Question

For all questions, assume that you have genes for beak color, tail-feather length, and feather color all linked (located) on the same chromosome, but you don't know the order of the genes on the chromosome or the distance between them. The alleles for these genes are beak color: Y for yellow beak (YY and Yy) and y for black beak (yy) tail-feather length: L for long tail (LL and Ll) and l for short tail (ll) feather color: B for blue feathers (BB and Bb) and b for white feathers (bb) Total score: ____ of 22 points Part 1: Crossing Beak Color and Tail-Feather Length Cross a YyLl (heterozygous parent with dominant traits) with yyll (homozygous parent with recessive traits). Look at the number of genotypes of the F1 generation: YyLl: 400 Yyll: 100 yyLl: 100 yyll: 400 (Score for Question 1: ___ of 2 points) Which offspring are the recombinant offspring in this cross? Answer: Type your answer here. (Score for Question 2: ___ of 2 points) How far apart are Y and L? Give your answer in map units. (Hint: Add the numbers of the two recombinant types, divide by the total number of offspring, and multiply by 100.) Answer: Type your answer here.

Answer 1

1. If we cross YyLl (heterozygous parent with dominant traits) with yyll (homozygous parent with recessive traits)

P:  YyLl  x  yyll

F1 generation: YyLl: 400 Yyll: 100 yyLl: 100 yyll: 400

Recombinant offspring are those children whose genes contain a non-parental allele combination  (neither allele group is directly inherited from either parent). This happens when genes are located on the same chromosome but are so far apart from one another so that their alleles get crossed over during meiosis. In this case, Yyll and yyLl are the recombinant.

2. Calculation of distance between Y and L.  

The numbers of the two recombinant types is 200 (100 Yyll + 100 yyLl) and 800 parental offspring.

Total number of offspring is 1000.

So: 200/1000*100=20 map units.

There is 20 percent recombinant offspring frequency.