Staphylococcus aureus has become resistant to penicillin, methicillin, and now possibly to vancomycin. Since bacteria do not sex
ually reproduce, genetic recombination does not introduce new alleles into the genome. Bacteria do mutate at a high rate and can also swap genetic material through horizontal gene transfer. How did S. aureus acquire resistance in each instance?
1 answer:
Answer:
S. aureus acquire resistance by :
1) Beta-lactamase production.
2) Changes in penicillin-binding protiens (PBP) (mecA genes encode these altered PBPs )
3) Genes encoding the enzymes substituting D-lactate for D-alanine in the peptidogylcan.
Explanation:
More than 90% strains of Staphylococcus aureus contains plasmids which code for <u>beta- lactamases</u> which cause the enzymatic hydrolysis of the beta lactam ring resulting in the loss of antibacterial activity. (Resistance against penicillins )
Some of the strains of S.Aureus are resistant to beta lactamase resistant penicillins such as methicilin (MRSA) and nafcillin (NRSA) by virtue of <u>changes in the penicillin-binding protien (PBP)</u> in their cell membrane.
Some strains of S.Aureus are resistant to Vancomycin (VRSA) and they acquire this resistance by some genes located in a transposon on a plasmid and encode the enzymes that<u> substitute D-lactate for D-alanine in the peptidoglycan.</u>
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A.
The genotype frequencies can be determined by dividing the number of individuals with a genotype by the total number all the individuals.
FF genotype frequency:
98÷200=0,49. There are approximately 5 <span>homozygous dominant mice in 10.
Ff genotype frequency:
84</span>÷200=0,42. There are approximately 4 <span>heterozygous mice in 10.
ff genotype frequency:
18</span>÷200=0,09. There are approximately 1 <span>homozygous recessive mouse in 10.
B.
To determine the frequency of an allele we can divide the number of times that the allele appears by the total number of alleles.
F allele frequency:
</span>
<span>=0.7
There are 7 F alleles in 10 random alleles.
f </span>allele frequency:
=0.3
There are 3 f alleles in 10 random alleles.<span>
</span>
The genotype frequencies can be determined by dividing the number of individuals with a genotype by the total number all the individuals.
FF genotype frequency:
98÷200=0,49. There are approximately 5 <span>homozygous dominant mice in 10.
Ff genotype frequency:
84</span>÷200=0,42. There are approximately 4 <span>heterozygous mice in 10.
ff genotype frequency:
18</span>÷200=0,09. There are approximately 1 <span>homozygous recessive mouse in 10.
B.
To determine the frequency of an allele we can divide the number of times that the allele appears by the total number of alleles.
F allele frequency:
</span>
There are 7 F alleles in 10 random alleles.
f </span>allele frequency:
There are 3 f alleles in 10 random alleles.<span>
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