Genetics-i Miscellaneous Easy Questions and Answers | Page - 1

Genetics-i Miscellaneous

Of the two diploid species, species I has 36 chromosomes and species II has 28 chromosomes. How many chromosomes would be found in an allotriploid individual?

1. 42 or 54
1. 46 or 50
1. 74 or 86
1. 84 or 108

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Given, Species I has 36 (2n = 36) chromosomes and species II has 28 (2n = 28) chromosomes. Now, species I and II are
crossed to obtain an allotriploid individual. That means it has a total chromosome of 3n. When the two diploid species are crossed, the chromosome set of one species would not undergo reduction division for the purpose of obtaining the allotriploid hybrid. That means the 3n would be contributed by 2n and n. So, the 2n can be from species I or II and so the n can be from species II or I respectively. Hence, the required chromosome number would be :
Species I (2n) + Species II (n) = 3n
⇒ 36 + 14 = 50
Or Species I (n) + Species II (2n) = 3n
⇒ 18 + 28 = 46

Correct Option: B

Given, Species I has 36 (2n = 36) chromosomes and species II has 28 (2n = 28) chromosomes. Now, species I and II are
crossed to obtain an allotriploid individual. That means it has a total chromosome of 3n. When the two diploid species are crossed, the chromosome set of one species would not undergo reduction division for the purpose of obtaining the allotriploid hybrid. That means the 3n would be contributed by 2n and n. So, the 2n can be from species I or II and so the n can be from species II or I respectively. Hence, the required chromosome number would be :
Species I (2n) + Species II (n) = 3n
⇒ 36 + 14 = 50
Or Species I (n) + Species II (2n) = 3n
⇒ 18 + 28 = 46

In a genetic cross between the genotypes WWXX and wwxx, the following phenotypic distributions were observed among the F₂ progeny: WX, 562; wx, 158; Wx, 38; and wX, 42. Likewise, a cross between XXYY and xxyy yielded the following results: XY, 675; xy, 175; Xy, 72; and xY, 78. Similarly, a cross between WWYY and wwyy yielded: WY, 292; wy, 88; Wy, 12; and wY, 8. In all the genotypes, capital letters denote the dominant allele. Assume that the F₁ progeny were self-fertilized in all three crosses. Also, double cross-over does not occur in this species. Which of the following is correct?

1. Relative position: W-X-Y Distances: W-X = 5 map units, X-Y = 17 map units
1. Relative position: X-Y-W Distances: X-Y = 15 map units, Y-W = 11 map units
1. Relative position: Y-W-X Distances: Y-W = 5 map units, W-X = 11 map units
1. Relative position: X-W-Y Distances: X-W = 5 map units, W-Y = 10 map units

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In the various genetic crosses, it can be seen that W and Y are more close to each other than X and Y. At the same time, X and W are close to each other while Y is farthest from X. So, the relative positioning of the genes are Y-W-X and y-w-x.
Now, Relative distances of genes with respect to each other are as follows :
Distance of W gene from Y, Y - W = 5 units and W from X, W - X = 11 units

Correct Option: C

In the various genetic crosses, it can be seen that W and Y are more close to each other than X and Y. At the same time, X and W are close to each other while Y is farthest from X. So, the relative positioning of the genes are Y-W-X and y-w-x.
Now, Relative distances of genes with respect to each other are as follows :
Distance of W gene from Y, Y - W = 5 units and W from X, W - X = 11 units

A dioecious plant has XX sexual genotype for female and XY for male. After double fertilization, what would be the genotype of the embryos and endosperms?

1. 100% ovules will have XXX endosperm and XX embryo
1. 100% ovules will have XXY endosperm and XY embryo
1. 50% ovules will have XYY endosperm and XY embryo
1. while other 50% will have XXY endosperm and YY embryo

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Dioecious plants have male (staminate) flowers on one plant, and female (pistillate) flowers on another plant. Some of these plants are polygamo-dioecious, with some male flowers on female plants and some female flowers on male plants. A dioecious plant having XX femal genotype and XY mal genotype undergoes double fertilization. So, on double fertilization, we have :
50% ovules : XXX endosperm and XX embryo And rest 50% ovules : XXY endosperm and XY embryo

Correct Option: D

Dioecious plants have male (staminate) flowers on one plant, and female (pistillate) flowers on another plant. Some of these plants are polygamo-dioecious, with some male flowers on female plants and some female flowers on male plants. A dioecious plant having XX femal genotype and XY mal genotype undergoes double fertilization. So, on double fertilization, we have :
50% ovules : XXX endosperm and XX embryo And rest 50% ovules : XXY endosperm and XY embryo

Consider a population of 10,000 individuals, of which 2500 are homozygotes (PP) and 3000 are heterozygotes (Pp) genotype. The frequency of allele p in the population is ______.

1. 3.671
1. 0.671
1. 5.71
1. None of the above

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We have , according to Hardy -Weinberg's law of population dynamics and equilibrium ,'
P² + 2PQ + Q² = 1
Where , P is the allelic frequency of ' p ' .
Now , Total population = 10000
Therefore, P² = 2500 / 10000 = 0.25
and 2PQ = 3000 / 10000 = 0.3
Therefore, Q² = 1 – P² – 2PQ
= 1 – 0.25 – 0.3 = 0.45
So, Q = 0.671 which is the frequency of the allele p .

Correct Option: B

We have , according to Hardy -Weinberg's law of population dynamics and equilibrium ,'
P² + 2PQ + Q² = 1
Where , P is the allelic frequency of ' p ' .
Now , Total population = 10000
Therefore, P² = 2500 / 10000 = 0.25
and 2PQ = 3000 / 10000 = 0.3
Therefore, Q² = 1 – P² – 2PQ
= 1 – 0.25 – 0.3 = 0.45
So, Q = 0.671 which is the frequency of the allele p .

Hypophosphatemia is manifested by an X-linked dominant allele.What proportion of the offsprings from a normal male and an affected heterozygous female will manifest the disease?

1. ½ sons and ½ daughters
1. all daughters and no sons
1. all sons and no daughters
1. ¼ daughters and ¼ sons

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Given, Hypophosphatemia is caused due to X-linked dominant allele. Now, when a normal male (XY) and a diseased female (X^HX) are crossed together, the F1 generation will have offspring : XX, X^HX, XY and X^HY. So from this it can be seen that half of the male offspring and half of the female offspring bear the diseased allele.

Correct Option: A

Given, Hypophosphatemia is caused due to X-linked dominant allele. Now, when a normal male (XY) and a diseased female (X^HX) are crossed together, the F1 generation will have offspring : XX, X^HX, XY and X^HY. So from this it can be seen that half of the male offspring and half of the female offspring bear the diseased allele.