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A man with normal red blood cells and a woman who is a carrier of sickle cell anemia have children. What are the possible genotypes and phenotypes of their offspring? Describe the process of sublimati

Biology

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2 parts
c(ii)

1.5 Ω\text{1.5 } \Omega

c(iii)

3 A\text{3 A}

Step-by-step solution

7 steps

Part a· 3 steps

Let HbA represent the allele for normal red blood cells and HbS represent the allele for sickle cell. Sickle cell anemia is a recessive disorder, meaning HbS HbS results in the disease, while HbA HbA is normal and HbA HbS is a carrier with normal red blood cells.

  1. Determine the genotypes of the parents.

    The man has normal red blood cells and is not stated to be a carrier, so his genotype is HbA HbA.
    The woman is a carrier of sickle cell, so her genotype is HbA HbS.

  2. Construct a Punnett square to show the possible genetic compositions of their children.

    GametesHbA (man)HbA (man)HbA (woman)HbA HbAHbA HbAHbS (woman)HbA HbSHbA HbS\begin{array}{|c|c|c|} \hline \text{Gametes} & \text{HbA (man)} & \text{HbA (man)} \\ \hline \text{HbA (woman)} & \text{HbA HbA} & \text{HbA HbA} \\ \hline \text{HbS (woman)} & \text{HbA HbS} & \text{HbA HbS} \\ \hline \end{array}

  3. Determine the possible genetic compositions (genotypes) and their corresponding phenotypes.

    • HbA HbA: Normal red blood cells, not a carrier. (50% probability)
    • HbA HbS: Normal red blood cells, but a carrier of the sickle cell trait. (50% probability)

Part b(ii)

Three substances that can undergo sublimation are

i) Sublimation is the process where a substance changes directly from a solid state to a gaseous state without passing through an intermediate liquid phase. This occurs when the substance absorbs enough energy to overcome intermolecular forces in the solid state, but the conditions (temperature and pressure) do not allow for a stable liquid phase.

  • Dry ice (solid carbon dioxide)
  • Iodine
  • Naphthalene (found in mothballs)

Part c(i)

A circuit diagram for two lamps (resistors) connected in parallel to a battery

The diagram would show a battery symbol (a long line and a short line) with two parallel branches extending from its terminals. Each branch would contain a lamp symbol (a circle with a cross inside or a zigzag line for a resistor). The two branches would then rejoin before connecting back to the other terminal of the battery.

Part c(ii)

1.5 Ω\text{1.5 } \Omega

Calculate the effective resistance of the lamps.

  1. Identify the resistance of each lamp and the connection type.

    Each lamp has a resistance R=3 ΩR = 3 \, \Omega. They are connected in parallel.

  2. Use the formula for effective resistance in a parallel circuit.

    For two resistors in parallel:

    1Reff=1R1+1R2\frac{1}{R_{eff}} = \frac{1}{R_1} + \frac{1}{R_2} 1Reff=13 Ω+13 Ω\frac{1}{R_{eff}} = \frac{1}{3 \, \Omega} + \frac{1}{3 \, \Omega} 1Reff=23 Ω\frac{1}{R_{eff}} = \frac{2}{3 \, \Omega} Reff=32 ΩR_{eff} = \frac{3}{2} \, \Omega Reff=1.5 ΩR_{eff} = 1.5 \, \Omega

    The effective resistance is 1.5 Ω\boxed{\text{1.5 } \Omega}.

Part c(iii)

3 A\text{3 A}

Calculate the total current in the circuit.

  1. Identify the total voltage and effective resistance.

    The battery voltage V=4.5 VV = 4.5 \, V.
    The effective resistance Reff=1.5 ΩR_{eff} = 1.5 \, \Omega.

  2. Use Ohm's Law to calculate the total current.

    I=VReffI = \frac{V}{R_{eff}} I=4.5 V1.5 ΩI = \frac{4.5 \, V}{1.5 \, \Omega} I=3 AI = 3 \, A

    The total current in the circuit is 3 A\boxed{\text{3 A}}.

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