D is the answer. A gene is a part of a chromosome.
Answer:
The DNA sequence that produced the mRNA sequence uracil, guanine, cytosine, guanine adenine uracil adenine adenine during transcription is adenine, cytosine, guanine, cytosine, thymine, adenine, thymine, thymine.
Explanation:
Transcription of DNA into messenger RNA (mRNA) is a process in which the specific nucleotide sequence is transferred from one molecule to another, through the synthesis of mRNA molecules from a DNA strand.
The complete transcription process is done by nitrogenous base complementarity, where mRNA receives the sequence of complementary bases according to the DNA sequence:
- <em>Adenine</em><em> is complemented with uracil, since in RNA the thymine is substituted by this uracil.
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- <em>Thymine</em><em> is complemented with adenine.
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- <em>Cytosine</em><em> is complemented with guanine.
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- <em>Guanine</em><em> is complemented with cytosine.
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Thus an <u>uracil mRNA sequence, guanine, cytosine, guanine adenine uracil adenine U-G-C-G-A-U-A-A comes from a DNA sequence adenine, cytosine, guanine, cytosine, thymine, adenine, thymine, thymine or A-C-G-C-T-A-T</u>.
Answer:
D) B/b;S/s (x) b/b;s/s
Explanation:
Parent 1 : belted syndactylous sow
Since it is showing dominant phenotype for both the traits, it can either be BBSS or BbSs
Parent 2: unbelted cloven-hoofed
Since it is showing recessive phenotype for both the traits, it can only have bbss genotype
If we assume parent 1 to be BBSS all the resulting progeny with bbss will have dominant phenotype which is not the case.
If we assume parent 1 to be BbSs:
BbSs X bbss =
BbSs : belted syndactylous
bbSs : unbelted syndactylous
Bbss : belted cloven
bbss : unbelted cloven
The progeny will be produced in 1:1:1:1 ratio which means that each of them will make 25% of the population.
Hence, parent 1 will have BbSs genotype and parent 2 will have bbss genotype
