In: Biology
An alien species was found with the following characteristics Flat feet dominant over round feet A vs a 10 toes dominant to 15 toes B vs b Curved ears dominant to straight ears C vs c A flat footed, curved eared alien with 10 toes was crossed with a round footed, 15 toed, straight eared individual and produced the following offspring
Flat, curved, 10 toes 588
Round, straight, 15 toes 599
Flat, 15 toes, straight 165
Flat 15 toes, curved 159
Round, 10 toes, curved 145
Round, 10 toes, straight 163
Flat, straight, 10 toes 15
Round, curved, 10 toes 23
What gene is in the middle?
How far apart are the genes?
Dominant allele A = Flat feet
Recessive allele a = Round feet
Dominant Allele B = 10 toes
Recessive allele b = 15 toes
Dominant allele C = Curved ears
Recessive allele c = Straight ears
a)
Two frequencies of offspring phenotypes in highest number are parental types and two frequencies of offspring phenotypes in the lowest number are double cross overs
ABC (Flat, 10 toes, curved) = 588 (Parental type - No cross over)
abc (Round, 15 toes, straight) = 599 (Parental type - NCO)
ABc (Flat, 10 toes, straight) = 15 (double cross overs)
abC (Round, 15 toes, curved ) = 23 (Double cross over)
When parental types compared with double cross over,
ABC (parental type) when compared to ABc (DCO), the uncommon gene is C/c
abc (parental) when compared to abC (DCO), the uncommon gene is C/c
The uncommon gen comes in the middle, So the gene C comes in the middle
The order of genes on the chromosome is
A.....C....B
b).
Parents: ACB/acb * acb/acb
Offspring :
ACB (Parental type (NCO)) = 588
acb (parental type (NCO)) = 599
aCB (Single cross over between genes A and C) = 145
Acb (Single cross over between genes A and C) = 165
ACb (Single cross over between genes C and B) = 159
acB (Single cross over between genes C and B) = 163
AcB (Double cross over) =15
aCb (Double cross over) = 23
Total progeny = 1857
Map distance between two genes = (SCO between two loci + DCO)/ Total progeny ) * 100
Map distance between A and C = ((145 + 165 + 15 +23 )/1857 ) * 100
= (348/1857) * 100 = 0.1874 * 100 = 18.74m.u
Map distance between C and B = ((159 + 163 + 15 +23)/ 1857) * 100
= (360/1857) * 100 = 0.1939 *100 = 19.39 m.u
Therefore, the distance between genes are
A.......18.74 m.u.....C................19.39..............B
..............................38.13 m.u...........................