In the example, no RNA is made during the process. What would need to occur for lacZ to be expressed in mRNA?
1 answer:
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.2. Offspring Genotypes will be Cc or cc.
Offspring phenotypes : Cleft chin or no cleft chin.
% chance child will have cleft chin: 50%
3. % chance child will have arched feet: 25%
4. % chance child will have blonde hair: 50%
5. % chance child will have normal vision: 25%
Explanation:
CASE 1 :
Dominant trait: cleft chin (C)
Recessive trait: lacks cleft chin (c)
Father’s gametes: cc
Mother’s gametes: Cc
There are two possible combination of Gametes ,
C fom mother and c from father= Cc
c from mother and c from father = cc
Gametes of Cc Parents= ........(i)
Gametes of cc parents =<u> </u> .........(ii)
Combining (i) and (ii) we get,
There fore offspring Genotypes will be Cc or cc
Offspring phenotypes :
Genotype Cc then phenotype= Cleft chin
Genotype cc then phenotype = Lacks cleft chin.
percentage chance child will have cleft chin = ×100
Therefore the chance is 50%.
CASE 2 :
Dominant trait: flat feet (A)
Recessive trait: arched feet (a)
Mother’s gametes: Heterozygous (Aa)
Father’s gametes: Heterozygous (Aa)
There are four possible combination of genotypes are =AA , Aa, Aa and aa
i.e. A from mother, A from father= AA
A from mother, a from father =Aa
a from mother, A from Father = Aa
a from mother, a from father = aa
Gametes of Aa parent =
Gametes of other Aa parent =
<u>..................................................................................</u>
+ +
<u>..........................................................................................</u>
<u></u>
Offspring Genotypes will be: AA or Aa or aa
Offsprings phenotype will be:
Genotype AA then phenotype will be Flat feet
Genotype Aa then phenotype will be flat feet
Genotype aa then Phenotype will be arched feet.
Percentage chance child will have arched feet = × 100 = 25%
CASE 3:
Dominant trait: Brown hair (B)
Recessive trait: Blonde hair (b)
Mother’s gametes: Homozygous recessive (bb)
Father’s gametes: Heterozygous (Bb)
This case is very similar to the case 1 as one parent is homozygous recessive and other parent is heterozygous.
Resulting in half Bb and halve bb combination.
Genotypes will be Bb or bb
Phenotypes will be :
Genotype Bb then phenotype Brown hair
Phenotype bb then Phenotype bb.
% chance child will have blonde hair: 50%
CASE 4:
Dominant trait: farsightedness (F)
Recessive trait: normal vision (f)
Mother’s gametes: Heterozygous (Ff)
Father’s gametes: Heterozygous (Ff)
This Case is similar to case 2
it will result in one-fourth FF , half Ff and one-fouth ff combination.
Therefore Genotypes will be: FF, Ff and ff
Phenotypes:
Genotype FF then phenotype farsightedness
Genotype Ff then phenotype farsightedness
Genotype ff then phenotype normal vision.
% chance child will have normal vision: 25%