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4.Monohybrid Cross: Count the yellow and purple kernels for 3 ears of corn from the “3:1” collection. These are the result ...

ion. These are the result of a monohybrid cross (two heterozygous parents) and we expect a ratio of 3 dominant phenotypes to 1 recessive phenotype.
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5.In mice, grey coat colour, G, is dominant to white, g, and long tail, T, is dominant to short tail, ...

t. What is the genotypic and phenotypic ratio if a female mouse that is heterozygous for colour and short-tailed is crossed with a male mouse that is homozygous dominant for colour and is heterozygous for tail length?
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6.I was looking at my notes on protein structure and I am trying to understand quaternary structures for proteins. I ...

or proteins. I understand that primary, secondary, and tertiary structures are encoded by one gene each. However, I am not entirely sure if quaternary structures are encoded by one or multiple different genes. The reasons why I am a little confused is for two reasons. Firstly, quaternary structures are made up of more than one protein subunit (i.e. multiple polypeptides). Secondly, as I understand, Hemoglobin, for example, has different subunits, each of which is encoded by a different gene. Does this necessarily mean that all quaternary structures are composed of proteins encoded from different, separate genes? If quaternary subunits are encoded by different, separate genes, can those different genes be located on different loci, or are all of the subunits necessarily encoded by the different gene but its mRNA molecule is spliced differently?
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7.Monohybrid Cross: Count the yellow and purple kernels for 3 ears of corn from the “3:1” collection. These are the result ...

ion. These are the result of a monohybrid cross (two heterozygous parents) and we expect a ratio of 3 dominant phenotypes to 1 recessive phenotype. II. Test Cross: Count the yellow and purple kernels for 3 ears of corn from the “1:1” collection. These are the result of a test cross (two heterozygous parents) and we expect a ratio of 1 dominant phenotype to 1 recessive phenotype. III. Dihybrid Cross: Count the kernels for 3 ears of corn from the “9:3:3:1” collection. These are the result of a dihybrid cross (two heterozygous parents for two traits) and we expect a ratio of 9 dominant/dominant: 3 dominant/recessive: 3 recessive/dominant: 1 recessive/recessive.
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Nucleic acids: 


They are large biomolecules that are made up of nucleotides. The two main nucleic acids are DNA and RNA. The DNA contains deoxyribose and RNA contains ribose. They make up the genetic material of the cell and carry information that transfers to the next generation.


Nucleic acids Sample Questions:


Question 1: What is the type of replication of DNA in prokaryotes?


Answer: The replication of DNA in prokaryotes is theta type.

Explanation: There is a single origin of replication in the case of prokaryotes and replication goes bi-directionally and thus circular daughter DNA is formed at the end and this type of replication is called theta replication. 



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Question 2: Topoisomerase breaks which type of bonds in the DNA strand?


Answer: Phoshodiester bonds.

Explanation: The topoisomerase is a nuclease that breaks phosphodiester bonds in DNA and this reaction is reversible. The function of topoisomerase is to relax the DNA and also helps in knotting and unknotting the DNA.



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Question 3: What is the function of DNA polymerase alpha in eukaryotic replication?


Answer: Primer formation.

Explanation: The DNA polymerase alpha helps in initiating the strand formation has a very great activity of primer formation which helps in initiating the DNA replication but it doesn’t possess the 3’ to 5’ exonuclease activity.


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Question 4: What is the inhibitor of mammalian nuclear DNA polymerases?


Answer: Aphidicolin, a diterpenoid is an inhibitor.

Explanation: This diterpenoid inhibits the eukaryotic DNA polymerases which are polymerase-,,,,but it doesn’t affect mitochondrial DNA polymerases.



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Question 5: When we need to start DNA replication, there is a need to unwind the DNA first so what is the name of the enzyme that is used to unwind the DNA in prokaryotes?


Answer: DNA-B is a helicase that use to unwind the DNA.

Explanation: During the DNA replication, the DNA-A firstly binds with the origin called Ori-c of DNA and facilitates the initial strand separation and helps the binding of helicase called DNA-B which act as helicase, and DNA-C is called helicase loader.



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Question 6: What is the reason that ATP hydrolysis is required for the process of unwinding?

Answer: Energy is needed for the breakage of stacking interactions and hydrogen bonds.

Explanation: The ATP is very important to unwind the DNA and meting of DNA and the hydrogen bonds are present between the bases of a double helix and stacking interactions are also present that break with the help of ATP hydrolysis.


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Question 7: Which of the following is an antibiotic is used to inhibit the translocation process by binding to EF-G factor?


Answer: Fusidic acid.

Explanation: The translation is the process that is inhibited by various toxins and antibiotics, they inhibit the protein synthesis in both prokaryotes and eukaryotes such as fusidic acid inhibit translocation.



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Question 8: The genetic code is degenerate what does that mean?


Answer: It means one amino acid can be specified by more than one triplet codon.

Explanation: The genetic code show degeneracy of the codon because the property of more than one codon specifies one amino acid, for example, UUU and UUC code for the same amino acid called phenylalanine.



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Question 9: In eukaryotes, there is no process of gene regulation with the attenuation process why?

Answer: Because in eukaryotes the transcription and translation don’t occur simultaneously.

Explanation: In eukaryotes, the transcription and translation process occurs at different places, the transcription occurs in the nucleus and translation occur in the cytoplasm so there is no process of attenuation in eukaryotes.                                                                                                                                                              



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Question 10: What is the role of SSB in DNA replication?


Answer: It inhibits the reannealing of DNA strands after strand separation.

Explanation: The SSB is called a single strand binding protein that binds with DNA after unwinding of DNA and prevents the reannealing so that the process of DNA replication can be continued.


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Frequently Asked Questions



Why are nucleic acids are important?

Nucleic acids consist of DNA, RNA that transmit genetic information from one generation to the next. The study of nucleic acids helps us to understand the function of the genes and their importance to bring evolution. The understanding of nucleic acids is also important to know about diseases that are associated with genes.


What are nucleic acid and its function?

Nucleic acids are very large molecules and polymers. They consist of DNA and RNA, that carry out genetic information. They undergo replication and form proteins and proteins are very important for the growth and development of the body. Nucleic acid is named because they found in the nucleus.  


What is another term for nucleic acids?

Another term used for nucleic acid is deoxyribonucleic acid and ribonucleic acid. They are macromolecules that contain phosphate, sugar, and nitrogenous base. The nitrogenous bases are purine and pyrimidine. They contain genetic information in them. 


What is the monomer of nucleic acids?

Nucleic acids are made up of nucleotides, that are consist of sugar, phosphate, and nitrogenous base and the nitrogenous bases are adenine, guanine, cytosine, and thymine. In the case of RNA, in place of thymine, uracil is present. The nucleotides are present in repetitive units so form polymeric nucleic acids



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