Ask question

Ask question

All categories

3 years ago

15

You place a cell in a solution that contains 12 more water molecules than the inside of the cell. How will you know when the cel

l reaches equilibrium?

1 answer:

DiKsa [7]3 years ago

7 0

Answer:

B

Explanation:

the water molecules will stop moving across the

cell membrane because the cell membrane has

the same amount of water as in the solution.

This process is called osmosis

You might be interested in

In Oompa Loompas, gray face (G) is dominant to orange recessive face (g). If two gray faced Oompas have an offspring with an ora

Neko [114]

Answer:

The genotype for each of the parents must be

parent 1 : Gg

parent 2 : Gg

Explanation:

Please note that a dominant trait is a trait that is expressed phenotypically in a heterozygous state, while a recessive trait is a trait that can only be expressed in a homozygous state.

Now, since gray face (G) for Oompa Loompas is dominant, and orange face (g) is recessive, for an offspring to be orange faced, it means that the genotype of the offspring must be 'gg'. Also, since both parent contribute an allele in the pair of alleles in the offspring, both parents must have the recessive (g) in their genotype. Moreover, we are told that both parents are gray-faced, meaning that their genotypes were 'Gg' and 'Gg'. To confirm, let me do the cross

G g

G GG Gg

g Gg gg

from the cross above, we find out that out of 4 offspring, 3 were gray face (GG, Gg ) while one was orange face (gg).

3 0

3 years ago

Which of the following describes how the body uses molecules like the one illustrated above?

Ymorist [56]

Answer:

B

Explanation:

5 0

2 years ago

What kind of consumer is a grass carp

irina [24]

The grass carp is herbivorious

8 0

3 years ago

Read 2 more answers

Greenhouse climate on earth was associated with relatively high atmospheric levels of which gas?

STALIN [3.7K]

D.carbon dioxide gas

4 0

3 years ago

Duchenne muscular dystrophy is an X-linked recessive disease. A phenotypically normal couple wants to start a family. The woman’

Andreyy89

Answer:

$\frac{1}{8}$

Explanation:

From the question: Duchenne muscular dystrophy is an X-linked recessive disease.

Now for an X-linked recessive disorder to be affected by a male individual; it only requires the presence of only one copy of the recessive allele of the disease to be present BUT in a female individual. both copies of the recessive allele must be present.

Let the X-linked recessive disease(i.e Duchenne muscular dystrophy) be = (ⁿ)

However, we are told that the couples are normal and are unaffected. Therefore;

the male partner (XY) will be: $X^NY$

the female partner (XY) will be: $X^NX^N$

Similarly, the question proceeds by telling us that: the woman's brother has the disease.

Definitely, it's possible that this unaffected woman is a carrier of the disease.

So, if she is a carrier; we have her traits to be: $X^NX^n$

NOW, if a cross exist between these couples; we have

$X^NY$ × $X^NX^n$

$X^N$ $Y$

$X^N$ $X^NX^N$ $X^NY$

$X^n$ $X^NX^n$ $X^nY$

So, we have offspring as follows:

$X^NX^N$ = normal unaffected female

$X^NY$ = normal unaffected male

$X^NX^n$ = female carrier for the Duchenne muscular dystrophy disease

$X^nY$ = affected male with the Duchenne muscular dystrophy disease

So, the probability of the child to be affected with the disease = $\frac{1}{4}$

Also, the probability that the first child will be a male or a female = $\frac{1}{2}$

∴

the probability that the couple’s first child will be affected = $\frac{1}{4}*\frac{1}{2}$

= $\frac{1}{8}$

4 0

3 years ago