All categories

3 years ago

Commitment, challenge, and control are components of which stress fighting personality trait?

1 answer:

7 0

The answer is Hardiness. This is the stress fighting personality trait that is associated with stress and consists of three components:commitment, challenge, and control. Some common symptoms of stress are high blood pressure, headaches, backaches, skin rashes, indigestion, fatigue, and constipation.

You might be interested in

According to the cladogram, which is more closely related to caimans—hares or parrots?

irakobra [83]

We have to be able to see the cladogram to give you that answer. With out it we can’t.

6 0

3 years ago

In the transition from rest to moderate exercise, ________________ supplies almost all of the energy. A) glycogen stored in the

zaharov [31]

Answer:

glycogen stored in the liver

6 0

3 years ago

When cartilage is produced at the epiphyseal side of the metaphysis at the same rate as bone is deposited on the opposite side,

katrin [286]

Answer:

grow longer

Explanation:

In a long bone, the epiphyseal is the growing region. In young bones, bone formation occurs in a layer of hyaline cartilage. The epiphyseal plate forms cartilage on the epiphyseal end. Cartilage is calcified on the diaphyseal side, and the diaphysis lengthens.

3 0

1 year ago

Read 2 more answers

In animals, the inability to make the pigment melanin results in albinism, a recessive autosomal condition. Two unaffected paren

Fittoniya [83]

Answer: 1/16, or approximately 6.25% (see explanation below)

Explanation:

Answering this question requires two steps.

First, we need to figure out the probability that this couple will have a child with albinism in the first place. We know the following:

- Both parents are unaffected.

- The couple has already had one affected child.

- Albinism follows an autosomal recessive inheritance pattern.

Let ( M = normal gene ) and ( m = mutated gene ). Since the condition is recessive, the affected child can be assumed to have a “mm” genotype. Barring the possibility of a de novo mutation (which are assumed to be rare), the affected child must have inherited one ”m” allele from each parent. Since both of them are unaffected, however, we can assume that they are both carriers (genotype “Mm”). In conclusion, 1/4 of their offspring (25%) for any given pregnancy may be expected to have albinism. See the resulting Punnett square:

| M | m

M | MM | Mm

m | Mm | mm

Note that the question asks about the probability that not one but two consecutive births result in affected children. Since it can be assumed that both events are independent (meaning: the outcome of a pregnancy does not influence the outcome of following ones), we may apply the rule of multiplication for probabilities. The final answer is therefore 1/4 * 1/4 = 1/16.

7 0

3 years ago

14). During the early 1700’s, a small group of pacifist Protestants fled Germany to avoid religious persecution. This group, the

abruzzese [7]

A) In the Dunker population, the frequency of IB allele is 0.3 and the frequency of i allele is 0.4. In the general population, the frequency of IB allele is 0.1 and the frequency of i allele is 0.5.

If:
$I^{A}$ - the frequency of IA allele
 $I^{B}$ - the frequency of IB allele
 $i$ - the frequency of i allele

Then:
 $I^{A} I^{A}$ + $I^{A} i$ - the frequency of individuals with A blood type
 $I^{B} I^{B}$ + $I^{B} i$ - the frequency of individuals with B blood type
 $ii$ - the frequency of individuals with O blood type

Let's first take a look on the Dunker population:
$I^{A} = 0.3$
$ii=0.16
$

Since there is only one possible genotype for O individuals - ii - the frequency of the allele i is square root of the frequency of O individuals:
 $i= \sqrt{ii}$
⇒ $i = \sqrt{0.16}$
⇒ $i=0.4$

Now, we have the frequencies of two alleles ( $I^{A}$ and $i$ ). To calculate the frequency of $I^{B}$ allele, we will use the formula:
 $I^{A} + I^{B} + i = 1$
⇒ $I^{B} = 1- I^{A} - i$
⇒ $I^{B} = 1-0.3-0.4$
⇒ $I^{B} = 0.3$

Now, in the general population:
$I^{A} = 0.4$
$ii=0.25$

Similarly to the work for the Dunker population:
 $i= \sqrt{ii}$
⇒ $i = \sqrt{0.25}$
⇒ $i=0.5$

$I^{A} + I^{B} + i = 1$
⇒ $I^{B} = 1- I^{A} - i$
⇒ $I^{B} = 1-0.4-0.5$
⇒ $I^{B} = 0.1$


b) A founder effect is a result of geographical separation of a few individuals from the original population. Those founding individuals will form a new population. The Dunker population was not only geographically separated, but also genetically. The group interbreeding was present resulting in increasing those allele frequencies that were the most common in the founding population. In this case, the most individuals from the founding population had B blood type.

6 0

3 years ago