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Leona [35]
3 years ago
15

To trace our maternal lineage, we observe mutations that have accumulated in our mitochondrial dna (mtdna). in the dna isolation

and amplification lab, we amplified a highly variable and non-coding region in the hyper-variable segment i of our own mtdna using polymerase chain reaction. why did we choose to amplify a non-coding region rather than a coding region to determine our lineage?
Biology
1 answer:
Sergio [31]3 years ago
4 0
In the coding region, natural selection tends to eliminate all of the mutations because of the high importance these regions have. The coding region contains genes that synthesize proteins and the changes in the DNA sequence can have devastating effects on the cell. Therefore, there are very few differences in the sequences of coding regions that can help us trace the lineage.
On the other hand, in the non-coding regions, the mutations often accumulate because they have little effect on the cell and the adaptive value of the organism. This enables us to trace up the lineage by comparing the sequences and seeing the differences in the sequences. 
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A drug binds to a receptor in the nucleus. This binding turns on a gene to make a protein. This drug is ____. lipophilic and hyd
Serjik [45]

The drug that to a receptor in the nucleus. This binding makes on a gene to make a protein. This drug is actually lipophilic and hydrophobic, the correct option is A.

What is receptor?

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As the drug is crossing the lipid membrane so it should be lipophilic and hydrophobic to be able to bind the receptor as in the given situation.

Thus, the correct option is A.

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4 0
2 years ago
In beans the pod texture and growth form genes are located on different chromosomes. There are two varieties of beans in a farme
irina1246 [14]

Answer:

a. Genotype in F1 - TtBb

   Phenotype in F1 - tough pods, bush-type growth

b. Proportion of F2 that are heterozygous for bush-type growth form = 5/8

c. Proportion of the F2 with tender pods =  1/4

d. Proportion of F2 that are heterozygous for growth form and have tender     pods = 1/8

Explanation:

Since the two traits are located on different chromosomes, they will independently assort during gametes formation. T is dominant over t and B is dominant over b.

Two true-breeding varieties were crossed at the beginning of the program:

TTBB   x   ttbb

All F1 plants will have the genotype TtBb with tough pods and bush-type growth form.

Advancing the cross to F2:

TtBb   x   TtBb

Progeny

1 TTBB   -   tough pods, bush-type

2 TTBb     -  tough pods, bush-type

2 TtBB    -   tough pods, bush-type

4 TtBb     -   tough pods, bush-type

1 TTbb     -    tough pods, climbing-type

2 Ttbb     -    tough pods, climbing-type

1 ttBB       -    tender pods, bush-type

2 ttBb     -      tender pods, bush-type

1 ttbb      -      tender pods, climbing-type

<em>Proportion of F2 that are heterozygous for bush-type growth form = 10/16 = 5/8</em>

<em>Proportion of the F2 with tender pods = 4/16 = 1/4</em>

<em>Proportion of F2 that are heterozygous for growth form and have tender pods = 2/16 = 1/8</em>

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