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Whitepunk [10]
4 years ago
5

NEED HELP ASAP!! WILL MAKE BRAINLIEST - 25 POINTS

Biology
1 answer:
djverab [1.8K]4 years ago
3 0

Answer:

a and d

Explanation:

the birth and death rates are important, they are they  numbers in which a population is

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I NEED HELP ASAP (WILL MARK BRAINLIEST IF ITS THE RIGHT ANSWER) which of the following describes asexual reproduction
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<em>B. One parent organism and no egg or sperm.</em>

Explanation:

In asexual reproduction, an organism can produce offspring with no gametes involved (eggs or sperm), and does not need to undergo meiosis.

If an organism produces offspring through asexual reproduction, all of the DNA from the parent will go to the offspring. Since there is only one parent, meiosis will not occur since there is no need for genetic differentiation.

All of the organisms produced by the parent during asexual reproduction will be identical to the parent. They will have the same DNA and genetic make up.

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3 years ago
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The length of the classroom is 10 meters and the width is 8 meters. The population size of the classroom is 40 students. What is
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2

Explanation:

7 0
3 years ago
Please help for brainliest!!
9966 [12]

Answer is Genus.... <em>hope</em><em> it</em><em> helps</em><em>.</em><em>.</em><em>.</em><em> </em><em>pls</em><em> mark</em><em> brainliest</em>

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3 years ago
Describe the difference between a physical property and a chemical property
polet [3.4K]
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6 0
4 years ago
A woman with normal vision has a son. The woman’s maternal grandfather is color-blind. What is the probability that the son is c
Reptile [31]

Answer:

= \frac{1}{4}

Explanation:

Given that the woman’s maternal grandfather is color-blind, then it is possible that the woman is a carrier of color-blind allele.

Then probability that the woman is a carrier of color-blind allele is = \frac{1}{2}  

So if a carrier woman cross with a normal man

Let say the gene of the carrier woman is: XX^c

and the normal man = XY

so if we have a cross between  XY and XX^c; we have:

                     X                     Y

X                   XX                 XY

X^c                 XX^c                 X^cY

we can see the offspring to be :

XX =normal unaffected female

XY = normal unaffected male

XX^c = carrier female for color-blind

X^cY = carrier male for color-bind

Then probability that of  inheriting defective allele = \frac{2}{4} = \frac{1}{2}

Let's not forget that the probability that the woman is a carrier of color-blind allele is = \frac{1}{2}  

∴  the probability that the son is colorblind = \frac{1}{2}*\frac{1}{2}

= \frac{1}{4}

3 0
3 years ago
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