Question

In: Biology

An alien species was found with the following characteristics Flat feet dominant over round feet A...

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?

Solutions

Expert Solution

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

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...........................

gladiator answered 1 year ago

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