What is the probability that a sighted dog carries the gene for blindness if it is crossed with another such dog? - briefly
The probability that a sighted dog carries the gene for blindness when crossed with another similar dog can be determined by the genetic makeup of both parents and their offspring. In simplified terms, if both parents are heterozygous (carrying one copy of the dominant sighted gene and one recessive blind gene), there is a 50% chance that any given offspring will also carry the gene for blindness.
What is the probability that a sighted dog carries the gene for blindness if it is crossed with another such dog? - in detail
The probability that a sighted dog carries the gene for blindness, when it is crossed with another similar dog, is determined by the genetic makeup and inheritance patterns of the trait in question. In many cases, blindness in dogs can be associated with recessive genes. This means that an individual needs to inherit two copies of the recessive allele (one from each parent) to express the trait—in this case, blindness.
Given a sighted dog, it is likely heterozygous for the trait, meaning it carries one copy of the normal allele and one copy of the recessive allele responsible for blindness. In such cases, the probability that the dog carries the gene for blindness is high, as the dog indeed possesses at least one copy of the recessive allele.
When two heterozygous dogs are crossed, the potential outcomes for their offspring can be determined using a Punnett square or simple probability calculations:
- Normal (sighted) Alleles: Each parent can contribute either a normal allele or a recessive allele to their offspring. The probability of a sighted dog contributing a normal allele is 0.5, and the same applies to the recessive allele.
- Offspring Genotypes: The possible genotypes for the offspring are:
- Normal (AA) if both parents contribute a normal allele.
- Heterozygous carrier (Aa) if one parent contributes a normal allele and the other contributes a recessive allele.
- Blind (aa) if both parents contribute a recessive allele.
The probability of each outcome can be calculated as follows:
- Normal (AA): 0.5 * 0.5 = 0.25 or 25%
- Heterozygous carrier (Aa): 0.5 0.5 + 0.5 0.5 = 0.5 or 50%
- Blind (aa): 0.5 * 0.5 = 0.25 or 25%
Thus, when two heterozygous sighted dogs are crossed, there is a 50% probability that their offspring will also be carriers of the gene for blindness. This highlights the importance of genetic testing and careful breeding practices to manage the prevalence of recessive traits in dog populations.