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Neko [114]
3 years ago
15

Microscopic shells that are buried under the seabed are converted to organic sedimentary rocks. during this process, which prope

rty of the shells will remain constant?
a. mass of the shells
b. color of the shells
c. texture of the shells
d. chemistry of the she
Biology
2 answers:
Scorpion4ik [409]3 years ago
7 0
I think the correct answer would be mass of the shells. According to the law of conservation of mass, mass can never be created not destroyed. Therefore, mass here should be constant before the transformation and after the process. Hope this helps.
pshichka [43]3 years ago
3 0

Answer is A

hope it helps

:D

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Answer:

The number of chromosomes in sex cell must be haploid (1n) as compared to body cell which is diploid(2n).

Explanation:

Body cells are diploid while the sex cells are haploid. We can say that sex cells contain half number of chromosomes as compared to body cells.

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A locus that affects susceptibility to a degenerative brain disease has two alleles, V and v. In a population, 16 people have ge
Pie

Answer:

Yes, the population is evolving

Explanation:

Hardy-Weinberg equilibrium states that a population is not evolving if it is in hardy weinberg equilibrium as the alleles frequency do not change over time in generations.

In mathematical equation, the genotype frequency will be -P²+2Pq+q²=1 and allele frequency will be p+q=1, where p = dominant allele and q will be recessive allele.

In the question, since the observed genotype are-

P² - VV= 16

2Pq- Vv- 92

q²- vv- 12

<u>1. Calculate allele frequency</u>

Now we will calculate if the expected value is same or close or not.

We will calculate allele frequency

Population has 120 individuals which means it has - 240 alleles.

In which V (p) will be- 32 +92= 124 allele

p- dominant allele/ total allele=  124/ 240 = 0.52

P= 0.52

q will be 1-0.52= 0.48

<u>2. Expected genotype frequency</u>

Now we will calculate expected genotypic frequency.

P²- 0.52 X 0.52= 0.27

2Pq = 2 x 0.52x 0.48 =  0.5

q²= 0.48 x 0.48 =  0.23

<u>3. Expected progeny </u>

P²= 0.27 x 120= 32 VV

2Pq= 0.5 X 120= 60 Vv

q²= 0.23 x 120= 28 vv

Since the observed value is not equal to the expected value therefore the population is not in Hardy- Weinberg equilibrium and the population is evolving.

Thus, Yes, the population is evolving is correct.

4 0
3 years ago
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Are most mutations bad?
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Answer:

Mutational effects can be beneficial, harmful, or neutral, depending on their context or location. Most non-neutral mutations are deleterious. In general, the more base pairs that are affected by a mutation, the larger the effect of the mutation, and the larger the mutation's probability of being deleterious.

Explanation:

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