Question 1
Choose the off one out in case of duplicate recessive episasis.
A. A/A b/b
B. A/A B/B
C. A/a B/b
D. A/a B/-
View Answer
Answer: Option A
Explanation:
All other remaining locus will express the dominant phenotype as they have atleast one dominant allele in each locus, but the A/A b/b is recessive for B locus and will express recessive character.
Question 2
Flower colour in sweet pea shows duplicate epistasis, if purple colour is dominant what is the F2 ratio?
A. 15:1
B. 9:7
C. 3:1
D. 9:3:4
View Answer
Answer: Option B
Explanation:
The F2 ratio in recessive duplicate epistasis as seen in case of sweet pea flower colour is 9:7. Where the ratio corresponds to purple(9): white(7) respectively.
Question 3
If you cross two different strains of white flower you may get a purple flower in a sweet pea plant.
A. True
B. False
View Answer
Answer: Option A
Explanation:
In this case, the purple sweet pea flower due to duplicate epistasis needs atleast one dominant allele at each locus. Two white flowers may have homozygosity and heterozygosity at different locus and finally would give a purple flower.
Question 4
If you self cross purple flower in sweet pea, you will get__________
A. Purple flower in every occasion
B. White flower in every occasion
C. May be white or purple flower
D. A blend of white and purple flower
View Answer
Answer: Option C
Explanation:
If the purple flower is homozygous in both loci then we will get only purple flowers, but heterozygosity in both loci would give a 9:7 ratio of purple to white flower in the F1.
Question 5
In sweet pea a white substance is first converted to a ___________ substance by 1st gene and that to a ______________ substance by 2nd gene to give a purple flower.
A. Yellow, purple
B. Green, purple
C. Purple, yellow
D. White, yellow
View Answer
Answer: Option D
Explanation:
It is expected that in case of duplicate recessive epistasis of sweet pea plant a white substance is 1st converted to another white substance that is converted to purple substance in case of presence of two dominant genes. If any of these genes are absent this would give white flower.
Question 6
Self crossing a white flower of sweet pea would give a ____________________
A. White flower only
B. Equal probability of white or purple
C. More white flower than purple
D. Purple flowers only
View Answer
Answer: Option A
Explanation:
The white flower arises due to a homozygous recessive condition in atleast one of the two gene locus. In this case self crossing can never yield a combination of heterozygous or homozygous dominant at that gene locus, so it will only give white flowers.
Question 7
Shepherd’s purse has two shape heart and narrow. If you cross a true breeding heart to a true breeding narrow, F1 will give___________
A. Heart
B. Narrow
C. A mixture of heart or narrow
D. Disc shaped
View Answer
Answer: Option A
Explanation:
A heart shaped Shepherd’s purse is dominant over the narrow shaped one. Thus crossing the true breeding form of each will give a heterozygous condition in both the locus resulting in heart phenotype. Disc is a phenotype for summer squash gene interaction.
Question 8
The 15/16 of the duplicate dominant epistasis doesn’t correspond to _____________
A. A/a B/b
B. A/A b/b
C. a/a B/b
D. a/a b/b
View Answer
Answer: Option D
Explanation:
A recessive phenotype in both the locus will result in the recessive gene expression forming 1/15 of the duplicate epistasis ratio.
Question 9
What is the phenotypic ratio in case of duplicate dominant epistasis?
A. 9:3:4
B. 12:3:4
C. 15:1
D. 9:7
View Answer
Answer: Option C
Explanation:
In case of duplicate dominant epistasis the normal 9:3:3:1 ratio is modified to give the ratio of 15:1. In this case the 1/16 represents homozygous recessive condition.
Question 10
Which of the following condition in recessive duplicate epistasis will give the odd phenotype?
A. A/- B/-
B. A/- b/b
C. a/a B/-
D. a/a b/b
View Answer
Answer: Option A
Explanation:
Only when atleast one dominant allele is present in both the locus we get the expression of dominant phenotype. Otherwise the resulting cases give only recessive phenotype.
Question 11
You crossed two heart shaped Shepherd’s purse which is heterozygous at one locus and homozygous recessive at other gene locus, what would be the ratio of heart shaped and narrow in F1?
A. All heart
B. All narrow
C. 3:1
D. 15:1
View Answer
Answer: Option A
Explanation:
As the parent is homozygous recessive at one locus and heterozygous at the other, the combinations that are yield would give 3:1 ratio of heart to narrow.