If one parent is heterozygous for a dominant autosomal allele and the other parent does not carry the allele, a child of theirs has a________ chance of being heterozygous.a. 25 percentb. 50 percentc. 75 percentd. no chance; it will die

Key Concepts: Mendelian Genetics, Autosomal Dominant Traits, Heterozygous, Punnett Squares Explanation: In this scenario, we are considering a single autosomal dominant trait, where the heterozygous genotype is denoted as (Aa) and the homozygous recessive genotype is denoted as (aa). The dominant allele is denoted as (A) and the recessive allele is denoted as (a). \n Step by Step Solution: Step 1. First, let's determine the genotype of each parent. We know that one parent is heterozygous (Aa) and the other parent does not carry the allele (aa). Step 2. We can set up a Punnett square to determine the possible genotypes of their offspring. Step 3. The Punnett square will have two columns for each possible gamete from each parent and two rows for each possible gamete from the other parent. This will result in four boxes that represent the possible genotypes of the offspring. The Punnett square for this scenario is shown below:\n | | A | a |\n| - | - | - |\n| a | Aa | aa|\n Step 4. From the Punnett square, we can see that there are two possible genotypes for the offspring: Aa (heterozygous) and aa (homozygous recessive). The probability of each genotype is 50%. Step 5. Therefore, the chance of the child being heterozygous (Aa) is 50%. Final Answer: The chance of the child being heterozygous is 50%.