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sdas [7]
3 years ago
13

Individuals who have favorable traits will likely have ______ offspring. less

Biology
1 answer:
oksian1 [2.3K]3 years ago
5 0
If less is the only option then go with that like your question suggests but more feels like a better answer

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“how many eggs form in meiosis? what else forms and why “
kakasveta [241]

Answer:Meiosis contains two separate cell divisions, meaning that one parent cell can produce four gametes

Explanation:

because I am smart

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2 years ago
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What is a potential danger of excess cholesterol in the human body?
goblinko [34]
Atherosclerosis and the obvious danger of free floating cholesterol in the blood that can circulate and deposit anywhere, even in the arteries/veins while flowing which can also cause a blockage of arteries/veins and cause an embolism. Also fatty deposits in the cell's membrane that can make it more fluid than it should be causing the cell to lose its rigid membrane structure or even making it more rigid sometimes depending on the environment.
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3 years ago
What is theaw of superposition? How is it used to understand how old the Earth is?
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5 0
3 years ago
The zebra mussel is an invasive species. Which statement is most likely true if the zebra mussel is introduced to a new environm
crimeas [40]

Answer:

A.

Explanation:

when given that a species is <em>invasive</em> you can read through the answers and determine which description sounds invasive

7 0
3 years ago
Alkaptoneuria is a recessive genetic disorder found among North American populations at a frequency of about 1 in 500,000 births
Dafna11 [192]

Answer:

The correct answer is - 1/41,493

Explanation:

Let assume the frequency of the two possible same allele genotype (dominant and recessive) in an inbred population is p and q. Then the frequency of heterozygotes (H) is denoted as:

2pq + 2pqF.  ( where F is the inbreeding coefficient).

The frequency of the two different hoozygotes in inbred population can be calculated as:

p2 + pqF and q2 + pqF. (Where p and q are the allele frequency of the dominant and recessive phenotype.

Given: Frequency of Alkaptonuria (q 2) = 1:500, 000

=> q = 1/707

p = 706/707 ( Approx values)

solution:

Inbreeding coefficient (F) = 1/64

Therefore,

Frequency of Alkaptonuria in second cousins= q 2 + pqF

= 1/500, 000 + (706/707 x 1/707) x (1/64)

= 1/500, 000 + 1/45, 248

= 1/41,493 (approx)

5 0
3 years ago
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