In: Biology
1) Probability of having one son is 1/2=0.5
Probability of having next one male child is 1/2 x1/2
probability that two individuals will have five sons in a row = (1/2)5= 0.03125
2) F - Freckles (Dominant allele)
f - No freckles (recessive allele)
**Genotypes and phenotypes possible are
FF - Freckles
Ff - Freckles
ff - No freckles
*Man with freckles can have genotype FF or Ff
* Woman with no freckles can have genotype ff
**Children get one allele from each of the parents.
They have 2 children with freckles and one does not have freckles.
a) man's genotype will be Ff
Because they have one child with no freckles. If father has the genotype FF child will never have ff.
b) Gametes produces by man will have alleles F and f
c) Woman's genotype will be ff
d) Alleles produces by woman are f and f
e) Genotype of Children with freckles is Ff
F from father and f from mother.
f) Genotype of child with no freckles is ff
f from father and f from mother.
3. Free earlobes (F) dominant
Attached earlobes (f) recessive
FF - Free earlobes
Ff - Free earlobes
ff - attached earlobes
Chad has attached earlobes - Genotype is ff
Andrea has attached earlobes - Genotype is ff
All Children of theirs will have genotype ff
All children will have attached earlobes.
So it is not possible for them to have a child with free earlobes.