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Ymorist [56]
3 years ago
7

Name two ways pollination can occur

Biology
1 answer:
iogann1982 [59]3 years ago
8 0
Cross pollination through wind and other animals and self pollination through the flowers themselves.
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How do nerves pass along messages to and from the brain?
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The dendrites are covered with synapses formed by the ends of axons from other neurons. ... When neurons receive or send messages, they transmit electrical impulses along their axons, which can range in length from a tiny fraction of an inch (or centimeter) to three feet (about one meter) or more.
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The exhaustion stage of stress comes from the ___________ cycle.
devlian [24]

Answer: alarm and resistance

The exhaustion stage of stress originates from <span>the first two stages of the cycle of stress,  the alarm resistance. tress,  </span>and<span>Once alarm ( where the body system is flooded with adrenaline)  is ended, resistance sets in and finally, exhaustion takes over when the body runs out of energy.</span>

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Cell walls in plant cells provide support and structure for the cell, allowing plants to 'stand up' without a skeleton. What are
natta225 [31]

Answer:

Cellulose

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6 0
3 years ago
Read 2 more answers
A population of 150 individuals has an allele frequency of 0.2 for the dominant allele (H) and a frequency of 0.8 for the recess
Amanda [17]
Hardy-Weinberg Equation (HW) states that following certain biological tenets or requirements, the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimel. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = genotype frequency of homozygous dominant individuals, 2pq = genotype frequency of heterozygous individuals, and q^2 = genotype frequency of homozygous recessive individuals.
The problem states that Ptotal = 150 individuals, H frequency (p) = 0.2, and h frequency (q) = 0.8.
So homozygous dominant individuals (HH) = p^2 = (0.2)^2 = 0.04 or 4% of 150 --> 6 people
Heterozygous individuals (Hh) = 2pq = 2(0.2)(0.8) = 0.32 or 32% of 150
--> 48 people
And homozygous recessive individuals (hh) = q^2 = (0.8)^2 = 0.64 = 64% of 150 --> 96 people
Hope that helps you to understand how to solve these types of population genetics problems!
6 0
3 years ago
Which of the following best supports the claim that organisms of different domains share a common ancestor?
natita [175]
B. glycolysis occurs in both prokaryotic cells and eukaryotic cells.
4 0
2 years ago
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