Answer:
the answer is A. E. coli B
Explanation:
The multiplicity of infection (MOI) refers to the ratio between the numbers of viruses used to infect <em>E. coli</em> cells and the numbers of these <em>E. coli </em>cells. Benzer carried out several experiments in order to define the gene in regard to function. Benzer observed that <em>E. coli </em>strains with point mutations could be classified into two (2) complementary classes regarding coinfection using the restrictive strain as the host. With regard to his experiments, Benzer observed that rII1 and rII2 mutants (rapid lysis mutants) are complementary when they produce progeny after coinfect E. coli K (where neither mutant can lyse the host by itself). The rII group of mutants studied by Benzer does not produce plaques on <em>E. coli</em> K strains that carry phage λ (lysogenic for λ), but they produce plaques on <em>E. coli</em> B strains. This study showed that rIIA and rIIB are different genes and/or cistrons in the rII region.
<span>The
answer is more than 6 METs, MET measures by oxygen used by the body. Vigorous-intensity exercise is a physical
activity done with a large amount of effort. It is the intensity at which you
have a substantially higher heart rate and rapid breathing. </span>
Answer:
C)Parental: 41% Dr, 41% dR; recombinant: 9% DR, 9% dr.
Explanation:
The notation Dr/dR for genotypes means that one homologous chromosome has the alleles Dr and the other homologous chromosome has the alleles dR.
The heterozygous plant Dr/dR will produce 4 types of gametes: two identical to the chromosmes the individual has in its somatic cells (called parental), and two gametes which will be a mix of the alleles in the homologous chromosomes (called recombinant).
- Dr: parental
- dR: parental
- DR: recombinant
- dr: recombinant
To calculate the frequency of each type of gamete, we must use the formula:
Distance (map units) / 100 = frequency of recombination.
18 mu / 100 = 0.18.
The total frequency of recombination between the genes D and R is 0.18, but every time crossing over happens, two recombinant gametes are generated. Therefore, each recombinant gamete will have a frequency of 0.18/2=0.09 = 9%.
The frequency of parental gametes will be:
1 - frequency of recombinant gametes
1 - 0.18 = 0.82
But there are 2 parental gametes, so each of them will have a frequency of 0.82/2=0.41 = 41%.
