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A gene mutation is a permanent, heritable change in the DNA sequence of a single gene. These changes can alter the protein produced by the gene, potentially affecting an organism's traits or health.

Biology

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Answer

pedigree chart\text{pedigree chart}

3\text{3}

25%25\%

Step-by-step solution

5 steps

1.1:
A gene mutation is a permanent, heritable change in the DNA sequence of a single gene. These changes can alter the protein produced by the gene, potentially affecting an organism's traits or health.

1.2:
The type of diagram shown is a pedigree chart\boxed{\text{pedigree chart}}.

1.3:
There are 3\boxed{\text{3}} generations shown in the diagram.

1.4:
The gene for cystic fibrosis is a recessive allele because unaffected parents (individuals 1 and 2) have an affected child (individual 13). If the allele were dominant, at least one parent would have to be affected to pass on the trait. Since individuals 1 and 2 are unaffected but produce an affected child, they must both be heterozygous carriers, meaning the affected allele is recessive.

1.5:

  1. Determine the genotypes of individual 9 and individual 10.

    Let N represent the dominant allele (unaffected) and n represent the recessive allele (cystic fibrosis).
    Individual 9 and individual 10 are both unaffected (unshaded). However, they have an affected child (the first child between them, which is shaded). An affected child must have the genotype nn. For a child to have genotype nn, both parents must contribute an 'n' allele. Therefore, both individual 9 and individual 10 must be heterozygous carriers (Nn).

  2. Perform a genetic cross between individual 9 (Nn) and individual 10 (Nn).

    Parents: Nn ×\times Nn
    Gametes from Nn: N, n

  3. Construct a Punnett square.

    GametesNnNNNNnnNnnn\begin{array}{|c|c|c|} \hline \text{Gametes} & \text{N} & \text{n} \\ \hline \text{N} & \text{NN} & \text{Nn} \\ \hline \text{n} & \text{Nn} & \text{nn} \\ \hline \end{array}
  4. Determine the genotypes and phenotypes of the offspring.

    • NN: Unaffected
    • Nn: Unaffected (carrier)
    • nn: Affected (cystic fibrosis)
  5. Calculate the percentage chance of having a child with cystic fibrosis.

    From the Punnett square, there is 1 chance out of 4 for an offspring to have the genotype nn (cystic fibrosis).
    Percentage chance = 14×100%=25%\frac{1}{4} \times 100\% = 25\%.

    The percentage chance of them having a child with cystic fibrosis is 25%\boxed{25\%}.

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