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1. A) Explain what is sex linkage, and B) Apply the concept of sex linkage to...

1. A) Explain what is sex linkage, and B) Apply the concept of sex linkage to explain why color blindness is more prevalent in men than in women. A boy who is colorblind, which parent did he inherit this gene from? Explain.

2. In plants long pollen gain is dominant to short grain and purple flower color is dominant to white color. You are mating two plants. One is homozygous recessive for both traits, flower color and pollen grain. The other plant is heterozygous for both traits. A) What types of gametes do you expect each of these plants to produce (which allele combinations). B) What ratio do you expect for each allele combination in the gametes of these plants? C) What are the phenotype ratios you expect for the offspring? D) Which Medenlian assumption are you making when estimating those allele ratios?

3. After mating the two plants indicated in the example above, you find that 40% of the offspring have short pollen grain and purple flowers, 40% have long pollen grain and white flowers, 10% have short pollen and white flowers and the remaining 10% have long pollen and purple flowers. A) Are the genes for flower color and pollen grain assorting independently? B) What could you say about their location on the chromosome?

4. Following on the example above: A) Draw a pair of chromosomes showing the relative location of the pollen grain and the flower color genes. B) Which combination of alleles did the heterozygous plant inherit from each parent?

5. Explain the differences in transcription initiation (RNA polymerase binding to the promoter region and starting transcription) between prokaryotes and eukaryotes.

6. A) Where does transcription and translation occur in prokaryotes vs eukaryotes? B) Explain how having a nucleus affects what happens between transcription and translation in eukaryotes vs prokaryotes. C) List the differences in the modifications to the RNA transcript between prokaryotes and eukaryotes.

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Answer:

Question-5) answer-

Eukaryotic transcription: The three steps in eukaryotic transcription include the following.

1. Initiation; 2. Elongation; 3. Termination

Several of the transcription factors bind to the promoter region to recruit the appropriate RNA polymerase. The complex of RNA polymerase and transcription factors bind to the promoter region to form a pre-initiation complex (PIC).

A group of transcription factors assemble at the promoter region of DNA, and bind to TATA box to change its shape. This event exposes the binding sites of other transcription factors. After several of these events, RNA polymerase binds.

The binding of RNA polymerase to the promoter site starts the process of transcription. There are three different classes of RNA polymerases that bind to different promoter sites and transcribe different parts of genome. Of these, RNA polymerase II causes the transcription of mRNA. The terminator present at another end of the genome signals the completion of transcription, thus terminating the programme.

Operon is a functioning unit of genomic DNA (deoxyribonucleic acid), which is a group of genes whose transcription is under the control of a single promoter. The rate of transcription is affected by proteins called transcription factors (repressors and activators), which bind directly to DNA at “promoter sites.”

Operator is a part of operon, which is a DNA binding site for a repressor, binding of repressor to the operator site terminates the process of transcription in prokaryotes. Repressor has two binding sites for operator and inducer respectively.

Lactose is not all a direct inducer of lac operon. Initially beta galactosidase converts lactose to allolactose and then induces the operon system.

Prokaryotes: In the prokaryotic protein synthesis, the following initiation factors enable in finding out the intiation codon AUG that codes for methionine. They are IF1, IF2, and IF3. Mainly

1. IF2, is the factor that enable fMet tRNAi to bind to the ribosomal smaller subunit.

2. IF3 is the other crucial factor that ensure that an initial AUG codon only to used as as a starting codon in the cytosolic translation. IF3 examines the dependability of the initiation codon selection. Eukaryotes, In these organisms, predominantly the smaller subunit of the ribosome is considered to bind to the 5 prime end of the messenger RNA. This type of binding is potentially mediated by the 7-methylguanosine cap.

A variety of proteins such as initiation factors bind to this 7-methylguanosine cap, polyadenyl tail of the 5 prime regions and finally enable the small ribosomal subunit biding to the messenger RNA. This initiation along mRNA can identify initiation codon AUG, this meticulous process if defined as “scanning”. Identification of the AUG mainly based mediated through base pair interactions of AUG and anticodon in fMet tRNA.

tRNA that recognizes 5’ UGC 3’ on mRNA transcript then

A. the anticodon sequence will be 3’ ACG 5’

B. 5’ UGC 3’ on mRNA transcript codes for Cysteine

C. aminoacyl-tRNA synthetase

Protein synthesis:

Elongation:

Protein synthesis occurs in three steps namely, initiation, elongation and termination. During initiation, the mRNA, tRNA comes together at the ribosome and the translation initiates at the start codon (AUG, codes for methionine).

During elongation, amino acids are added to the growing polypeptide chain according to the mRNA sequence. The aminoacyl tRNA molecules are picked up by the elongation factors in the presence of GTP, Which enters the A site. On the ribosome, the mRNA codon bound at the A-site will be matched with appropriate anticodon of aminoacyl tRNA. The translational machinery plays a key role in this selection and proof reading.

After binding of correct codon and anti codon, the new amino acid is linked to the growing polypeptide chain in the P –site (peptidyl site) by a peptide bond. This process is catalysed by an enzyme peptidyltransferase located at the peptidyl transferase center of the large ribosomal subunit. A new peptidyl-tRNA now occupies the A site. The empty tRNA now moves to the E site and this process is facilitated by another elongation factor (eEF 2) and GTP. The new peptidyl tRNA moves to the P site, and the cycle repeats until it reads a stop codon. The following picture depicts the various stages in elongation.

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