The first heart chamber that the dye would enter is right atrium.
This will be followed by
left ventricle
right ventricle
left atrium
coronary valve
Answer: 4%
Out of 5000kcal, the owl loses 2300kcal which mean it could only process 2700kcal. Out of 2700kcal processed, 2500 kcal is used for cellular respiration so there is 200 kcal used to make body cells. The production efficiency would be: 200kcal/5000kcal= 4%
A. The following statements are hypotheses:
1. Glucose may cause an increase in bacterial growth.
2. Increased glucose may lead to capsule formation in bacteria.
B. The following statements are observations:
1. The bacterial colony grown without glucose do not have capsules around their cells.
2. Bacteria growth in the glucose environment increase rapidly and then stopped after sometime.
Full question found from other source
The F2 generation phenotypes for each cross are shown in Table 1. (See attachment) Which of the following is the mean number per cross of F2 generation offspring that are the result of crossing over?
Answer:
B, 2.2
Explanation:
The parental genotypes are long and black vs short and white. Therefore the phenotypes that result from crossing over are long and white, and short and black. (middle two rows of the table). If we add up the total number of offspring with these genotypes we get 6 long whites, and 5 short blacks.
The total is 11 from 5 crosses, so the mean is 11/5 = 2.2
Answer: as the question lack attachment i have added the link to complete answer in section ask for details.
a.
6ATP
6 ADP
b.
6 NADPH
6 NADP+
c.
6Pi
d.
2P
e.
3 ADP
3 ATP
Explanation: i have added the picture illustrating the complete reaction see the attachment.
