Question

The son, Seif, is born with sickle cell anemia. This means both parents carry the allele for sickle cell anemia.

The father, Ahmad has a heterozygous genotype as told by the doctor.

The mother can also be proved heterozygous for the trait since her mother was anemic. This means one of the affected alleles was inherited from the mother.

So, the genotypes of Ahmad is Ss and that of Amina is Ss.

Gametes formed by Ahmad are S and s.

Gametes formed by Amina are S and s.

1 offspring having genotype SS does not carry the affected allele and will be phenotypically normal.

2 offsprings having genotype Ss carry the affected allele but will be phenotypically normal.

1 offspring having genotype ss have both affected alleles and will be sickle cell disease.

X	S	s
S	SS	Ss
s	SS	ss

Question:

Fill the tables by doing the calculations and transforming your results into one ratio, fraction, and percentage for each genotype and phenotype

As such:

ratio 1:2:1, Genotypes: BB ½ , 50%,Bb ¾ 75%                                                  

Phenotype: wrinkled 75%,Smooth 25%

Ratio:
Genotypes
Number
Fraction Form
Percentage

Ratio:
Phenotypes
Number
Fraction Form
Percentage

Answer 1

1. Note: Genotypes from the given table are SS( homozygous not anemic), Ss( heterozygous not anemic), ss ( homozygous anemic).[ Genotype is genetic character].

The number of SS offsprings - 2, number of Ss offsprings - 1, number of ss offsprings-1.

From the above note:

(a) Ratio of genotypes: SS:Ss:ss = 2:1:1.

(b) Number of genotypes: 3 types.

(c) Fraction form: There are 4 offsprings so faction for each offspring is 1/4.

(That is fraction = No. of offsprings in a genotype/ total No. of offsprings in all genotypes).

So, fraction forms for SS is 2/4= 1/2; Ss is 1/4; ss is 1/4.

(d) Percentage: SS = 50%, Ss = 25%, ss = 25%

Calculation: 100% for 4 offsprings then

For SS - 2 offsprings →2× 100÷ 4=50%.

For Ss - 1 offspring →1×100÷4=25%.

For ss - 1 offspring→ 1×100÷4=25%.

2. Note: From the above table Phenotypes are SS,Ss both are non anemic under one phenotype & ss anemic under other phenotype. ( Phenotype is external character).

From the above note:

(a) Ratio of phenotypes: 3:1( SS& Ss-2+1= 3 and ss-1).

(b) Number of phenotypes: 2 types.

(c) Fraction form of phenotypes: 1/4 for each offspring ( That is No. of offsprings in a phenotype / total No of offsprings in all phenotypes).

So, fraction for SS& Ss - 3/4, and for ss - 1/4.

(d) Percentage for SS& Ss is 75% and for ss is 25%.

Calculation: 100% for 4 offsprings.

So, For SS& Ss - 2+1= 3 offsprings →3×100÷4=75%.

For. ss - 1 offspring →1×100÷4=25%.