During cyclic photophosphorylation, electrons are cycled continuously out of and back into the reaction center of PSI. It does not involve PSII and plastoquinone. During non cyclic photophosphorylation, the reaction center of plant photosystem II passes electrons to plastoquinone. The reduced plastoquinone carries electrons to the cytochrome b6f complex. Since PSII and plastoquinones are not included in cyclic photophosphorylation, dichlorophenyldimethylurea would not affect the process.

5 0

2 years ago

To investigate whether roots exert a pushing force strong enough to cause water to move up the stems plants,my question is state

polet [3.4K]

The stem of the plant should be cut at least three to four inches above the surface level.
The glass capillary should be tightly fixed with the stem.
The initial water level and the final water level should be carefully noted.
The droplets on the leaves should be observed and conclude that these drops are not from transpiration.

7 0

3 years ago

When corn plants are too tall, they lodge (topple over) prior to harvest. This is bad since the grain is often lost when it cann

Shalnov [3]

Answer:

The phenotypic variance= 2113

The eenvironmenal variance= 644

The genetic variance= 1469

The broad-sense heritability = 0.695

Explanation:

To determine the phenotypic variance, we need to know the calculate the environmental variance and the genetic variance first;

because Phenotypic variance = V(p) such that;

V(p) = V(e) + V(g)

where V(e) = environmental variance and V(g) = genetic variance

Therefore, Environmental variance can be defined as a portion of phenotypic variance due to differences in the environments to which the individuals in a population have been exposed. The total amount of variance observed is dependent on the genetic component, determined by the variation that is inherited.

This implies that V(e)= initial variance which was there in the parents ( i.e Variance of inbred line 1 = 311 and Variance of inbred line 2 = 333 )

V(e) = (311 + 333) = 644

Genetic Variance simply refer to the variance that occurs or preferrably , that is observed after inbreeding (i.e Variance of F1 = 316 and Variance of F2 = 1153)

V(g) = (316 + 1153) = 1469

Now, back to phenotypic variance = V(e) + V(g)

substituting the parameters we have; 1469 + 644 = 2113

Broad-sense heritability can be calculated as the ratio of total genetic variance to total phenotypic variance.

= V(g)/ V(p)

= 1469/ 2113

= 0.695

4 0

2 years ago