Answer:
Recessive phenotype plant
Explanation:
If the plant has recessive genotype then it will be easy to identify because it will have the recessive phenotype. However if the plant has dominant phenotype it can either have homozygous genotype or heterozygous genotype. To confirm if the plant is homozygous or heterozygous, a test cross can be done with plant having recessive phenotype.
If our test plant is homozygous dominant, it will pass one dominant allele to the next generation and all the offspring would have dominant phenotype. If our test plant is heterozygous dominant it will pass one dominant allele to half of the offspring and one recessive allele to another half so 50% of next generation will have dominant phenotype and other 50% will have recessive phenotype.
Hence by test cross (cross with recessive phenotype plant) it is possible to determine the genotype of the uncharacterized pea plant.
Answer:
A. 1' ... 5'
Explanation:
The pentose sugars present in nucleotides have five-carbon structures. Other components of a nucleotide are a phosphate group and a nitrogenous base. The nitrogenous base in a nucleotide can be a purine or pyrimidine. The purine bases are covalently joined to the 1' carbon atom of the pentose sugar at their N-9. The bond between the nitrogenous base and a pentose sugar is called the N-beta glycosyl bond.
Similarly, N-1 of a pyrimidine base is covalently linked to the 1' carbon of the pentose sugar. The phosphate group of a nucleotide is bonded to the 5' carbon of the pentose sugar.
Answer:
With 6 we should multiply gross photosynthesis (as we were measuring gross photosynthesis for 10 min).
Explanation:
According to this question:-
Net photosynthesis- 3mm/20min- 1.5mm/10min
Cellular respiration- 1mm/15min- 0.66mm/10min
Gross photosynthesis = Net photosynthesis + Cellular respiration = 1.5+0.66 = 2.16 (mm/10min)
Rate of photosynthesis = (gross photosynthesis)6 = 2.16 *6 = 12.96 mm/hr
As we gage the photosynthesis rate in hr.