Question

1. In humans, red/green color blindness is sex-linked.  The gene for normal vision (N) is dominant to its recessive allele for color blindness (n). A normal vision woman marries a man with red/green color blindness. Both of their fathers are color blind. What % of their sons would have normal vision?

A. 50%

B. 75%

C. 100%

D. 0%

2. The alleles for ABO blood types are designated by I^A, I^B & i  (also written I^o). A woman with type B blood whose parents were both AB marries a man with blood type A blood whose father has type O blood. What is the genotype of the man?

A.

I^Bi

B.

I^AI^A

C. I^Ai

D. I^BI^B

3. The alleles for ABO blood types are designated by I^A , I^B  & i  (also written I^o). A woman with type B blood whose parents were both AB marries a man with blood type A blood whose father has type O blood. What is the genotype of the woman?

A. I^AI^B

^B. I^A I^A

C. I^Bi

D. I^B I^B

4. Give the number of possible genetic combinations in gametes from the genotype SsPpIiDDEeRr.

A. 32

B. 6

C. 10

D. 64

5. What is the primary blue-green pigment that all photosynthetic organisms contain?

A. chlorophyll a

B. chlorophyll b

C. xanthophyll

D. carotene

6. The process of Spermatogenesis results in:

A. 4 sperm

B. 1 ovum and 2 sperm

C. 3 polar bodies and 1 ovum

D. 3 sperm and 2 polar bodies

7. In humans, red/green color blindness is sex-linked.

The gene for normal vision (N) is dominant to its recessive allele for color blindness (n). A normal vision woman marries a normal vision man. Both of their fathers are color blind.

Use this scenario to answer the question below.

What are the genotypes of the man and the woman?

		man = XnY, woman = XNXn
		man = XNY, woman = XNXN
		man = XnY, woman = XnXn
		man = XNY, woman = XNXn

8. In humans, red/green color blindness is sex-linked.

The gene for normal vision (N) is dominant to its recessive allele for color blindness (n). A normal vision woman marries a man with red/green color blindness. Both of their fathers are color blind.

Use this scenario to answer the question below.

What are the genotypes of the man and the woman?

		man = XNY, woman = XNXN
		man = XnY, woman = XnXn
		man = XnY, woman = XNXn
		man = XNY, woman = XNXn

9. This blue roan horse is the offspring of a cross between a Black stallion (B’B’ male) and a White mare” (BB female). Both phenotypes are fully expressed in the heterozygous offspring (B’B), which produces a foal (young horse) that has both black and white hairs. If the roan horse (B'B) is breed with a another roan (B'B) horse what percentage of the offspring would be white in color?

A. 0%

B. 50%

C. 25%

D. 100%

10. The pink carnation (Rr) is the result of a cross between the red (RR) and white (rr) carnation. If you were to cross two pink carnations from the F1 generation, what would be the

expected phenotypic ratio of the F₂ offspring?

A. 2 red, 1 pink, 1 white

B. 3 red and 1 white

C. 1 red, 2 pink, 1 white

D. All offspring are pink

11.

The alleles for ABO blood types are designated by I^A, I^B & i  (also written I^o).

A man with type B blood whose parents were both AB marries a wmona with blood type B blood whose father has type O blood.  Use this scenario to answer the question below.

What is the genotype of the woman?

		IAi
		IBi
		IAIB
		IBIB

Answer 1

1.

The correct option is A - 50%

Parent cross: X^NXⁿ (normal vision woman) * XⁿY (colorblind man)

Gametes of female parent: X^N, Xⁿ

Gametes of male parent: Xⁿ, Y

offspring: X^NXⁿ(normal vision female), X^NY (normal vision male), XⁿXⁿ (colorblind female), XⁿY (colorblind male)

Phenotypic ratio : 1:1:1:1

50% of the daughter will be normal

50% of the daughters will be colorblind

50% of the sons will be normal

50% of the sons will be colorblind

2.

The correct option is C - I^Ai

The man has A blood group, receiving one allele from the father, and one allele from the mother. The father of the man has an O blood group (ii). Since the man will inherit I^O (i) allele from the father, the genotype of man is I^Ai

3.

The correct option is D - I^BI^B

Both the parents of the woman have an AB blood type. So the genotype of both the parents is I^AI^B. The woman has B blood type receiving one I^B allele from the mother and one I^B allele from the father. Hence, the genotype of the woman is I^BI^B

4.

The correct option is A - 32

Number of gametes produced = 2^n

n = number of heterozygous genes

SsPpIiDDEeRr has 5 heterozygous genes

Number of gametes produced = 2^5 = 2 * 2 * 2 * 2 * 2 = 32

5.

The correct option is A - Chlorophyll a

Chlorophyll a is the primary photosynthetic pigment found in all photosynthetic organisms. Chlorophyll a has blue-green color. Chlorophyll a absorbs blue-violet and red light. Chlorophyll a is found in plants, algae, and cyanobacteria

6.

The correct option is A - 4 sperm

Spermatogenesis is the process by which spermatogonia is transformed into sperms. Spertogenes occurs in the male reproductive organ testis. Spermatogenesis starts with puberty. Spermatogenesis has three phases. They are mitosis, meiosis, and spermiogenesis. At the end of spermatogenesis, four mature sperms (haploid) are produced from the diploid immature germ cells  (spermatogonia)

7.

The correct option is d - Man = X^NY, Woman = X^NXⁿ

Males have one X chromosome and one Y chromosome. A man inherits an X chromosome from the mother and Y chromosome from the father. In sex-linked inheritance, There is no male to male transmission. The females have two X chromosomes inheriting one X chromosome from the mother and one X chromosome from the father. The man is normal, so his genotype is X^NY. The father of the woman is colorblind (XⁿY). So the genotype of the woman is X^NXⁿ (carrier), inheriting the allele X^N from the mother and the allele Xⁿ from the father

8.

The correct option is c - Man = XⁿY, Woman = X^NXⁿ

In sex-linked inheritance, there is no male to male transmission. The man is colorblind, so the genotype of man is XⁿY. Since the males have one X chromosome and one copy of the recessive allele is sufficient to cause the disorder in males. The woman has a colorblind father (XⁿY). So the woman inherits the Xⁿ allele from the father. So the genotype of the woman who has a normal vision is X^NXⁿ inheriting the allele X^N  from the mother and the allele Xⁿ from the father

9.

The correct option is C - 25%

The given problem is an example of Codominance. In codominace, both the traits are expressed in heterozygotes

Parent cross: B'B * B'B

Gametes of parent 1 = B', B

Gametes of parent 2 = B', B

Offspring: B'B' (black), B'B (roan), B'B (roan), BB (white)

Genotypic ratio : 1 (B'B') : 2(B'B) :1(BB)

phenotypic ratio : 1 (black) :2 (roan) : 1(white)

Therefore, black = 25%, roan = 50%, white = 25%

10.

The correct option is C - 1 red, 2 pink, 1 white

The given problem is an example of incomplete dominance. In incomplete dominance, neither allele is completely dominant and nor recessive. In incomplete dominance, heterozygous individuals exhibit a third phenotype which is a mixture of two alleles. The cross between homozygous red and homozygous white phenotypes produces a third phenotype pink (heterozygous) which is the mixture of two phenotypes.

Parent cross : RR (red) * rr (white)

F1 offspring: Rr (pink)

F1 cross: Rr (pink)* Rr (pink)

Gametes F1 male = R, r

Gametes of F1 female = R, r

F2 progeny: RR (red), Rr (pink), Rr (pink), rr (white)

F2 Genotypic ratio:1RR: 2Rr: 1rr

phenotypic ratio: 1(red) : 2(pink): 1(white)

11.

The correct option is B - I^Bi

The woman has B blood type. The father of the woman has O blood type (ii). The woman inherits allele I^O (i) from the father. So the genotype of the woman is I^Bi inheriting allele I^B from the mother and allele I^O(i) from the father