Albinism is a recessive trait. A man and woman who both have normal pigmentation have one child out of three who has albinism (without melanin pigmentation). The genotypes of this child's parents: (C) Both parents must be heterozygous.
Albinism is an autosomal and not a sex-linked recessive trait.
Only homozygous recessive (aa) offspring will be albino, both homozygous dominant (AA) and heterozygous dominant (Aa) offspring will possess the normal phenotype.
Since it is given that both the father and the mother have normal pigmentation, thus they must both have the genotype Aa.
From the Punnett square it can be observed that a cross between the parents would result in the creation of three offspring with the normal genotype (one with AA and two with Aa) and one offspring with the genotype aa (albino).
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Answer:
0.153
Explanation:
We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.
So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.
ρVg = ρ'V'g + ρ"V"g
ρV = ρ'V'g + ρ"V"
Its new body volume = volume of water displaced, V = V' + V"
ρ(V' + V") = ρ'V' + ρ"V"
ρV' + ρV" = ρ'V' + ρ"V"
ρV' - ρ"V' = ρ'V" - ρV"
(ρ - ρ")V' = (ρ' - ρ)V"
V'/V" = (ρ - ρ")/(ρ' - ρ)
= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)
= (0.9988 g/cm³ ÷ 0.18 g/cm³)
V'/V" = 5.55
Since V = V' + V"
V' = V - V"
(V - V")/V" = 5.55
V/V" - V"/V" = 5.55
V/V" - 1 = 5.55
V/V" = 5.55 + 1
V/V" = 6.55
V"/V = 1/6.55
V"/V = 0.153
So, the fish must inflate its air sacs to 0.153 of its expanded body volume
Answer:
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Explanation:
As the temperature increases, the average kinetic energy increases, making the gas particles hit the walls of the container more (therefore increasing pressure). Hopefully this helps! Good luck :)
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