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fomenos
2 years ago
13

If an ecological disaster would occur, some members of a ______________1 _________________ population would have a greater chanc

e of surviving than members of a(an) _____________2 _____________ population.
Biology
1 answer:
bearhunter [10]2 years ago
4 0

If an ecological disaster would occur, some members of a genetically diverse, sexually reproducing population would have a greater chance of surviving than members of a(an) genetically identical, asexually reproducing population.

<h3>What is mean by genetically diverse?</h3>
  • Diversity is the possession of a wide variety of items.
  • The variety of various inherited features within a species is referred to as genetic diversity.
  • A species with a lot of genetic variation would have a lot of individuals with a variety of different features.
<h3>What is mean by sexual and asexual reproduction?</h3>
  • Asexual reproduction and sexual reproduction are the two main types.
  • An organism that reproduces sexually incorporates the genetic material from both of its parents and is genetically distinct.
  • Asexual reproduction creates genetically identical kids by having one parent replicate itself.

Learn more about population here:

brainly.com/question/27991860

#SPJ4

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People who have the o blood type possess
lesantik [10]
No because the type o blood is a universal type of blood

4 0
3 years ago
The _____ demonstrates the cell theory because it develops cells that fuse to form one giant cell, making up the whole organism.
saul85 [17]
It isn't Bacteria or Archea. They're just unicellular. Cork is dead.

Slime mold is the correct answer.




7 0
3 years ago
Read 2 more answers
If there is no light at the bottom of the deep sea, why do animals have well developed<br> eyes?
IRISSAK [1]

Answer:

They adapt and survive because of Natural Selection

Explanation:

The sea animals die off if they cannot survive under certain conditions so they adapted to having eyes that can see, and other necessary survival tools for the bottom of the deep sea.

5 0
4 years ago
Which equation best represents conservation of matter?
worty [1.4K]

Answer:

A 6H2O + 6CO2 → C6H12O6 + 6O2

Explanation:

The law of conservation of matter states that matter can neither be created nor destroyed in a system. In other words, when elemental matter undergoes a reaction, the amount of elements used must be equal the amount produced.

The equations in this question depicts the overall photosynthetic process where CO2 combines with H2O to produce C6H12O6 and O2. In order for this equation to obey the "conservation of matter", each individual element must be equal on both the reactant and the product side i.e. the equation must be balanced.

The equation that fulfills this is:

6H2O + 6CO2 → C6H12O6 + 6O2

Hence, it correctly describes the conservation of matter.

4 0
3 years ago
Karen and Steve each have a sibling with sickle-cell disease (recessive disorder). Neither Karen nor Steve nor any of their pare
Talja [164]

To first understand your question, here are some vocabulary terms I will be using throughout this problem.

  • Homozygous dominant: (BB) <em>This </em><u><em>is not</em></u><em> having sickle cell disease.</em>
  • Homozygous recessive: (bb) <em>This </em><u><em>is</em></u> <em>having sickle cell disease.</em>
  • Heterozygous: (Bb) <em>This is being a </em><u><em>carrier</em></u><em> for sickle cell disease.</em>

We know that having sickle cell disease is a recessive disorder so it will be homozygous recessive (bb).

We also know that Karen and Steve each have a sibling with sickle cell disease, but they don't. This means that <em>both</em> of their parents have to be heterozygous (Bb), or a carrier for the disease.

If their parents are both heterozygous (Bb), there is a 1/4 chance that Karen and Steve are completely healthy (BB) and a 2/4 chance that they are carriers (Bb). We know for a fact that they aren't sick so that is out of the question.

This means that the probability of Karen and Steve being a carrier for this disease is 2/3.

Now, we can make a Punnett square for the probability of their children, which I have attached below. Since being heterozygous is most likely, I made both Karen and Steve heterozygous in the Punnett square.

This means that their children are 1/4 likely to be completely healthy (BB), 2/4 likely to be a carrier (Bb), and 1/4 likely to have sickle cell disease (bb).

<u>Now we can multiply the probability of both of Karen and Steve being a carrier for sickle cell, which is </u><u>2/3 each</u><u>, with the probability of their children having it, which is </u><u>1/4</u><u>.</u>

2/3 × 2/3 = 4/9

4/9 × 1/4 = 1/9

The final probability of Karen and Steve having a child with sickle cell disease is <u>1/9</u>.

3 0
3 years ago
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