number of PCR reactions by pooling individual YACs together. First, yeast colonies containing the YACs are grown individually in the wells of ten 96-well plates. Suppose the plates are numbered I to X and the wells of each plate are arrayed in an 8-row x 12 grid. The rows are coded A-H and the columns coded 1-12. This allows the position of a single YAC to be specified uniquely by a code (e.g., II-C6 specifies the YAC clone from plate II, row C, column 6). In one pooling scheme, all YACs from each row of a plate are pooled into one well (e.g., those on plate II, A1-A12 are pooled together into a well designated II-A) and all YACs from each column of a plate are pooled into one well (those on plate II, A7-H7 are pooled together into a well designated II-7).
i. How many different pools would now have to be screened with each STS?
ii. How many PCRs would have to be performed?
iii. If the II6 and IIF pools had a positive result with STS #6239, what is the code of the YAC containing this STS?
iv. How would you interpret a result where only the IV3 pool was positive for a particular STS?
e) Construct a YAC contig based on the results in the following table
f) Devise a method to combine the YACs pooled in (c) to further reduce the number of PCRs. In your method, how many pools are there and how many PCRs must be performed?
25. The average size of fragments, in base pairs, observed after genomic DNA from 8 different species was individually cleaved with each of 6 different restriction enzymes, is shown below:
| | | | | | | |
| Species | ApaI | HindIII | SacI | SspI | SrfI | NotI |
| | GGGCCC | AAGCTT | GAGCTC | AATATT | GCCCGGGC | GCGGCCGC |
| Escherichia coli | 68000 | 8000 | 31000 | 2000 | 120000 | 200000 |
| M. tuberculosis | 2000 | 18000 | 4000 | 32000 | 10000 | 4000 |
| S. cerevisiae | 15000 | 3000 | 8000 | 1000 | 570000 | 290000 |
| A. thaliana | 52000 | 2000 | 5000 | 1000 | No sites | 610000 |
| C. elegans | 38000 | 3000 | 5000 | 800 | 1.11M | 260000 |
| D. melanogaster | 13000 | 3000 | 6000 | 900 | 170000 | 83000 |
| Mus musculus | 5000 | 3000 | 3000 | 3000 | 120000 | 120000 |
| Homo sapiens | 5000 | 4000 | 5000 | 1000 | 120000 | 260000 |
continuing the table...
| | |
| Species | NotI |
| | GCGGCCGC |
| Escherichia coli | 200000 |
| M. tuberculosis | 4000 |
| S. cerevisiae | 290000 |
| A. thaliana | 610000 |
| C. elegans | 260000 |
| D. melanogaster | 83000 |
| Mus musculus | 120000 |
| Homo sapiens | 260000 |
a) Under the assumption that each genome has equal amounts of A, T, G, and C, and that on average these bases are evenly distributed, what average fragment size is expected following digestion with each enzyme?
b) How might you explain each of the following?
i. There is a large variation in the average fragment sizes when different genomes are cut with the same enzyme.
ii. There is a large variation in the average fragment sizes when the same genome is cut with different enzymes that recognize sites having the same length (e.g., ApaI, HindIII, SacI, and SspI).
iii. Both SrfI and NotI, which each recognize an 8-bp site, cut the Mycobacterium genome more frequently than SspI and HindIII, which each recognize a 6-bp site.
c) Based on this data, which enzymes would be good choices for constructing a restriction map of a chromosome (or a large segment of a chromosome) in each organism? Explain your choices.
26. STS mapping has been useful to generate clone contig maps. Perform the following exercise to consider the logistics of locating STSs using PCR.
a) A plasmid library contains 500-bp inserts generated from randomly sheared mouse DNA. How would you identify clones harboring STRs with the dinucleotide repeat (AT)N or the trinucleotide repeat (CAG)N?
b) How would you use these STRs as STSs in a mapping experiment?
c) Nusbaum and colleagues generated a YAC-based physical map of the mouse genome by localizing 8,203 STSs onto 960 YAC clones. If each of the STSs were assayed in each of the 960 YAC clones, how many different PCRs would need to be analyzed?
d) Although PCRs can be performed robotically, each reaction consumes time and material resources, and with each there is a certain chance of a false positive result or other error. It is therefore advantageous to reduce the number of PCR reactions by pooling individual YACs together. First, yeast colonies containing the YACs are grown individually in the wells of ten 96-well plates. Suppose the plates are numbered I to X and the wells of each plate are arrayed in an 8-row x 12 grid. The rows are coded A-H and the columns coded 1-12. This allows the position of a single YAC to be specified uniquely by a code (e.g., II-C6 specifies the YAC clone from plate II, row C, column 6). In one pooling scheme, all YACs from each row of a plate are pooled into one well (e.g., those on plate II, A1-A12 are pooled together into a well designated II-A) and all YACs from each column of a plate are pooled into one well (those on plate II, A7-H7 are pooled together into a well designated II-7).
i. How many different pools would now have to be screened with each STS?
ii. How many PCRs would have to be performed?
iii. If the II6 and IIF pools had a positive result with STS #6239, what is the code of the YAC containing this STS?
iv. How would you interpret a result where only the IV3 pool was positive for a particular STS?
e) Construct a YAC contig based on the results in the following table
| STS marker | Positive YAC pools |
| 63 | II-6, II-A |
| 210 | II-6, II-A, IV-C, IV-3 |
| 522 | VII-E, VII-12, X-G, I-C, I-8 |
| 713 | I-C, I-8 |
| 714 | VII-E, VII-12 |
| 719 | X-H, X-9, IV-C, IV-3 |
| 991 | X-H, X-9, VII-E, VII-12 |
| 1071 | II-6, II-A, IV-C, IV-3, X-H, X-9 |
| 2631 | II-6, II-A |
| 3097 | VII-E, VII-12, I-C, I-8 |
| 4630 | VII-E, VII-12, I-C, I-8 |
| 5192 | X-H, X-9, IV-C, IV-3 |
| 6193 | X-H, X-9, VII-E, VII-12 |
| 6892 | II-6, II-A, IV-C, IV-3 |
f) Devise a method to combine the YACs pooled in (c) to further reduce the number of PCRs. In your method, how many pools are there and how many PCRs must be performed?
MBB 141 Questions 21-23
21. A cDNA library is made with mRNA isolated from liver tissue. When a cloned cDNA from that library is digested with the enzymes EcoRI (E), HindIII (H), and BamHI (B), the restriction map shown above, part (a) is obtained. When this cDNA is used to screen a cDNA library made with mRNA from brain tissue, three identical cDNAs with the restriction map show in the following figure, part (b) are obtained. When either cDNA is used to synthesize a uniformly labeled 32-P-labeled probe and the probe is allowed to hybridize to a Southern blot prepared from genomic DNA digested singly with the enzymes EcoRI, HindIII, and BamHI, an autoradiograph shows the pattern of bands in the figure part (c). When either cDNA is used to synthesize a uniformly labeled 32-P-labeled probe and used to probe a northern blot prepared with poly(A) RNA isolated from live and brain tissues, the pattern of bands in part (d) of the figure is seen. Fully analyze these data and then answer the following questions:
a) Do these cDNAs derive from the same gene?
b) Why are different-sized bands seen on the northern blot?
c) Why do the cDNAs have different restriction maps?
d) Why are some of the bands seen on the whole-genome Southern blot different sizes than some of the restriction fragments in the cDNAs?
22. Draw the pattern of bands you would expect to see on a DNA sequencing gel if you annealed the primer 5’-CTAGG-3’ to the following single-stranded DNA fragment and carried out a dideoxy sequencing experiment. Assume the dNTP precursors were all labeled.
3’-GATCCAAGTCTACGTATAGGCC-5’
23.
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