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RSB [31]
3 years ago
10

A 9-month old baby with a macular rash is brought to the hospital. parents report that the rash started on the face, spread to t

he rest of the body, and was preceded with coldlike symptoms. there are small reddish spots in the baby's mouth. what is the diagnosis? a 9-month old baby with a macular rash is brought to the hospital. parents report that the rash started on the face, spread to the rest of the body, and was preceded with coldlike symptoms. there are small reddish spots in the baby's mouth. what is the diagnosis? smallpox rubeola chickenpox shingles
Biology
1 answer:
Katen [24]3 years ago
3 0
<span>The answer would be rubeola
</span><span>
The small reddish spots in the baby's mouth could be </span><span>Koplik's</span><span> spots that were pathognomonic for rubeola. </span>The infection also causes reddish spots/rash that starts from face to the body with flu-like symptoms. All of them is the signs of rubeola(also known as measles) infection.
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Two autosomal genes control horn color in dragons. Pure-breedinggold-horned dragons were mated to pure-breeding silver-horneddra
horrorfan [7]

Answer:

The genetically based phenotypic  ratio for of Gold: Silver: Bronze = 9:1:6

The expected number of gold-horned dragons is = 144

The expected number of silver-horned dragons is = 16

The expected number of bronze-horned dragons is = 96

Explanation:

Given that; F2 generation consisted of:

147 gold

17 silver

92 bronze.

Total autosomal genes control horn color in dragons = 256

When we conducted the punnet square for the Pure-breedinggold-horned dragons were mated with the  pure-breeding silver-horneddragons, then intercrossed them for the F2 generation, It is seen that;

The genetically based phenotypic  ratio for of Gold: Silver: Bronze = 9:1:6

∴

The expected  number of gold-horned dragon = 256*\frac{9}{16}

= 144

The expected number of silver- horned dragon = 256*\frac{1}{16}

= 16

The expected number of Bronze-horned dragon = 256*\frac{6}{16}

= 96

Null Hypothesis = The observed ratio of Gold:Siver:Bronze

=144:16:96

=9:1:6

Observed          Expected               (OF - EF)²                 \frac{(OF-EF)^2}{EF}

Frequency         Frequency

(OF)                    (EF)

147                       144                           9                                0.062500                    

17                         16                              1                                0.062500

92                       96                             16                               0.166667

Total:                                                                                      0.291667

Since Degree of Freedom = 2

Our P-Value = 0.864

∴ Our P. Value is very high that we accept the null hypothesis

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