Explanation:
Animals lived in the water during the Mesozoic and on land in the Paleozoic. Animal complexity increased during the Paleozoic while the first flowering plants appeared in the Mesozoic era. ...
The Paleozoic era, not the Mesozoic era, had the first dinosaurs. The first mammals emerged in the Paleozoic era, not the Mesozoic era. The Mesozoic era, not the Paleozoic era, had the first animals with shells.
Answer:
(a) 1/2; (b) no
Explanation:
Glucose-6-phosphate dehydrogenase deficiency (G6PD) is an X-linked recessive disorder and the woman's father was diseased so it means that woman is a carrier of the allele but has normal phenotype. It means that she will have XXᵇ genotype.
In contrast to this, her husband is diseased so his genotype will be XᵇY.
The Punnett square diagram related to the cross is attached.
(a) Proportion of their sons expected to be G6PD is 1/2:
They both may give birth to 4 progeny with genotypes XXᵇ, XᵇXᵇ, XY and XᵇY. It means they both may have 2 sons out of which one with genotype XᵇY will be diseased while the one with genotype XY will be healthy. So the proportion of their sons having G6PD is 1/2 or 50%.
(b) If the husband were G6PD deficient, the answer will not change.
The reason behind this is that this disease is caused by an allele located in X chromosome. But father contributes only Y chromosome to his son not X chromosome. The X chromosome will affect the genotype of his daughter not son that is why answer will not change. It means they will still have 1/2 of their sons diseased.
I am pretty sure it is glycogenesis<span>. Please tell me if I'm wrong.</span>
They also need energy to move and to stay alive
Answer:
Probability that the son will be:
color-blind = 1/2
blue eyed = 1/8
blue eyed and color-blind = 1/16
Explanation:
Mother : BbXcX
Father : BbXY
Colorblindness is X linked recessive trait hence will be caused when the son has XcY genotype.
XcX X XY :
Xc X
X XcX XX
Y XcY XY
Half of the sons will have normal vision and other half will be colorblind hence, there is 1/2 probability that the son will be colorblind.
Eye color is autosomal trait and blue color will be produced when the child has bb genotype.
Bb X Bb :
B b
B BB Bb
b Bb bb
1/4 of the children will have blue eyes. Since half of the children are expected to be males, 1/4 * 1/2 = 1/8 sons are expected to have blue eyes.
Probability of a son being blue eyed and color blind will be :
1/8 * 1/2 = 1/16
