Answer:
1/8 (12.5%)
Explanation:
An autosomal recessive disease is an inherited disease in which an individual need to receive both defective alleles at the same gene <em>locus</em> to be expressed in the phenotype. In this case, both parents are carriers of the recessive mutant allele associated with the sickle cell anaemia trait, thereby both parents are heterozygous, ie., each parent has one copy of the normal allele 'H' and one copy of the defective mutant allele 'h' associated with this condition. In consequence, their first child has a 1/4 (25%) chance of having sickle-cell anaemia. Moreover, the chance of having a girl is 1/2 and the chance of having a boy is 1/2, thereby the final chance of having a girl sickle cell anaemia individual is 1/4 x 1/2 = 1/8 (12.5%).
- Parental cross for sickle cell anaemia trait = Hh x Hh >>
- F1 = 1/4 HH (normal); 1/2 Hh (normal); 1/4 hh (sickle cell anaemia) >>
- Sex proportion of sickle cell anaemia individuals = 1/8 female sickle cell anaemia individuals + 1/8 male sickle cell anaemia individuals (1/8 + 1/8 = 1/4)
<span>Well active transport absolutely moves substances against their concentration gradient (answer C), although I dislike how the other answers could probably be argued for. This seems to be a case of pick the "best" answer</span>
Answer:
Canis represents the genus :D
The probability of having a second child presenting the desease is 25% (i think so)
