Question in photo **Genetic Linkage and Probability Study****Exercise 2: Gene Linkage in Twitch and Freckle Traits**You are studying two genes that are linked: The Twitch and Freckle genes. There are two alleles of each gene:- **T** is dominant to **t**, where T results in twitching behavior and t results in no twitching.- **F** is dominant to **f**, where F results in freckles and f results in no freckles.The T and F loci are 36 map units apart.**a. Probability of Twitchy and Freckled Offspring**We have a heterozygous individual (Tf/tF); their replicated chromosomes are pictured on the right. The heterozygote mates with a homozygous recessive individual (tf/tf). What is the probability they will have a child that is twitchy and freckled?**b. Hypothetical Offspring Phenotypes**If the individuals from part a had 25 children, what do you expect their phenotypes to be, and in what proportions? How does this differ from what you would expect if there was no linkage between these loci?**Diagram Explanation**- The diagram on the right illustrates a pair of homologous chromosomes.- The map distance between the **T (Twitch gene)** and **F (Freckle gene)** loci is marked as 36 map units.- The upper line represents a chromosome with alleles T and f, and the lower line represents another chromosome with alleles t and F.- This linkage map shows the physical proximity of the two gene loci on the chromosome, indicating the likelihood of recombination.**Considerations for Analysis:**- Understand the concept of genetic linkage and how it affects inheritance patterns.- Use recombination frequency (36%) for calculating expected offspring phenotypes.- Compare linked versus unlinked inheritance scenarios to highlight the impact of genetic linkage.

If genes are located on the same chromosome then they are classified as linked genes. In this condition the crossing over frequency between the two genes will be reduced. This crossing over frequency is used to calculate the distance between the two genes.The distance between the two genes is 36 m.u. So, the crossover frequency will be 36% or 0.36 So, the total number of crossover products from the heterozygous individuals (Tf/tF) will be 36% So, the crossover products are- TF (twitching and freckled) = (36÷2) =18% tf (no twitching, no freckled) = (36÷2) = 18% So, the total number of non-crossover products (NCO) are (100-36) = 64% The number of Tf (only twitching) = (64÷2) = 32% and the number of tF (only freckled) = (64÷2) = 32% a) So, the probability of having a child with both twitching and freckled is 18% or 18/100 or 9/50b) If the genes are not linked, then the gametes will be produced in 1:1:1:1 ratio. So, the proportion of all types of gametes will be same. If the total number of offsprings is 25 then each phenotypic classes will be produced in (1/4×25) = 6.25 numbers. The different phenotypic classes are: Only twitching = 6.25 Only freckled = 6.25 Both twitching and freckled = 6.25 No twitching and no freckled = 6.25