Question

Colour blindness is a recessive trait. A couple with normal vision have two sons, one with colour blindness and one with a normal vision. If the couple had daughters, what proportion of them will have a normal vision? Explain with the help of a pedigree diagram.OR Where do transcription and translation take place in a prokaryotic cell? Describe the three steps in transcription.

Answer 1

Answer:

Solution :

From the given situation , it is evident that mother is a carrier for colour blindness. Hence of the two sons, one was normal and the other was colourblind. As the gene for colour blindness is located on X chromosome, in daughters it would occur only in heterozygous state. Hence of the two daughters, one is normal and other is a carrier for colour blindness.

Both transcription and translation occur in cytoplasm in prokaryotes. The three steps of transcription are:

Initiation

• RNA polymerase binds to the specific site of DNA called promoter.

• Promoter of the DNA is recognized by initiation factor or sigma

• RNA polymerase along with initiation factor binds to the promoter.

Elongation

• RNA polymerase unzipped the DNA double helix and forms an open loop.

• It uses ribonucleoside triphosphates as substrate and polymerizes in a DNA template following the rule of complementarity.

• Only a short stretch of polymerized RNA remains binds with the enzyme.

• The process of polymerization continued till the enzyme reaches the terminator gene.

Termination

• RNA polymerase recognizes the terminator gene by a termination-factor called rho (ρ) factor.

• The RNA polymerase separated from the DNA and also the transcribed RNA.

Answer 2

The chance of the daughter being colourblind is zero percentage.

• Colourblindness is a recessive trait and is a sex-linked disorder.
• In the first generation, the father is with normal vision so he is 44 + XY and the mother is also with vision but needs the carrier so she is 44 + XX°
• In the second generation, one son is colour blind so he has 44 + X°Y
• The second son is normal he has XY
• Now in the case of the daughter can be 44 + XX or 44 + XX° that is either she can be a carrier or not.
• But the daughter cannot have the colourblind vision.

The pedigree diagram is given below.