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V125BC [204]
3 years ago
13

Consider a cross between a man with an X-linked recessive disorder and a woman that's a carrier for the disorder. Approximately

what percent of their children are predicted to express the disorder?

Biology
1 answer:
Vadim26 [7]3 years ago
6 0

Answer:

50%

Explanation:

Let's assume that the X linked allele "a" gives the disorder. Therefore, the genotype of the affected man would be X^aY while the genotype of the carrier woman would be X^aX. A cross between X^aY man and X^aX woman would produce progeny in following phenotype ratio: 1 affected daughter: 1 affected son: 1 normal but carrier daughter: 1 normal son.

Therefore, there are 50% chances that their children can express the disorder.  

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This type of tissue is associated with diffusion of water, electrolytes, gases, etc. it lines the body's outer surface and inter
stiks02 [169]

The correct answer is <em>epithelial tissue. </em>

There are different types of epithelial tissue depending on the type of cells its made of: squamous, columnar, and cuboidal. It can also be divided into simple and stratified depending on the number of layers. Functions of epithelial cells are wide such as secretion, absorption, protection..


7 0
3 years ago
A planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three phenotypic characters
Zigmanuir [339]

Answer:

For the first experiment, the recombination frequency is shown below. This experiment comprises the cross between the Tall and the Antennae.  

Tall-antennae - 46% - Expected

Dwarf-no antennae - 42% - Expected

Dwarf-antennae - 7% - Recombinant

Tall-no antennae - 5% - Recombinant

Total = 100%

For the recombinants, the recombination frequency is:  

= 7 + 5 = 12%

Thus, the recombination frequency between the T and A genes is 12%.  

For the second experiment, the cross takes place between the heterozygous antennae and the upturned snout. The results are shown below:  

Antennae-upturned snout - 47% - Expected

No antennae-downward snout - 48 % - Expected

Antennae-downward snout - 2% - Recombinant

No antennae-upturned snout - 3% - Recombinant

Total = 100%

The recombinant frequency is = 2 + 3 = 5%

Thus, the recombination frequency between the A and S genes is 5%.

4 0
3 years ago
Read 2 more answers
How does carrying capacity affect population growth?
raketka [301]

Answer:

Carrying capacity is like the population's limit. If the carrying capacity is at it's limit, the population can not grow anymore.

Explanation:

5 0
3 years ago
The commitments of the convention on biological diversity include:
nika2105 [10]
Evolution of birth food chain. etc
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3 years ago
Species in which dark brown fur (D) is dominant relative to light brown fur (d). Using the Punnett square below, predict the fre
Debora [2.8K]

Answer:

<h3>Genotype frequencies</h3>

25% DD

50% Dd

25% dd

or

1 DD: 2 Dd: 1 dd

<h3>Phenotype frequencies</h3>

75% Dark brown, 25% light brown

or 3 dark brown: 1 light brown

Explanation:

Two heterozygous parents have the genotypes Dd. Therefore, the cross is Dd x Dd.

           D                   d

D        DD                 Dd

d        <em>Dd                  dd</em>

<em />

25% (1:4) of the organisms have the genotype DD, and therefore the phenotype is dark brown,  since it has two copies of the dark brown fur allele (DD)

50% (1:2) of the organisms have the genotype Dd, and therefore, the phenotype is dark brown since dark brown fur (D) is dominant to light brown  fur (d).

25% (1:4) of the organisms have the genotype dd, and therefore the phenotype is light brown, since it has two copies of the light brown fur allele (DD)

3 0
3 years ago
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