Answer:
The last one
Explanation:
They are both found in the middle and everything else is false
The given question is incomplete, the complete question is:
During germination, peas sprout and grow. The data table shows the carbon dioxide produced during the germination period of peas under different conditions. Condition Rate of carbon dioxide produced (mL/min) Germinating peas, 10ºC 0.01 Germinating peas, 20ºC 0.02 What is the best conclusion? The rate of cellular respiration in germinating peas is exactly one thousand times the rate of ATP production. Germinating peas at 10ºC create carbon dioxide at a rate of 0.01 mL/min during ATP production. Germinating peas at 20ºC have a higher rate of cellular respiration than germinating peas at 10ºC. The rate of cellular respiration cannot be measured without knowing the rate of ATP production.
Answer:
The correct statement is that at 20 degree C, the germinating peas exhibits a higher rate of cellular respiration in comparison to the germinating peas at 10 degree C.
Explanation:
The process of respiration results in the production of carbon dioxide, respiration refers to a chemical reaction that generates water, carbon dioxide, and energy by undergoing oxidation of the glucose molecules. This phenomenon plays an essential role in the life of the organisms for obtaining energy from the food they consume to perform daily activities.
From the question, it is evident that the peas, which were germinating at 20 degrees C exhibit a higher rate of cellular respiration as they are generating 0.02 milliliters of carbon dioxide in a minute, while on the other hand, the germinating peas at 10 degrees C are giving rise to 0.01 milliliters of carbon dioxide in a minute.
Answer:
a closed reduction procedure
Explanation:
The procedure basically takes about 30 minutes. The doctor administer anaesthesia to the patient to reduce pain. The dislocated bone is pulled to its normal position.
Physical therapy is recommended by the doctor after healing to prevent inflammation that may occur around the soft tissue region that had been injured.
Answer:
The ratio of blue to white offspring in the progeny is 4 blue : 12 white.
Explanation:
<u>Available data:</u>
- The dominant allele K is necessary to synthesize blue flower pigment
- K is inhibited by the dominant allele D
- Plants with the genotype K- D- will not produce pigment (and their flowers will be white)
Cross: testcross for (Kk Dd) plants
Parental) KkDd x kkdd
Gametes) KD kD Kd kd
kd kd kd kd
Punnet square) KD Kd kD kd
kd KkDd Kkdd kkDd kkdd
kd KkDd Kkdd kkDd kkdd
kd KkDd Kkdd kkDd kkdd
kd KkDd Kkdd kkDd kkdd
- Whenever D is present, it inhibits the expression of the K gene, so every plant with the dominant D allele will be white. This plants´ genotype is kkD- or K-D-.
- Whenever D is absent and K is present, every plant with genotype K-dd will be blue.
- The recessive form for K and D genes will express white-flowered plants, with genotype ddkk
F1) Progeny genotype: 4/16 KkDd, white-flowered plants
4/16 Kkdd, blue-flowered plants
4/16 kkDd, white-flowered plants
4/16 kkdd, white-flowered plants
The ratio of blue to white offspring in the progeny is 4 blue : 12 white.
Blue-flowered plants: 4 Kkdd
White-flowered plants: 4 KkDd + 4 kkDd + 4 kkdd
Answer:
(B) The organism lives and replicates despite protease and DNase treatment, but the organism dies when treated with RNase.
Explanation:
DNA is the genetic material in present day organisms while some of the proteins serve as enzymes and catalyze the metabolic reactions.
An organism having RNA with the ability to serve as genetic material would not require DNA to survive. Likewise, if RNA molecules in these organisms also serve as the enzyme, proteins would not be required for survival.
Therefore, when treated with DNase and protease that digest the DNA and proteins respectively, these organisms would be able to survive. However, treatment with RNase enzymes that digest the RNA would kill them as RNA was their biocatalyst and genetic material.
