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TEA [102]
3 years ago
6

EASY QUESTION. WILL MARK BRAINIEST

Biology
1 answer:
Andrei [34K]3 years ago
6 0

Answer:

A

Explanation:

a. Bioengineers have developed an artificial sharkskin that does not allow resistant bacteria to grow on it.

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These structures allow sperm cells to move through the style:
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Which cells we don’t see no answer choices to really give a answer
8 0
3 years ago
Which of the following is NOT true about restriction enzymes? Group of answer choices They are used to determine whether or not
larisa [96]
They are used to directly amplify segments of dna
5 0
3 years ago
Iven the equation representing a nuclear reaction in which x represents a nuclide: which nuclide is represented by x?
kaheart [24]

The nuclide represented as X is thorium and this is an alpha decay where an alpha particle is removed off.

<h3>What is nuclide?</h3>

A kind of atom or nucleus which is represented by a specific number of protons and neutrons.

The equation given represents an alpha decay. In an alpha decay, an alpha particle is removed off.

The atomic number of the parent nuclide is more than the daughter nuclide by two units and the mass number of the parent is more than that of the daughter nuclide by four units.

Thus, the x is thorium.

For more details regarding nuclide, visit:

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8 0
2 years ago
Scientists have created trains that use magnets to make the trains float above the tracks as they travel. These trains float bec
Arada [10]

Answer:

the poles repel

Explanation:

i believe this is the best answer

7 0
2 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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