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antiseptic1488 [7]
3 years ago
13

List developmental changes in animals

Biology
2 answers:
Basile [38]3 years ago
3 0
Embryonic Phase, a Larval Stage, Metamorphosis and an Adult Phase to Reach Eventually Sexual Maturity
IgorC [24]3 years ago
3 0
Early development. Plant development. Human development. Prenatal development. Regeneration. Morphogenesis. Ontogeny.
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the gemone of an organism is its total genetic material what aspect of the gemone can and cannot be determined through karyotypi
blondinia [14]
Karyotyping can give information on a persons sex and chromosomal disorders. It cannot give information on a persons trait's and how severe a disorder is.
4 0
3 years ago
Place the following events in order to sequence the evolution of life on Earth. (A)Prokaryotes (B)Land Plants (C)Multicellular E
3241004551 [841]

Answer:

The correct sequence would be E, A, D, C, B

Explanation:

For the evolution of life to begin, the Earth formed as Earth is the only planet which can support life and has favourable conditions. Life first began from prokaryotic organisms which engulfed smaller prokaryotes and gave rise to unicellular eukaryotes. With the passage of time, the unicellular eukaryotes evolved to form multicellular eukaryotes. The life on Earth then started after the formation of multicellular eukaryotes in the oceans and seas.

7 0
3 years ago
What keeps intracellular receptors from binding to dna before a hormone binds to the receptor?.
Gemiola [76]

Answer:

chaperone proteins

Explanation:

The chaperone protein blocks the DNA binding site until a hormone binds at the hormone binding site.

hope this helps

5 0
3 years ago
A fruit fly population has a gene with two alleles, A1 and A2. Tests show that 70% of the gametes produced in the population con
IceJOKER [234]

Answer:

Option C

Explanation:

Given ,

A1 allele is carried by 70 % people

Let us assume A1 s dominant genotype

This means p= \frac{70}{100} = 0.7

Thus, frequency of allele in the given population is 0.7

It is also given that the population is in Hardy-Weinberg equilibrium thus

p+q=1\\0.7+q=1\\q = 1-0.7\\q= 0.3

Frequency of fruit fly with genotype A2A2 will be

q^2\\= (0.3)^2\\= 0.09

As per Hardy-Weinberg's second equation of equilibrium

p^{2} + q^{2} + 2pq = 1\\0.49+0.09+2pq = 1\\2pq = 1-0.49-0.09\\2pq= 0.42

Hence, option C is correct

7 0
3 years ago
The difference between the images projected onto each eye is referred to as ____.
Sonja [21]
I do believe it is C hope this helps its been a while since i was leaning about this.....
3 0
4 years ago
Read 2 more answers
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