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liberstina [14]
3 years ago
9

A scientist compares two cells, Cell A and Cell B, and notes that Cell B has a higher phospholipase C content than Cell A. Assum

ing no other factors are affecting this pathway, what can be said about the likely concentration of Ca2+ in Cell B compared to Cell A?
Biology
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0
No cell B is not compared to cell A
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Which best describes tbe reactants and products of photosynthesis?​
ankoles [38]

Answer:

The reactants for photosynthesis are light energy, water, carbon dioxide and chlorophyll, while the products are glucose (sugar), oxygen and water. you start with reactants and end with products

Explanation:

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What the best way to embalmingprocedure
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Chemicals are the best way
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Consider a locus with two alleles - B and b. B is dominant, while b is recessive. There is no mutation. B has a selective advant
Marat540 [252]

Answer:

0.09

Explanation:

<h3><u>Before selection</u></h3>

Total number in population = 1000

Genotype frequencies

Genotype frequency of BB = 500/1000 = 0.5

Genotype frequency of Bb = 250/1000 = 0.25

Genotype frequency of bb = 250/1000 = 0.25

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.5 + (0.25/2) = 0.625

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.25 + (0.25/2) = 0.375

<h3><u>After selection</u></h3>

We are told that after selection, the genotype frequency of bb is changed as they become 50% less fit. This means the frequency of bb individuals changes from 250 to 125 individuals (50% reduction).

Now the total number of individuals is 500 + 125 + 250 = 875.

Genotype frequencies

Genotype frequency of BB = 500/875 = 0.57

Genotype frequency of Bb = 250/875 = 0.29

Genotype frequency of bb = 125/875 = 0.14

Allele frequencies

Allele frequency of B = BB genotype frequency + half of the Bb genotype frequency = 0.57 + (0.29/2) = 0.715

Allele frequency of b = bb genotype frequency + half of the Bb genotype frequency = 0.14 + (0.29/2) = 0.285

<h3><u>Change in frequency of B after 1 generation</u></h3>

0.715 - 0.625 = 0.09

7 0
3 years ago
The offspring of a sexually-reproducing organism receives *
arlik [135]

Answer:

Lack of ability to adapt to changes in the envrionment

Explanation:

Asexual reproduction produces offspring that are genetically identical to the parent and no DNA is mixed, so they would not be able to adapt because there would be no genetic variation.

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3 years ago
Mitochondrial dna (mtdna) recovered from a fossil hominin from sima de los huesos in northern spain shows connections to the mtd
Fofino [41]

Mitochondrial DNA (mtdna) recovered from a fossil hominin from sima de los huesos in northern spain shows connections to the mtdna of the Denisovans.

The Sima de los Huesos showed distinctive proof of DNA survival over an extended period of time. Based on the enzymatic amplification of a few short mitochondrial DNA (mtDNA) fragments from Middle Pleistocene cave bear bones. Given that the Sima de los Huesos fossils have Neanderthal-derived characteristics (such as in their dental, mandibular, midfacial, supraorbital, and occipital morphology), it is unexpected that the mtDNA from these fossils has a common origin with Denisovan rather than Neanderthal mtDNAs.

To learn more about Mitochondrial DNA click here

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8 0
2 years ago
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