reproduction, heredity, and DNA.
Please Use the overview and other internet sources to answer the following.
Part 1: Genetics – From Genes to Proteins, Mutations (Chapter 10)
Overview: Genetic information in DNA is transcribed to RNA and then translated into the amino acid sequence of a Protein.
A) Step 1 – Transcription: During the process of transcription, the information in the DNA codons of a gene is transcribed into RNA.
Suppose that gene X has the DNA base sequence 3′-TACCCTTTAGTAGCCACT-5′.
Question: What would be the base sequence of RNA after transcription occurs?
(In this particular example, assume that the RNA product does not require processing to become mRNA. In other words, the transcribed RNA becomes the mRNA sequence.)
B) Step 2 – Translation: During protein synthesis at the ribosome, the base sequence of the mRNA codons is translated to the amino acid sequence of a protein.
Question: Using the mRNA that you transcribed above, use the genetic code table to determine the resulting amino acid sequence. Turn this in.
And, turn in the answer to these questions:
What is the significance of the first and last codons? What meaning do these codons have for protein synthesis?
C) Mutations: A mutation is defined as a change in the base sequence of DNA. This may occur as a “mistake” in DNA replication, for example.
Suppose that during DNA replication, two mutant DNA sequences are produced as shown below.
For the 2 mutated DNA sequences, you will investigate how these changes might affect the sequence of amino acids in a protein.
Question: For each of the two, you will need to first transcribe the mRNA, and then use the genetic code table to determine the amino acid sequence.
Turn these in, and state whether the protein sequence changes for each.
Question: Then, explain why a change in amino acid sequence might affect protein function. Turn in your answer.
Here is the original sequence, followed by two mutated sequences, 1 and 2:
Original sequence 3′- TACCCTTTAGTAGCCACT-5′
Mutated sequence 1) 3′-TACGCTTTAGTAGCCATT-5′
Mutated sequence 2) 3′-TAACCTTTACTAGGCACT-5′.
Part 2: Inheritance of Traits or Genetic Disorders (Chapter 12)
Bob and Sally recently married. Upon deciding to plan a family, both Sally and Bob find out that they are both heterozygous for cystic fibrosis, but neither of them has symptoms of the disorder.
Set up and complete a Punnett Square for cystic fibrosis for this couple; turn in the Punnett square.
When doing the Punnett Square, C = normal allele; and c = allele for cystic fibrosis.
Note: You can use the Table function in MS Word to create and fill in a Punnett Square.
Based on the Punnett square, calculate chances (percentages) for having a healthy child (not a carrier), a child that is a carrier for the cystic fibrosis trait, and a child with cystic fibrosis? Turn in these percentages.
Part 3: Cell division, sexual reproduction and genetic variability (Chapter 11)
Eukaryotic cells can divide by mitosis or meiosis. In humans, mitosis produces new cells for growth and repair; meiosis produces sex cells (gametes) called sperm and eggs.
Although mutations are the ultimate source of genetic variability, both meiosis and sexual reproduction also can contribute to new genetic combinations in offspring.
Question: How do both meiosis and sexual reproduction (fertilization) produce offspring that differ genetically from the parents? Be sure to talk about steps in meiosis that increase variability as well as the process of fertilization.
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