By producing progeny cells that include plasma cells and memory cells
Answer:
Day 10
Explanation:
If the number of bacteria increases by a factor of two each day (i.e. it doubles), it will reach 5120 by day 10. At this rate, the colony likely started with 10 bacteria on day one and doubled each day to reach 320 by day 6.
Day 6- 320 bacteria
Day 7- 320×2= 640 bacteria
Day 8- 640×2= 1280 bacteria
Day 9- 1280×2= 2560 bacteria
Day 10- 2560×2= 5120 bacteria
Answer:
0.2404
Explanation:
The genes R/r and E/e are linked and there is 4% recombination between them.
<u>The possible genotypes and phenotypes are:</u>
- RR or Rr: Rh+ blood type
- rr: Rh- blood type
- EE or Ee: elliptocytosis
- ee: normal red blood cells
Tom and Terri each have elliptocytosis (they are E_), and each is Rh+ (they are R_).
Tom's mother has elliptocytosis (E_) and is Rh- (rr), so she has the genotype Er/_r. His father is healthy (ee) and has Rh+ (R_), so he has the genotype eR/e_. Tom must have inherited his E allele from his mother and his R allele from his father, so he has the genotype eR/Er.
Terri's father is Rh+ (R_) and has elliptocytosis (E_), while Terri's mother is Rh- (rr) and is healthy (ee) with the genotype er/er. Terry could only receive the chromosome <em>er </em>from her mother, and because she is heterozygous for both genes the dominant alleles were both received from her father. Terri's genotype is ER/er.
The frequency of recombination is 4%, so 4% of the produced gametes will be recombinant. There are two possible recombinant gametes, so each will appear 2% of the times (a frequency of 0.02).
<u />
<u>Tom will produce the following gametes:</u>
- eR, parental (0.48)
- Er, parental (0.48)
- er, recombinant (0.02)
- ER (recombinant (0.02)
<u>Terri will produce the following gametes:</u>
- ER, parental (0.48)
- er, parental (0.48)
- Er, recombinant (0.02)
- eR, recombinant (0.02)
A child Rh- with elliptocytosis has the genotype rrE_. This can happen from the independent combination of the following gametes from Tom and Terri respectively:
- Er (0.48) × er (0.48) = 0.2304 Er/er
- Er (0.48) × Er (0.02) = 0.0096 Er/Er
- er (0.02) × Er (0.02) = 0.0004 er/Er
And the total probability of having a rrE_ child will be 0.2304 + 0.0096 + 0.0004 = 0.2404
Answer:
<u>Nutrition and health benefits</u> are the most likely motivation ,if a person decides to switch from drinking freshly squeezed orange juice to calcium-enriched orange juice after a yearly checkup,
Explanation:
As the Calcium enriched orange juice contains calcium almost equal to that of milk , hence it is an essential mineral which helps in building and maintaining strong bones and teeths.
It is also an excellent source of vitamin D helping the body to absorb more calcium . Thus providing excellent nutrition and health benefits.
Hence <u>Nutrition and health benefits</u> are the most likely motivation
