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gtnhenbr [62]
3 years ago
13

Which relational dialectic is at work for the following scenario

Biology
1 answer:
Tpy6a [65]3 years ago
6 0

Answer:

To answer this question, narrow the passage down to a cause and effect scenario, or a simple chain of events.

1. Lisa and her mom get along > 2. Lisa separates from her mother > 3. The mother works to repair their relationship

This can then be shortened to:

Lisa seeks independence > Her mother rebuilds their connection

Of the answers you listed, this most fits the "autonomy versus connection" dynamic.

Explanation:

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How does a frog get rid of urea/urine? Question 4 options: gills lungs heart kidneys
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Did birds evolve from the gliding reptiles called pterosaurs?
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Where would the air contain the most moisture? a) over hawaii b) over arizona c) over the archive circle d) over the rocky mount
balu736 [363]

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A: over Hawaii

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3 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
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