Can someone please explain how they got the answers in a clear and simple way. 2

Question

Can someone please explain how they got the answers in a clear and simple way.

27. (18 points) The following pathway produces pigments in butterfly wings, a diploid organism. The 5 genes for these enzymes are all on different chromosomes. Show your work. 4 Orange Red White Yellow 2 Blue Enzyme #1 converts a white precursor to a yellow pigment. Enzyme #2 converts yellow to blue pigment, and enzyme #3 converts yellow to orange pigment. Both enzymes #4 and #5 can convert orange pigment into red pigment. Wild-type butterflies have functional copies of all 5 enzymes, make both red and blue pigments, and so have a purple color. Butterflies that make blue pigment plus orange pigment (but not red) are brown a) You have butterflies each with loss-of-function mutations in two genes: a 1-1- 2-2- butterfly and a 2-2- 3-3-butterfly. Each is homozygous wild-type for all other genes. You cross them to make F1 progeny, then cross F1 siblings to make F2 progeny. What colors do you expect among F1 and F2 generations? red (2 pts) Color of F1 generation individuals: Fraction of F2 generation of each color (if 0, leave blank): White: 4/16 Purple Yellow: 3/16 Orange: Red: 9/16 Blue: Brown: b) You cross a true-breeding blue butterfly to a true-breeding orange butterfly. The blue butterfly is homozygous recessive for only one gene. You cross them to make F1 progeny, then cross F.1 siblings to make F2 progeny. What colors do you expect among F1 and F2 generations? Color of F1 generation individuals: purple(2 pts) Fraction of F2 generation of each color (if 0, leave blank): White: - Yellow: 16/256 (1/16 Orange: 3/256 Red: 45/256 Blue: 48/256 (-3/16) Purple: 135/256 Brown: 9/256

Explanation / Answer

For understanding purpose let us consider the enzyme-coding genes 1, 2, 3, 4, 5 as A, B, C, D and E

a) In the cross between two mutants for first 3 genes, 1-1-2-2  X 2-2-3-3 can be considered as

aabbCCDDEE X AAbbccDDEE

F1 generation= AabbCcDDEE (here enzyme 1 and enzyme 3 mutations are complimented due to the other gamete)

=The butterfly in F1 generation able to produce all enzymes except enzyme 2(represented here by B).

So F1 butterfly will be in red colour.

If F1 are intercrossed to produce F2.

AabbCcDDEE x AabbCcDDEE

gametes= AbCDE, abCDE, AbcDE, abcDE ( here bDE are common in all gametes)

AACCbbDDEE

Red

AACcbbDDEE

Red

AaCCbbDDEE

Red

AaCCbbDDEE

Red

AACcbbDDEE

Red

AAccbbDDEE

Yellow

AaCcbbDDEE

Red

AaccbbDDEE

Yellow

AaCcbbDDEE

Red

AaCcbbDDEE

Red

aaCCbbDDEE

White

aaCcbbDDEE

White

AaCcbbDDEE

Red

AaccbbDDEE

Yellow

aaCcbbDDEE

White

aaccbbDDEE

White

9/16= red, Yellow = 3/16, White = 4/16

b) If we cross true breeding blue butter fly with orange

AABBccDDEE (blue homozygous for one gene) X AAbbCCddee (orange)

F1= AABbCcDdEe (purple colour)

If F1 intercrossed to produce F2

AABbCcDdEe X AABbCcDdEe

the proportion of different coloured butterflies in F2

white - No white butterfly possible because of the AA genotype in both F1

Yellow= produced by AAbbccddee genotype only so= It is in 16/256 proportion

Explanation: In tetrahybrid cross, it is tedious to build a punnet square board for evaluating all phenotypes or genotypes. It can be simply solved by multiplying the monohybrid ratios of the different locus to give a phenotipic ratio of tetrahybrid.

Yellow colour can be produced by phenotype

AAbbccddee offspring= so if we multiply all monohybrid ratios then

1X 1/4X 1/4x 1/4x11/4=1/256

Like wise

Purple= AAB-C-D-E-= 1x 3/4x 3/4x3/4x3/4=81/256

Yellow= AAbbccddee= 1X 1/4 X1/4x 1/4x 1/4= 1/256

orange= AAbbC-ddee=1 X 1/4X3/4x1/4x 1/4=3/256

Red is produced by 4 or 5 enzymes

so we can add all possiblilities

AAbbC-D-E- + AAbbC-D-ee + AAbbC-ddE- = 45/256

AAbbC-D-E- = 1x 1/4x 3/4x 3/4 x3/4=27/256

AAbbC-D-ee= 1x 1/4x 3/4x 3/4x1/4=9/256

AAbbC-ddE-= 1x 1/4x 3/4x 1/4x 3/4= 9/256

Blue= Blue can be produced by

AAB-ccD-E-+ AAB-ccD-ee + AAB-ccddE-+AAB-ccddee= 48/256

AAB-ccD-E-= 1x 3/4x1/4x3/4x3/4=27/256

AAB-ccD-ee =1x 3/4x 1/4x 3/4x 1/4=9/256

AAB-ccddE-= 1x3/4x1/4x1/4x3/4=9/256

AAB-ccddee=1x 3/4x1/4x1/4x1/4=3/256

Brown can be produced by

AAB-C-ddee= 1x3/4x 3/4x 1/4x1/4=9/256.

ACbDE AcbDE aCbDE acbDE ACbDE

AACCbbDDEE

Red

AACcbbDDEE

Red

AaCCbbDDEE

Red

AaCCbbDDEE

Red

AcbDE

AACcbbDDEE

Red

AAccbbDDEE

Yellow

AaCcbbDDEE

Red

AaccbbDDEE

Yellow

aCbDE

AaCcbbDDEE

Red

AaCcbbDDEE

Red

aaCCbbDDEE

White

aaCcbbDDEE

White

acbDE

AaCcbbDDEE

Red

AaccbbDDEE

Yellow

aaCcbbDDEE

White

aaccbbDDEE

White

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