Answer:
The correct answer is A: Simple cuboidal epithelium.
Explanation:
Simple Cuboidal Epithelium: These epithelia are formed by a single layer of cuboidal cells, that are <em>as </em>tall as they are wide. They have a rounded, central nucleus. Occasionally these cells present cilia, flagella or <em>microvillus</em> in their apical surface, lining the lumen. Microvillus are fingerlike projections. This epithelium can be found for example in the proximal convolutes renal tubule.
Answer:
Because each subject in an experiment has to have equal conditions to ensure the most accurate results. And water temperature is known to impact plant growth. An experiment using different water temperatures is a terribly inaccurate experiment.
The intensity of light had greater impact on the rate of photosynthesis. It was observed that the jar in which the intensity of light was high, large amount of oxygen was produced as compared to the jar in which the intensity of light was low.
In the process of photosynthesis, oxygen is produced from the carbon dioxide. As the oxygen in the jar increases, the leaf disk rises with in the jar which also signifies the higher oxygen production with higher rate of photosynthesis in presence of high intensity of light.
Mitochondria is the correct answer
Answer:
C)Parental: 41% Dr, 41% dR; recombinant: 9% DR, 9% dr.
Explanation:
The notation Dr/dR for genotypes means that one homologous chromosome has the alleles Dr and the other homologous chromosome has the alleles dR.
The heterozygous plant Dr/dR will produce 4 types of gametes: two identical to the chromosmes the individual has in its somatic cells (called parental), and two gametes which will be a mix of the alleles in the homologous chromosomes (called recombinant).
- Dr: parental
- dR: parental
- DR: recombinant
- dr: recombinant
To calculate the frequency of each type of gamete, we must use the formula:
Distance (map units) / 100 = frequency of recombination.
18 mu / 100 = 0.18.
The total frequency of recombination between the genes D and R is 0.18, but every time crossing over happens, two recombinant gametes are generated. Therefore, each recombinant gamete will have a frequency of 0.18/2=0.09 = 9%.
The frequency of parental gametes will be:
1 - frequency of recombinant gametes
1 - 0.18 = 0.82
But there are 2 parental gametes, so each of them will have a frequency of 0.82/2=0.41 = 41%.
