The correct answer for this question is:
A farmer planted legumes and cabbage in the same field that is devoid of fertilizers. The yield from this field is better than the cabbage planted inanother field without legumes. The reason for this is because (A) <span>nitrogen-fixing present in the roots of legumes aid enrichment of nitrogen in the soil.</span>
Answer:
0%
Explanation:
Round = Dominant
Wrinkled = recessive
Recessive = r
Dominant = R
homozygous dominant = RR
homozygous recessive = rr
Heterozygous= Rr
Since both seeds are homozygous recessive (rr)
In your punnet square, one lower case r will go on the top and the left side
r r
r
r
Something like that.
Go across your four boxes and you will end up with;
rr, rr
rr, rr
In order for it to be round, at least one box would need to have to be a heterozygous(Rr) or be a homozygous dominant.
<span>If you locate an animal that has segmentation and cephalization as well as deuterostome embryonic development, the phylum that this animal will belong to is A. chordates.
The remaining options do not have these characteristics that only chordates, such as fish, mammals, amphibians, reptiles, and birds have.
</span>
For eukaryotes to produce the next generation through sexual reproduction, two cells must contribute genetic material.
Sexual reproduction, the process through which two individuals of different sexes combine their genetic information to create new species. In the majority of organisms, reproductive cells called gametes, which mate to create a diploid zygote, have chromosomes in their nuclei that carry the genetic material. The zygote grows into a new person. The most common method of reproduction among living things is sexual reproduction. A remarkable variety of offspring, each with a genetic make-up different from that of its parents, may result from sexual reproduction because it allows the rearranging of genetic material both within and between members of one generation. Sexual reproduction is involved.
Learn more about Sexual reproduction here
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Answer:
Explanation:
1. Over time more and more molecules of DCPIP would be decreased by electrons taken from plastoquinone (Pq). This statement is true in the sense that DCPIP is an artificial electronic acceptor which is decreased by taking electrons from plastoquinone.
2.) Water splitting by Photosystem II would occur at higher levels than if the drug were administered without DCPIP. This statement is true.
3.)Events in Photosystem I would occur just as they would in the absence of drug and DCPIP. This statement is false because the drug obstructs the electron transport from plastoquinone to cytochrome B. So the events are abnormal because no protein transfer is involved in this mechanism.
