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USPshnik [31]
3 years ago
11

Loss-of-function mutations in the coding region of the homologous gene in humans result in loss of hair, teeth, and sweat glands

, as in the toothless men of Sind (India). What does this suggest about hair, teeth, and sweat glands in humans? Select the two correct answers. a) The phenotype of loss of Eda function in humans indicates that Eda has been co-opted by evolution to control different aspects of development, as it does in sticklebacks. b) The phenotype of loss of Eda function in humans indicates that human Eda protein and stickleback fish Eda protein contain structurally similar DNA-binding domains. c) The phenotype of loss of Eda function in humans indicates that Eda controls multiple aspects of human development, either directly by controlling each tissue type independently or by controlling a tissue that is a developmental precursor to the affected tissue types. d) The phenotype of loss of Eda function in humans indicates that human hair and teeth might be homologous to stickleback fish bony plates. e) The phenotype of loss of Eda function in humans indicates that hair, teeth, and sweat glands are homologous structures.
Biology
1 answer:
Veronika [31]3 years ago
7 0

Answer:

c) The phenotype of loss of Eda function in humans indicates that Eda controls multiple aspects of human development, either directly by controlling each tissue type independently or by controlling a tissue that is a developmental precursor to the affected tissue types.

e) The phenotype of loss of Eda function in humans indicates that hair, teeth, and sweat glands are homologous structures.

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For a series of experiments, a linkage group composed of genes W, X, Y and Z was found to show the following gene combinations.
Gelneren [198K]

Complete question:

For a series of experiments, a linkage group composed of genes W, X, Y and Z was found to show the following gene combinations. (All recombinations are expressed per 100 fertilized eggs). Construct a gene map. Determine the sequence of genes on the chromosome.

  • w-x = 5
  • w-y = 7
  • w-z = 8
  • x-y = 2
  • x-z = 3
  • y-z = 1

Answer:

The sequence of genes on the chromosome is:

----W-------------------------X-----------Y------------Z---

Explanation:

First, we need to know that 1% of recombination frequency = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.  

The map unit is the distance between the pair of genes for which every 100 meiotic products, one results in a recombinant one.  

The recombination frequencies between two genes determine their distance in the chromosome, measured in map units. So, if we know the recombination frequencies, we can calculate distances between the four genes in the problem and we can figure the genes order out. This is:

Recombination frequencies:  

1% of recombination frequency = 1 map unit (MU)

  • w-x = 5 MU
  • w-y = 7 MU
  • w-z = 8 MU
  • x-y = 2 MU
  • x-z = 3 MU
  • y-z = 1 MU

Now that we know the distances, we just need to analyze them to find out the correct order of the genes. First, we can look for the biggest distance, which tells us which genes are located in the extremes. w-z distance is the biggest one, so these two genes are in the extremes of the chromosome segment. ---W----------------------------------------------Z---

                     ∫---------------------8 mu-------------------∫

The rest of the genes are located in the middle between these two.

The second biggest distance is between w-y (7 mu). Y is also 1mu distant from Y. 7 mu + 1 mu = 8 mu. So, Y is located closer to Z.

---W-------------------------------------Y------------Z---

    ∫-----------------------7 mu---------∫∫---1 mu--∫

    ∫---------------------8 mu-------------------------∫

w-x = 5 mu, and x-y = 2mu, so x is located between w and y. The sum of these distances equals the distance w-y ( 5 mu + 2 mu = 7 mu). So,

---W-------------------------X----------Y------------Z---

    ∫-----------5 mu -------∫∫--2mu--∫

    ∫-----------------------7 mu---------∫∫---1 mu--∫

    ∫---------------------8 mu-------------------------∫

We know that the distance between x-y equals 2, and the distance between y-z equals 1. Also, the distance between x-z equals. This leads us to assume that Y is located between X and Z.

----W-------------------------X-----------Y------------Z---

    ∫-----------5 mu -------∫∫--2mu--∫∫----1mu---∫

                                     ∫------ 3 mu-----------∫

    ∫-----------------------7 mu---------∫∫---1 mu---∫

    ∫---------------------8 mu--------------------------∫

   

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The adenine of the start codon is located at the +1 position. Use the DNA Mutations Interactive to determine which statements de
IgorC [24]

Answer:

Question contains 6 statements.

1. a substitution of G to U at position +13 in the sense strand

2. a transition at position +12 in the antisense strand

3. a transversion of C in the sense strand of the histidine codon

4. a G nucleotide is inserted after position +1 in the antisense strand

5. a deletion of the A at position +7 in the antisense strand

6. an insertion of A between positions +7 and +8 on the sense stands

Among these only 2 statements are correct.

Below are the correct statements.

Explanation:

2. a transition at position +12 in the antisense strand

5. a deletion of the A at position +7 in the antisense strand

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Yes.

The kidney is the part of the internal organ that help in the production of urine. And a normal urine should be yellowish in color.

Therefore, in a situation whereby the urine is seen to contain some red tiny droplets, specifically means the presence of an infection to the kidney.

In this case, a cloudy urine might mean the presence of white blood cell which stipulates the presence of infection.

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Answer:

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Explanation:

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Answer: I wish I could help pero yo no me la se

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