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 pedigreechart.
1.3:
There are 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:
Step 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).
Step 2: Perform a genetic cross between individual 9 (Nn) and individual 10 (Nn).
Parents: Nn × Nn
Gametes from Nn: N, n
Step 3: Construct a Punnett square.
GametesNnNNNNnnNnnn
Step 4: Determine the genotypes and phenotypes of the offspring.
- NN: Unaffected
- Nn: Unaffected (carrier)
- nn: Affected (cystic fibrosis)
Step 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 = 41×100%=25%.
The percentage chance of them having a child with cystic fibrosis is 25%.
That's 2 down. 3 left today — send the next one.