Question

The following figure represents experimental results from study designed after the Beadle and Tatum experiment.  

Assume the mutants are homozygous for recessive alleles causing their phenotypes.

	Nutrient A	Nutrient B	Nutrient C	Nutrient D
Mutant 1	No	Yes	Yes	No
Mutant 2	Yes	Yes	Yes	No
Mutant 3	Yes	Yes	Yes	Yes
Mutant 4	No	No	Yes	No
Mutant 5	No	Yes	Yes	No

a. Create an ordered pathway of nutrients and mutants (pull down the correct mutant for M1-M4 and the correct nutrient for Nut1-Nut4):

#1. M1

a). Mutant 1

b.) Mutant 2

c.) Mutant 3

d.) Mutant 4

e.) Mutant 5

f.) Mutant 1 and 2

g.) Mutant 2 and 5

h.) Mutant 2 and 3

i.) Mutant 1 and 5

#2. M2

a). Mutant 1

b.) Mutant 2

c.) Mutant 3

d.) Mutant 4

e.) Mutant 5

f.) Mutant 1 and 2

g.) Mutant 2 and 5

h.) Mutant 2 and 3

i.) Mutant 1 and 5

#3. M3

a). Mutant 1

b.) Mutant 2

c.) Mutant 3

d.) Mutant 4

e.) Mutant 5

f.) Mutant 1 and 2

g.) Mutant 2 and 5

h.) Mutant 2 and 3

i.) Mutant 1 and 5

j.) Mutant 3 and 4

k.) Mutant 4 and 5

#4. M4

a). Mutant 1

b.) Mutant 2

c.) Mutant 3

d.) Mutant 4

e.) Mutant 5

f.) Mutant 1 and 2

g.) Mutant 2 and 5

h.) Mutant 2 and 3

i.) Mutant 1 and 5

j.) Mutant 3 and 5

k.) Mutant 1 and 2

l.) Mutant 1 and 3

#5. Nut 1

a.) Nutrient A

b.) Nutrient B

c.) Nutrient C

d.) Nutrient D

#6. Nut 2

a.) Nutrient A

b.) Nutrient B

c.) Nutrient C

d.) Nutrient D

#7 Nut 3

a.) Nutrient A

b.) Nutrient B

c.) Nutrient C

d.) Nutrient D

#8 Nut 4

a.) Nutrient A

b.) Nutrient B

c.) Nutrient C

d.) Nutrient D

b. If you crossed mutants 1 and 5, what would the nutrient requirements of the offspring be?

c. If you crossed mutants 1 and 4, what would the nutrient requirements of the F₁ be?

d. If you conducted an F₁xF₁ cross of mutants 1 and 4, what would the phenotypic ratios of the F₂ be?

Answer 1

The logic behind beadle and tatum experiment is that a mutant that lacks the gene to synthesize nutrient will grow in that nutrient .

The biochemical pathways are sequential

Eg

A,B,C,D are nutrients and 1,2,3,4,5 are the genes.

So now we construct a table helping us to understand experimental result better

	A	B	C	D
1	-	+	+	-
2	+	+	+	-
3	+	+	+	+
4	-	-	+	-
5	-	+	+	-

A,B,C,D are nutrients and 1,2,3,4,5 are the genes.

Logically the nutrient with the least no of mutated genes should be in the beginning of pathway.

When a gene gets knocked up ,the no of mutants increases .So the next nutrient in the pathway will be the one with higher mutations than earlier one and lower mutation than the later.So the nutrients can be arranged in the increasing order of mutants.

in this the order is

D→A→B→C→

Now coming to the position of genes

D→A we see the mutation happened at gene 2

A→B we see the mutation happened at genes 1 and 5

B→C we see the mutation happened at gene 4

and gene 3 is the final resulting mutant(product)

So the results are

a)

Nut 1- D

Nut 2- A

Nut 3- B

Nut 4- C

M1- Mutant 2

M2- Mutant 1 and 5

M3- Mutant 4

M4- Mutant 3

b) If you crossed mutants 1 and 5, what would the nutrient requirements of the offspring be?

Both mutants 1 and 5 are incapable of metabolizing B and C and also if you see in the pathway both the genes 1 and 5 are required for the pathway to proceed from A-B.

When a cross happens between the two mutants both the genes 1 and 5 will be present and the biochemical pathway wil proceed from A-B.

So all the nutrients are synthesized by the offsprings.So nutritional requirement is needed.

c) If you crossed mutants 1 and 4, what would the nutrient requirements of the F1 be?

Parent bbcc *BBcc or Bbcc

F1 BbCc or bbcc

Mutants 1 and 4 both are incapable of metabolizing nutrient C.Also mutant 1 is incapable of metabolizing B but mutant 4 can. So when a cross occurs the resulting offspring(Bb) will inherit the gene for metabolizing B (homozygous recessive is responsible for phenotype ) and offspring (bb) will be deficent in metabolizing B.Both The offsprings  will still be deficent in metabolizing nutrient C.So nutrient C is required for F1 offspring.

d) If you conducted an F1xF1 cross of mutants 1 and 4, what would the phenotypic ratios of the F2 be ?

F1 BbCc x  bbcc

F2 Bbcc,BbCc,bbcc,bb