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3 years ago

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Cells located in the _ synthesize oxytocin and vasopressin and transport these hormones to the _.

1 answer:

schepotkina [342]3 years ago

5 0

<span>The answer to this question would be: A) hypothalamus; posterior pituitary

The hypothalamus produces many kinds of a hormone that will activate a cascade of metabolism. One of them is vasopressin and oxytocin. Vasopressin will end up in kidney and heart, causing the body to keep water and lowering the urine output. Oxytocin is used for releasing breastmilk.</span>

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The domestic cat genome contains 2.9×109 base pairs. The length of linker DNA in mammals is 50 base pairs. Approximately how man

IgorLugansk [536]

Answer:

Number of nucleosomes in $2.9 * 10^9$ bp is equal to $1.47 * 10^7$

Explanation:

For wounding one nucleosome, total length of DNA required is equal to $146$ bp

The length of linker DNA in mammals is equal to $50$ bp

Thus , the total length of DNA that confides between two nucleosome is equal to the sum of wounding length of DNA and the linker length

$= 146 + 50\\= 196$ bp

Thus, in $196$ bp length of DNA, the total number of nucleosomes is equal to $1$

Thus, number of nucleosomes in $2.9 * 10^9$ bp is equal to

$\frac{2.9* 10^9}{196} \\1.47 * 10^7$

8 0

3 years ago

Water crosses the plasma membrane through special channels known as _____.

Vinil7 [7]

Answer:

<em>aquaporins.</em>

Explanation:

4 0

3 years ago

If a genetic mistake (i.e., a mutation) occurred during replication, and the new DNA strand has the sequence CAG, what amino aci

erica [24]

The CAG codon codes for Glutamine

3 0

3 years ago

What do you call these 2 groups of ribs?

zavuch27 [327]

In the first image, those are fake ribs that are attached to the sternum through the costal cartilage

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3 0

3 years ago

3. In centaurs, the allele for curly tails (T) is incompletely dominant to the allele for straight tails (t). The hybrid genotyp

Reptile [31]

Answer:

a. Determine the allele frequencies of:

f(T) = 0.485 ≅ 0.48

f(t) = 0.515 ≅ 0.52

b. Determine the genotype frequencies of the exposed generation

F (TT) = 0.21
F (Tt) = 0.55
F (tt) = 0.24

the genotype frequencies of the next generation

F(TT) = 0.23
F(Tt) = 0.5
F(tt) = 0.27

Explanation:

Due to technical problems, you will find the complete answer and explanation in the attached files

3 0

3 years ago