Question

A homozygous \(A A B B\) mouse was crossed to a homozygous recessive aabb mouse. The \(\mathrm{F}_{1}\) were backcrossed to a homozygous recessive aabb mouse, and the offspring were analyzed for \(A\) and \(B\). The genetic distance between \(A\) and \(B\) was found to be \(10 \mathrm{cM}(\mathrm{mu})\). If 200 mice were analyzed, deduce the phenotypes observed in the offspring.

Short answer

Answer: The phenotypes observed in the offspring will include both dominant traits (A-B + a-b) in approximately 180 mice and recombinant traits (Ab + aB) in approximately 20 mice.

Step-by-step solution

Determine the genotype of F1 offspring

In the first cross (AA BB) x (aabb), the offspring will inherit one A allele and one a allele, and one B allele and one b allele. Therefore, the genotype of F1 offspring will be AB Ab aB ab.

Calculate recombination frequency

Given that the genetic distance between A and B is 10 cM, this indicates that there is a 10% recombination frequency between these genes. This means that in 10% of the offspring, the alleles A and B will be separated during gamete formation due to crossing over.

Identify possible genotypes and calculate probabilities from recombination frequency

The possible genotypes of the offspring are: 1. A-B gametes: 45% (non-recombination) of 50% F1 genotype 2. Ab or aB gametes: 5% recombination frequency × 50% F1 genotype 3. a-b gametes: 45% (non-recombination) of 50% F1 genotype

Calculate expected genotype distribution in the offspring from the backcross

In the backcross with a homozygous recessive aabb mouse, the genotype probabilities are: 1. A-B + a-b: (45% × 50%) + (45% × 50%) = 90% 2. Ab + aB: (5% × 50%) + (5% × 50%) = 10%

Determine the phenotypes observed in the offspring

From expected genotype distribution, we can estimate the number of offspring with each genotype, out of the 200 mice analyzed: 1. A-B + a-b offspring: 90% × 200 = 180 2. Ab + aB offspring: 10% × 200 = 20 As a result, the phenotypes observed in the offspring will include both dominant traits (A-B + a-b) and recombinant traits (Ab + aB).

Key concepts

Genetic Distance

Genetic distance is a crucial concept for understanding genetic linkage, measured in centimorgans (cM). It indicates how frequently genes on the same chromosome are separated during the formation of gametes due to crossing over. In this case, the genetic distance between genes A and B is 10 cM. This means these genes are likely to recombine and produce different combinations 10% of the time. When genes are close together on a chromosome, they have a smaller genetic distance, which means they have a lower chance of recombination. Conversely, genes that are farther apart have a higher genetic distance and are more likely to undergo recombination. Understanding genetic distance helps in mapping the location of genes on chromosomes, as well as predicting the inheritance patterns in offspring.

Recombination Frequency

Recombination frequency is the percentage at which genetic recombination occurs between two genes. It is directly related to genetic distance and helps us understand how traits are inherited. Given that the genetic distance between A and B is 10 cM, we can infer a 10% recombination frequency. This means that out of 100 meiosis events, you would expect 10 to yield alleles where recombination has occurred between the two loci.
In terms of the mouse example, when we backcross the F1 generation to a homozygous recessive aabb mouse, this 10% recombination frequency predicts the percentage of offspring that will exhibit a combination of the recombinant alleles (Ab, aB) in contrast to the non-recombinant parental alleles (AB, ab). This helps in quantifying how often these genes will assort independently, thus influencing the phenotypic patterns we observe in the next generation.
Understanding recombination frequency allows geneticists to calculate the likelihood of particular genetic traits being passed on, making it a vital tool in genetic linkage studies.

Phenotype Distribution

Phenotype distribution refers to the observable characteristics in a population, determined by the underlying genotypic combinations. In genetic crossing studies, it's essential to predict how these phenotypes distribute among offspring. For our mouse example, we rely on recombination frequency to calculate expected phenotypic outcomes.
After the backcross with a homozygous recessive aabb mouse, we have two groups of phenotypes based on genotype: those with dominant traits (A-B + a-b) and those with recombinant traits (Ab + aB).

• Non-recombinant phenotypes (A-B and a-b) will constitute around 90% of the offspring, which is 180 mice, showing the expected parental type traits.
• Recombinant phenotypes (Ab and aB) will show up in about 10% of the offspring, equating to 20 mice, demonstrating the new combination of traits.

Therefore, phenotype distribution helps in determining the proportion of features in the offspring, reflecting the inheritance patterns decided by genetic distance and recombination frequency.