Ask question

Ask question

All categories

3 years ago

9

if you drop a piece of quartz it will break apart in an irregular manner which property dose quarz show in this situation

1 answer:

mezya [45]3 years ago

8 0

The property is fracture
when there is a pattern, it is cleavage

You might be interested in

Can someone answer this please :)

IrinaK [193]

The answer will be for you U

7 0

3 years ago

Which of these is an example of elemental carbon?Diamond,methane,or proteins?

Masteriza [31]

Diamonds, as it contains carbon atoms in tetrahedral

7 0

3 years ago

Explain how a human offspring 47 chromosomes could be produced

Reika [66]

If they were to be no separation <span>of a pair of chromosomes during meiosis resulting in formation of a sperm or ovum with 24 chromosomes instead of 23.</span>

4 0

3 years ago

What do ferromagnesian (mafic) minerals have in common? List three examples of mafic minerals?

saveliy_v [14]

Ferromagnesian minerals are minerals that have iron or magnesium in their structure. The greater specific gravity is between 3.2 and 3.6. Some examples are pyroxene, biotite, garnet, and olvine.

Most mafic minerals are dark in color, and common rock-forming mafic minerals include olivine, pyroxene, amphibole, and biotite. Common mafic rocks include basalt, diabase and gabbro. Mafic rocks often also contain calcium-rich varieties of plagioclase feldspar.

3 0

3 years ago

In guinea pigs hair straightness or curliness is thought to be governed by a single pair of alleles showing incomplete dominance

Nimfa-mama [501]

Answer:

Frequency of allele S is p $= 0.4939$

Frequency of allele C is q $= 0.666$

The population is not in Hardy-Weinberg equilibrium

Explanation:

Given -

Number of individuals with straight hair $= 244$

Number of individuals with curly hair $= 444$

Number of individuals with Wavy hair $= 312$

Let "p" represents the frequency for allele for straight hair and "q" represents the frequency for allele for curly hair

$p^2$ represents the frequency of genotype "SS"

$p^{2} = \frac{244}{1000} \\= 0.244$

$q^2$ represents the frequency of genotype "CC"

$p^{2} = \frac{444}{1000} \\= 0.444$

$2pq$ represents the frequency of genotype "SC"

$2pq = \frac{312}{1000} \\= 0.312$

Frequency of allele S is p

$= \sqrt{0.244} \\= 0.4939$

Frequency of allele C is q

tex]= \sqrt{0.444} \\= 0.666[/tex]

For being in Hardy Weinberg's equation-

$p+q=1\\$

Substituting the values in above equation, we get -

$0.4939+0.666\neq 1$

hence, the population is not in Hardy-Weinberg equilibrium

4 0

4 years ago