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Andrej [43]
2 years ago
15

A child is born to parents that both have hypophosphatemic rickets, a disease that is caused by a sex-linked dominant trait. Wha

t is the chance that the child will have the disease if the mother is heterozygous for the trait?
75%
50%
25%
100%
Biology
1 answer:
Sati [7]2 years ago
7 0

Answer:

75%

Explanation:

Hypophosphatemic rickets is a disorder characterized by hypophosphatemia (low level of phosphate in the blood), defective intestinal absorption of calcium, and rickets (impaired mineralization of cartilaginous growth plates)  or osteomalacia (impaired mineralization of the osteoid) unresponsive to vitamin D. It can be inherited or acquired.

When it comes to inheritable hypophosphatemic rickets, the disorder is most often inherited in an X-linked dominant manner. This means that the gene is inherited through the X chromosome and that one dominant allele is enough to cause it.

As the mother is heterozygous for the trait, she has one dominant and one recessive allele (XHXh). The father has one X chromosome, which he inherited from his mother, and as he has the condition, the dominant allele will be present on it (XHY).

This means that, if the child is a girl, she will definitely have the disorder. Girls inherit one X chromosome from each of their parents. Even if she inherited the one with the recessive allele from her mother, she would get the one with the dominant allele from her father.

If the child is a boy, he has a 50% chance to inherit the condition. Boys inherit the Y chromosome from their father and the X chromosome from their mother. As the mother is heterozygous, he can either get the chromosome without the dominant allele or the one with it. This means that there is a 50% chance of inheriting the disorder.

If we say that there are equal chances of this couple having female and male children, when we combine these results, we get a 75% chance of them inheriting hypophosphatemic rickets.

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A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 fe
seraphim [82]

Answer:

Half.

Explanation:

Autosominal dominant trait are express are able to express themselves in the homozygous dominant as well as in the heterozygous condition. The X-linked recessive traits are mostly expressed in males since they have only one X chromosome. Female shows X linked recessive trait when her both X chromosome are affected.

The achandroplastic dwarf man with normal vision is married with a color blind woman with normal height. The mother is affected with vision, she definitely pass this trait to his son. The male is affected with achondroplastic (may be homozygous dominant or heterozygous) and the female is normal (homozygous recessive). The probability can be calculated including both the achandroplastic and vision trait. Half of the males are color blind with normal height (multiplying the probability of vision and achandroplatic trait).

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2 years ago
A group of researchers wanted to sort different white blood cell types (monocytes, lymphocytes, and granulocytes) apart from eac
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Answer:

C. placing cells in an environment with a lower solute concentration than that in the cells.

Explanation:

In pathological or diagnostic laboratories, to perform the differential leukocytes count (DLC) or white blood cells (WBC) count which are granulocytes, lymphocytes and monocytes we should first lyse the erythorcytes or red blood cells (RBC) with RBC lysis buffer. When RBC's lysis occur then only WBC's remain in the solution. Then, we add the hypotonic solution or lower solute concentration solution. By adding the hypotonic solution the cells will swell and increase in size.

So, the researchers used the hypotonic solution or lower solute concentration solution to increase the size of cells to differentiate between them.

Note:

In hypotonic solution the cells are increased in size or swell.

In hypertonic solution the cells are decrease in size or shrink.

In Isotonic solution the cells size remain same as normal.

3 0
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What occurs when the heads of myosin contact actin?
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If you have a pet cockroach with a brown body, how could you find out if it is
aleksklad [387]

Answer:

by testcrossing with a homozygous recessive partner

Explanation:

<u>If a pet cockroach exists whose zygosity is unknown, this can be determined by a test cross. A test cross involves crossing an organisms whose zygosity is unknown with a partner that is homozygous recessive for the same trait.</u>

Let us assume that brown body is represented by the allele B, the dominant allele. The homozygous recessive version would be bb.

The genotype of a brown cockroach whose zygosity is not known can be denoted as B_, where '_' can be a 'B' or a 'b'.

When B_ is crossed with bb:

B_   x   bb

Progeny

2 Bb

2 _b

The phenotype of Bb would be brown (since B is dominant over b) while the phenotype of _b would depend on the zygosity level of the cockroach.

If the unknown genotype is BB, then _b becomes Bb and the phenotype will be a brown body. This means that all the progeny will appear brown. (<em>see the first attached image for the Punnet's square</em>)

In other word, if the unknown genotype is bb, then _b becomes bb and the phenotype will be a alternate color (non-brown) body. This means that 50% of the progeny will appear brown while the remaining 50% will be in the alternate color. (<em>attached</em>

3 0
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