Answer:
2% of the progeny will be double crossovers for the trihybrid test cross
Explanation:
By knowing the positions of genes, we can estimate the distances in MU between them per region.
- Genes A and B are 10 map units apart (Region I)
- Genes B and C are 20 map units apart (Region II)
- Genes A and C are 30 map units apart
----A-------10MU--------B-------------20MU-------------C---
Region I Region II
We can estimate the recombination frequencies by dividing each distance by 100.
• recombination frequency of A-B region = 10MU / 100 = 0.10
• recombination frequency of B-C region = 20MU / 100 = 0.20
Now that we know the recombination frequencies in each region, we can calculate the expected double recombinant frequency, EDRF, like this:
EDRF = recombination frequency in region I x recombination frequency in region II.
EDRF = 0.10 x 0.20 = 0.02
2% of the progeny will be double crossovers for the trihybrid test cross
Answer:
photosynthesis
Explanation:
the formula for photosynthesis is
6CO2 + 6H2O + Light energy → C6H12O6 (sugar) + 6O2 +ATP
Answer:
The correct answer is b glucose is broken down into two pyruvate molecules.
Explanation:
Krebs cycle is one of the most important step of aerobic respiraion.Krebs cycle occur within mitochondrial matrix.During krebs cycle the acetyl CoA generated from pyruvate reacts with oxalo acetate in a cyclic manner to regenerate 4 carbon compound oxaloacetate.
During krebs cycle pyruvate is broken down or oxidized to form 2 carbon compound known as acetyl CoA and carbon dioxide as byproduct.
But the breakdown of glucose into two pyruvate molecules occur during glycolysis.
1. <span>what is the amount of the bolus dose, in both milligrams and milliliters, that you will administer in the first minute?
</span>The doses is 0.9 mg/kg and the weight of the patient is 143 pounds. So, the total doses of drug needed will be:
Total doses= 0.9 mg/kg * 143 pounds * 0.453592 kg/pound= 58.37 mg.
10% of the doses will be given bolus for 1 min, so the amount would be:
Bolus doses= 10%*58.37 mg= 5.837 mg.
In mililiters, it would be: 5.837 mg * 1ml/mg= 5.837 ml.
<span>2. what is the amount of the remaining dose that you will need to administer?
The remaining dose would be 90% of the total dose. You can either calculate it directly or subtract the bolus doses from the total doses.
Remaining doses= total doses- bolus doses= </span>58.37 mg- 5.837 mg= <span>52.533mg</span>
Answer:
All results for the experimental groups involving wing surgery would be invalid.
Explanation:
The experimental group involving wing surgery would not lead to correct results because it is a direct parameter that is impaired. This has been already clearly stated in the question that these flies are prone to attack compared to those which are not treated. There could be several reasons for that such as: (1) the reattachment would not have been successful, and/or (2) the flies' health (fitness) status would have been compromised.
On the other hand, it is not correct that all experimental groups will lead to incorrect results because the untreated flies were not attacked (or less attacked). Similarly, there is no condition given for houseflies rather only zonosemata flies. So 3rd option is also not possible. Finally, the reliability of the results would definitely change because the treated flies would be more attacked and lead to a clear difference in results.
