Question

12. Theoretically, what is the expected result from the parental cross of a grey mouse and an albino mouse, and why does the expected result not match the result stated in the information given above?

a)Theoretically, 100% if the mice should have grey coats so all 38 offspring should have grey coats. There was an experimental error that can account for the deviation.

b)Theoretically, there should be a 3:1 ratio of grey mice to albino mice, or 29 grey coat mice and 9 albino. With more litters, the actual results would match the expected results more closely.

c)Theoretically, there should be a 1:1 ratio of grey mice to albino mice, or 19 grey coat mice and 19 albino mice. There was an experimental error that can account for the deviation

d)Theoretically, there should be a 3:1 ratio of grey mice to albino mice, or 28.5 grey coat mice and 9.5 albino mice. There was an experimental error that can account for the deviation.

26. A male and female have three children, all girls. The couple is expecting their fourth child. The probability their fourth child will be female is __i__, and the probability the couple will have four female children is __ii__.

The row that indicates the correct probabilities is

a)

i	ii
1/16	1/4

b)

i	ii
3/4	1/4

c)

i	ii
1/2	1/16

d)

i	ii
1/4	1/8

Use the following information to answer question 42.

Below is a simplified chart for polygenic trait if eye color. The amount of melanin in the iris is what determines the color of the eye. People with brown eyes have more melanin in their iris than people with blue eyes. The amount if melanin us controlled by how many dominant alleles a person has for eye color as follows: Number of Dominant Alleles Genotypes Phenotypes 4 AABB very dark brown eyes 3 AaBB or AABb brown eyes 2 AaBb, aaBB etc. light brown or hazel eyes 1 Aabb or aaBb dark blue eyes 0 aabb light blue eyes

42. Which of the following parental crosses explains how two brown-eyed parents can produce a blue-eyed chilf?

a) AaBB x AaBb

b) AaBB x AABb

c) Aabb x aaBB

d)AABB x AaBb

37. The coat of dogs is an example of an epistatic interaction. The common alleles are black (B) and brown (b). There is a separate gene on a separate chromosome that influences coat color; it prevents color information and causes a white coat.

L=prevents color formation and causes a white coat

l=allows color formation to occur

If the P1 generation is a white dog (LlBb) crossed with a black dog (IIBb), then which of the following represents the correct phenotypes of the F1 offsprings?

a)

Phenotypes	Phenotypes	Phenotypes
White	Black	Brown
3/8	1/2	1/8

b)

Phenotypes	Phenotypes	Phenotypes
White	Black	Brown
1/8	1/2	3/8

c)

Phenotypes	Phenotypes	Phenotypes
White	Black	Brown
1/2	3/8	1/8

d)

Phenotypes	Phenotypes	Phenotypes
White	Black	Brown
1/2	1/8	3/8

Answer 1

12) When a grey mouse is crossed with an albino mouse, option (b) is correct i.e. Theoretically there should be a ratio of 3:1 ratio of grey mice to albino mice (or 29:9 in number). There is deviation because here the cross is done in between very small population. With more litters the actual result would match with the expected result more closely.

26) The probability of their 4th child to be a female is 1/2. Whatever may be the number of child, the probability of being a male or female is always 50% or 1/2. Because it depends on the father that which sperm of him fertilizes the egg of the mother, as the father has heterozygosity i.e. one X and one Y chomosome in his germ cell.

And the probability the couple will have 4 female children is 1/16. Because as every time the possibility is 1/2, so for 4 children the probability will be (1/2 to the power 4) i.e. 1/16.

42) Here both the crosses (a) and (c) can produce blue-eyed child. But as per the question the cross should be between two brown eyed parents only. And in option (c) has one brown eyed and one blue eyed parent. So the correct answer here is OPTION (a). Let me explain how it works.

So in option (a) the gamets produced by parent AaBB are AB & aB. Similarly the gamets produced by parent AaBb are AB, Ab, aB & ab. Now cossing them in each case we will gets the childen having genotypes, AABB, AABb, AaBB, AaBb, AaAB, AaBb, aaBB, aaBb. As we can see here, the child with genotype "aaBb" will be blue-eyed. Hence the option (a) is correct.

37) Here the white dog (LlBb) will produce gamets LB, Lb, lB & lb. And the black dog llBb will produce gametes two lB and two lb. Now cossing ecah of them with the other we will have offsprings with genotypes LlBB (White), LlBb (White), llBB (Black), llBb (Black), LlBb (White), Llbb (White), llBb (Black) and llbb (Brown). So the ratio of White:Black:Brown dogs are 4/8: 3/8 : 1/8 i.e OPTION (C) 1/2: 3/8: 1/8 is correct.