female dogs are crossed with male dogs with bushy tails (b), small legs (l), and black snouts (s). there are a total of
Posted: Fri Jul 15, 2022 5:19 pm
female dogs are crossed with male dogs with bushy tails (b), small legs (l), and black snouts (s). there are a total of 3000 offspring:
black snouts+small legs = 926
black snouts+bushy tails = 6
small legs + bushy tails = 98
black snouts + small legs + black snouts = 198
small legs = 10
wild type = 214
black snouts = 110
bushy tails = 1028
1. what gene is in the middle? EXPLAIN how you got this answer by using the given info: the nonrecombinants are (black snouts+small legs) and (bushy tail). the double cross overs are the (black snouts + bushy tails) and (small legs)
2. calculate the distances between the middle gene with the genes on either end. (3 answers total)
3. draw/label the gene with the distances from question 2
4. what are the calculations of the coefficient of coincidence and interference?
5. is there crossover interference? explain.
black snouts+small legs = 926
black snouts+bushy tails = 6
small legs + bushy tails = 98
black snouts + small legs + black snouts = 198
small legs = 10
wild type = 214
black snouts = 110
bushy tails = 1028
1. what gene is in the middle? EXPLAIN how you got this answer by using the given info: the nonrecombinants are (black snouts+small legs) and (bushy tail). the double cross overs are the (black snouts + bushy tails) and (small legs)
2. calculate the distances between the middle gene with the genes on either end. (3 answers total)
3. draw/label the gene with the distances from question 2
4. what are the calculations of the coefficient of coincidence and interference?
5. is there crossover interference? explain.