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kherson [118]
3 years ago
12

Can some one help me please

Biology
1 answer:
SOVA2 [1]3 years ago
6 0
Observable traits- phenotype
the second one is recessive. good luck!
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What is natural selection?
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those best suited to demands of environment survive

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genetics and the environment both play roles in cell differentiation. which is the best example of an environmental factor influ
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Answer: etiolation of plant growth in shade, with fast growing cell without chloroplasts. Out of shade, cell differentiate again to produce photosynthesising cells

Explanation: It is a subjective decision as to what is ‘best’. Good examples are plant responses to changes in the nutrient supply.

when soil nitrogen is depleted some plants, such as legumes, grow nodules on their roots, with cells that release chemical signals to attract nitogen fixing bacteria.

Lack of light induces etiolation in many plants. New cells elongate and develop without chloroplasts, with rapid cell division exhibiting gravitropism, extending upward. When they grow beyond the shade area, cell differentiation changes again, to produce photosynthesising cells.

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2 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}

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