Ask question

Ask question

All categories

3 years ago

15

Ana is six weeks pregnant, and her baby is officially in the embryo stage. What does the baby now have?

2 answers:

Helga [31]3 years ago

8 0

If I'm not mistaken, I believe the baby has a heart beat.

hoa [83]3 years ago

4 0

The baby has a heart beat.

Hope this helps!:)

You might be interested in

What are some characteristics of polyploid plants

DENIUS [597]

They tend to be bigger and stronger

8 0

3 years ago

Each planet is at an equal distance from the sun at all points in its orbit.<br> True or False

natta225 [31]

The answer for this question is false.

7 0

3 years ago

Read 2 more answers

How can colonies of organisms be produced by budding?

marissa [1.9K]

<span>Many species of sponges and the Hydra perform external budding. Cells will quickly divide within them and develop on the surface of the organism. These cells will form branches or extensions of the parent as they develop into a new organism.
i wish this helped</span>

5 0

3 years ago

Question 1 of 20

swat32

Answer:

Show the client pictures that illustrate how a colostomy looks and functions.

Explanation:

4 0

3 years ago

Read 2 more answers

small group of 100 people decide to isolate themselves from the world and move to a small and remote deserted island. Out of thi

irina [24]

Answer:

Frequency of p $= 0.684$

Frequency of p $= 0.316$

Number of individuals with homozygous dominant (AA) $= 47$

Number of individuals with heterozygous (Aa) $= 43$

Number of individuals with homozygous recessive (aa) $= 10$

Explanation:

Out of 100 people, 10 have albino skin (aa)

So, the frequency of homozygous recessive individuals $(q^{2})$ is $\frac{10}{100} = 0.1$

Now, q will be

$= \sqrt{q^{2} } = \sqrt{0.1} \\= 0.316$

As per Hardy Weinberg's equation -

$p + q = 1$

Substituting the value of q in above equation, we get -

$p + 0.316 = 1p = 1 -0.316\\p = 0.684$

Now the frequency of homozygous dominant (AA) will be

$p^{2} = 0.684^{2} \\= 0.467$

Hence, out of 100 people $0.467 * 100 = 46.7 or 47$ people are homozygous dominant (AA)

Like wise out of 100 people $0.1 * 100 = 10$ people are homozygous recessive (aa)

As per As per Hardy Weinberg's equation-

$p^{2} + q^{2} + 2pq = 1\\$

Substituting the values in above equation, we get -

$0.467 + 0.316 + 2pq = 1\\2pq = 1 -( 0.467+ 0.1)\\2pq = 0.433$

So, out of 100 people $0.433 * 100 = 43.3 or 43$ people are heterozygous (Aa)

7 0

3 years ago