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rodikova [14]
3 years ago
13

Type your response in the box.

Biology
1 answer:
Nataliya [291]3 years ago
7 0

Answer:

nvjgblkbhb k jkbjn kb pohjb

Explanation:

m gjhjkkvijjbknnjoln i

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How does comparing species genomes show accurate evidence of evolution?
ira [324]

Answer:

DNA and the genetic code reflect the shared ancestry of life.

Explanation:

DNA comparison can show how related species are. Biogeography. The global distribution of organisms and the unique features of island species reflect evolution and geological change.

8 0
4 years ago
Which of the following factors limits the effectiveness of the Treaty on Plant Genetic Resources?
Debora [2.8K]

A it does not address the loss of biodiversity

8 0
4 years ago
Read 2 more answers
A piece of steel has a mass of 1170g and a volume of 150 cm3. What is the density of<br> steel?*
julsineya [31]

Answer:

<h3>The answer is 7.80 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass of metal = 1170 g

volume = 150 cm³

We have

density  =  \frac{1170}{150}  =  \frac{117}{15}   =  \frac{39}{5}  \\

We have the final answer as

<h3>7.80 g/cm³</h3>

Hope this helps you

6 0
4 years ago
Illustrate the cellular respiration pathway​
Paraphin [41]

Explanation:

Cellular respiration is a collection of three unique metabolic pathways: glycolysis, the citric acid cycle, and the electron transport chain. ... In order to move from glycolysis to the citric acid cycle, pyruvate molecules (the output of glycolysis) must be oxidized in a process called pyruvate oxidation.

5 0
3 years ago
For an autosomal recessive trait, if a normal male and an affected female have a daughter with the dominant phenotype, what are
Naddika [18.5K]

Answer:

The possible genotype of the normal male: AA or Aa

The genotype of affected female: "aa"

The probability of having a carrier girl child= 1/4

Explanation:

The given trait is autosomal recessive. Let's assume that the recessive allele "a" is responsible for the trait while the dominant allele "A" gives the dominant phenotype. The genotype of the normal male can be AA or Aa while that of the affected female would be "aa"

Their daughter has dominant phenotype and therefore, would be heterozygous for the trait with "Aa" genotype. A cross between carrier daughter "Aa" and her carrier male "Aa" would give progeny in following ratio= 1/4 AA: 1/2 Aa: 1/4 aa. Therefore, they have 1/2 probability of having a carrier child.

The probability of having a girl child is always 1/2.

So, the total probability of having a carrier girl child would be 1/2 Aa x 1/2 XX = 1/4

5 0
3 years ago
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