In a comparative genomics study, 40 strains of Escherichia coli were sequenced and their genomes annotated. Of the 40 strains, 20 were recovered from the urinary tract of patients with urinary tract infections (UTIs) and 20 isolates were recovered from the large intestines of humans with no symptoms of disease. All 40 strains had 3700 genes in common. The 20 UTI isolates had 4100 genes in common while the 20 gut isolates had 3900 genes in common. All isolates had genomes of between 4700 and 5000 genes and each isolate had at least 15 genes not found in any other isolate. What is the size of the pan-genome of the entire set of strains?Question 9Answera.3700b.>5100c.4100d.3900
Question
In a comparative genomics study, 40 strains of Escherichia coli were sequenced and their genomes annotated. Of the 40 strains, 20 were recovered from the urinary tract of patients with urinary tract infections (UTIs) and 20 isolates were recovered from the large intestines of humans with no symptoms of disease. All 40 strains had 3700 genes in common. The 20 UTI isolates had 4100 genes in common while the 20 gut isolates had 3900 genes in common. All isolates had genomes of between 4700 and 5000 genes and each isolate had at least 15 genes not found in any other isolate. What is the size of the pan-genome of the entire set of strains?Question 9Answera.3700b.>5100c.4100d.3900
Solution
The pan-genome size of the entire set of strains can be calculated by adding the number of common genes to the unique genes found in each strain.
First, we know that all 40 strains had 3700 genes in common.
Next, we know that each strain had at least 15 unique genes not found in any other strain. Since there are 40 strains, this adds up to 40 strains * 15 unique genes/strain = 600 unique genes.
Therefore, the size of the pan-genome of the entire set of strains is 3700 common genes + 600 unique genes = 4300 genes.
However, the question states that all isolates had genomes of between 4700 and 5000 genes. This suggests that there may be additional genes that are not common to all strains but are present in some strains.
The 20 UTI isolates had 4100 genes in common, which is 400 more than the 3700 common to all strains. The 20 gut isolates had 3900 genes in common, which is 200 more than the 3700 common to all strains.
Adding these additional genes to our previous total gives 4300 genes + 400 genes (from UTI isolates) + 200 genes (from gut isolates) = 4900 genes.
However, since the question states that each isolate had at least 15 genes not found in any other isolate, and we have already accounted for these unique genes in our calculation, the size of the pan-genome of the entire set of strains is likely greater than 4900 genes.
Therefore, the correct answer is b. >5100.
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