Solution:
4.2 x 10^6 bp/10^3 bp/seconds = 4.2 + 103 s which is 4200 seconds and equivalents to 70 minutes
In addition, assuming a pause of 2 seconds for re initiating after completing every okazaki fragment and assuming the okazaki fragments average 1000 nucleotide long.
4.2 x 10^6 bp/10^3 bp = 4200 okazaki fragments 4200 * 2 seconds = 8400 seconds which is 140 minutes or 2 hours 20 minutes of pauses alone.
Therefore, overall time would be pauses plus the 70 minutes so total time of 210 minutes. Assuming that the replisome completely disassociates from the DNA after every okazaki fragment and must spend one-minute rebinding.
4200 okazaki fragments. 60 seconds rebinding time per fragment: 4200 x 1 minute = 4200 minutes rebinding time plus 70 minutes for actual replication. 4200 minutes is 70 hours which is almost 3 days.
Answer:
The correct answer will be- Meiosis I, anaphase I
Explanation:
Alleles are the variant alternative forms of a gene which determines the trait of an organism. The trait or phenotype is expressed only when the two alleles controlling a single trait on separate chromosomes are aligned in a complementary position.
During gamete formation, the alleles controlling a trait segregate independently of each other. This segregation of alleles takes place during Anaphase I of meiosis I. During anaphase, the sister chromatids are separated by the mitotic spindles so that the chromosomes can reach the opposite poles.
Thus, Meiosis I, anaphase I am the correct answer.
