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 = 926black snouts+bushy tails = 6small legs + bushy tails = 98black snouts + small legs + black snouts = 198small legs = 10wild type = 214black snouts = 110bushy tails = 10281. 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 24. 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.
female dogs are crossed with male dogs with bushy tails (b), small legs (l), and black snouts (s). there are a total of
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female dogs are crossed with male dogs with bushy tails (b), small legs (l), and black snouts (s). there are a total of
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