Question

This is my review study guide for my biology final can someone please answer them so i can study. Thank you

1. Assuming only full-chromosome separations in meiosis, how many different genetic combinations are there for Kate and Jon’s offspring? Show an equation for calculating this and explain the math.
2. The phenotype for cheater-gene expression is recessive. Kate was homozygous for the cheater gene mutation, and thus was very promiscuous with partners outside her relationship. Jon was homozygous too, but lacked the cheater-gene mutation. What are the chances that their child will express the trait of the cheater gene (i.e. - what are the chances that their child will have the cheater-gene phenotype)? Draw a Punnett square to illustrate this concept.
3. What if Jon and Kate have two children, a boy and a girl, and those two children have children (yuck!). Genetically speaking, what are the chances that their offspring (the F2 cross) will be homozygous for the cheater gene mutation? Draw a Punnett square to illustrate this concept.
4. The DNA sequence containing the Cheater Gene is given below.
   Transcribe the DNA into mRNA (as if RNA polymerase was using the gene below as a template).
   T C A G T A C A C A G T A C T C C G A T G A C T T G C A A C T T C A G
5. Translate the mRNA sequence you created in question 11 into an amino acid sequence. Use the single letter abbreviations given for each amino acid given in the table below (example: use A for Ala, M for Met, etc.)

Answer 1

See the sentence :- the phenotype for chaeter gene expression is reccessive . This means cheater gene has three type of alleles.

1. dominnat :- GG:- phenotype normal
2. Recessive :- gg:- phenotype normal
3. mutant recessive :- gmgm :- phenotype cheater

if we consider the heterozygous condition for above allelic combination and its expression-

1. Gg:- normal phenotype
2. Ggm:- normal phenotype
3. ggm:- normal phenotype

that means the above sentence said that the phenotypic cheater individual is having gmgm combination.

The allelic combination of Kate :- gmgm [homozygous for mutatnt]:- so phenotypically express cheater.

The allelic combination of Jon :- gg [homozygous reccessive]:- so phenotypically normal.

• let us do the cross and produce F1 generation
• self crossing of F1 produce F2 genration.
• with punnett square and genotype and phenotype ratio.

for this see photo.

• all the progeny of F1 combination are heterozygous of reccessive normal and recessive mutant : ggm combination ; so phenotypically they are normal
• in F2 generation only oneoffspring is  homozygous mutant for cheater.

[note : for your understanding of male and female offspring I have wrote XX and XY combination . but you have to remove that sex chromosome when will be you copy down it . see F! self cross pattern , i have given it in normal way. Just write it like this. ]

  

1. transcription :- is a process in which the DNA template is used for the formation of m- RNA molecule which a complementary strand of DNA with opposite polarity.
2. translation:- in this process the three nucleotide codon [for example AUG] is translated on the ribosome molecule and t -RNA bring amino acid according to that codon.and form a polypeptide molecule. let us see-

the DNA templet is :- 3' T C A G T A C A C A G T A C T C C G A T G A C T T G C A A C T T C A G 5'

transcription [formation of m-RNA with the help of RNA polymerase ]

the m- RNA strand is :- 5' A G U C A U G U G U C A U G A G G C U A C U G A A C G U U G A A G U C 3'

translation [a chain of amino acids for each codo ]

sequence of amino acid :- S- H- C- S- STOP - G - Y- STOP- T- L- K-

s= serine; h= histidine; c= cystein; g= glycin; y= tyrosine; t= threonine; l= lusein; k= lysine . here UGA is a stop codon , no amino acid for it .