Genetics MCQ 2


21 Questions MCQ Test Mock Test Series of IIT JAM Biotechnology | Genetics MCQ 2


Description
This mock test of Genetics MCQ 2 for IIT JAM helps you for every IIT JAM entrance exam. This contains 21 Multiple Choice Questions for IIT JAM Genetics MCQ 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Genetics MCQ 2 quiz give you a good mix of easy questions and tough questions. IIT JAM students definitely take this Genetics MCQ 2 exercise for a better result in the exam. You can find other Genetics MCQ 2 extra questions, long questions & short questions for IIT JAM on EduRev as well by searching above.
QUESTION: 1

Suppose a mouse with a black coat breeds with an albino mouse and all of their offspring have a grey coat phenotype. What then does the gene for coat color in mice appear to be an example of?

Solution:

Since the genes for coat color do not show a dominant-recessive relationship and one gene is incompletely dominant over the other gene, they show a blended progeny. Thus, it is an example of incomplete dominance.

QUESTION: 2

What fraction of the offspring will express one of the two dominant alleles, but not both in a cross of AaBb x AaBb?

Solution:

When the genotypes of all the progeny is listed, 6 out of 16 possible genotypes have either of the dominant alleles.

QUESTION: 3

Which of the following is true for two genes undergoing independent assortment?

Solution:

Only those genes which are located on different chromosomes undergo independent assortment. Those genes which are located on the same chromosome are said to represent a linkage group and are inherited together. They experience independent assortment only if they are located far apart and crossing over can occur between them.

QUESTION: 4

Color blindness is an X-linked recessive trait. If a carrier mother and a color blind father have two children, then what is the probability that both the children will be male and both of them would be color blind?

Solution:

he probability of having a male child is ½ . The probability of a male child being colorblind is also ½. Since the child has to be a male and color blind at the same time, we apply the product rule. Since we need to find out the probability of both the children being male and colorblind, So, 1/2*1/2*1/2*1/2 = 1/16 or 6.25%.

QUESTION: 5

Which inheritance pattern does the following trait represents and the possible genotype of the father in the 2nd generation in the given pedigree chart?

Solution:

The trait skips a generation and both males and females are affected by it. So it is most likely an autosomal recessive trait. Since the mother in the 2nd generation is affected by the trait, she would have the genotype aa. One of their children is also affected, so the father must be heterozygous for the trait i.e. Aa.

QUESTION: 6

The brown allele for hair color (B) is dominant over the blond allele (b). If two brown-haired parents produce a child that is blonde haired, what is the approximate probability that at least one out of the next three children they produce will also have blonde hair?

Solution:

According to the question
B → Brown hair
b → bloude hair
The two mating partners are heterozygous

∴ Probability for a brown haired progeny (p) = 3/4 
Probability for a blonde haired progeny (q) = 1/4
According to the question, at least one of the next three progenies have blonde hair. Therefore, following combinations of binomials will satisfy the condition-
(1) Only one progeny of the three has blonde hair
(2) Two of the three progenies have blonde hair
(3) All of them have blonde hair
Using binomial expression. the Probabilities of each possible combination can be calculated as
(1) n = 3 = Total progenies
x = 1 = No. of progenies with blonde hair
y = 2 = No. of progenies with blonde hair
p= 1/4 = Probability of having blonde hair
q = 3/4 = Probability of having brown hair 




QUESTION: 7

Which of the following is not a true for an ideal population under the Hardy-Weinberg Equilibrium? 

Solution:

A large population size is necessary to accommodate rare phenotypes in the population which would otherwise go amiss.

QUESTION: 8

A plant of the genotype AaBb is selfed. The two genes are linked and are 50 map units apart. What proportion of the progeny will have the genotype aabb?

Solution:

Given that the genes A and B are linked by 50 map unit apart. So the frequency of recombinant is 50%. That is crossing the AaBb with aabb will results 50% recombinants and 50% parental gametes (AB : aB : Ab : ab : : 1 : 1 : 1 : 1). Thus the genes will seggregate by independent assortment as proposed by Mendel.
According to Mendelian dihybrid cross selfing of AaBb will result in aabbprogeneis in the ratio of 1/16.

QUESTION: 9

Humans with Phenylketonuria develop light colored hair because of reduced hair pigmentation. This condition is an example of : 

Solution:

Phenylketonuria is an example of plieotropy in which mutation in a gene can cause multiple unrelated phenotypic symptoms and reduced hair pigmentation is one of them. 

QUESTION: 10

Crossing two Drosophila flies with two different mutations yields almost all progeny with wild type traits. With this information, we can conclude that: 

Solution:

The genes complement each other when they are present on two different chromosomes and produce a wild type product. This can be seen using a complementation test. 

QUESTION: 11

A black Labrador homozygous for the dominant alleles (BBEE) is crossed with a yellow Labrador homozygous for the recessive alleles (bbee). On inter-crossing the F1, the F2 progeny was obtained in the following ratio, 9 Black : 3 Brown : 4 Yellow.This is an example of   

Solution:

Recessive epistasis (or supplementary genes interaction) is the condition when homozygous recessive of one gene mask the effect of other gene in the dominant or recessive condition. Presence of B allele is dominant against E or e, results in black and E produces brown in the absence of B allele, whereas recessive homozygous ‘ee’ is epistatic to B or b and thus, gives yellow. The cross can be explained as follows   

QUESTION: 12

On basis of statements given below the mode of inheritance is 
• Mostly males are sufferer of disease
• All male child developed from affected mother are diseased
• Female develope disease only when her father is diseased and mother is carrier

Solution:

Based on the given statements the inheritance of the trait can be explained as X-linked recessive. This is because.
1. Males are heterozygous X-chromosome (XY). So X linked trait usually shows in males due to single X- chromosome.
2. Affected mother (XaXa) gives all sons affected (XaY) and daughter as carrier(XaX).
Female progenies develop disease when the father is affected (XaY) and mother is earner (XaX) for the allele.


3. Males are heterozygous X-chromosome (XY). So X linked trait usually shows in males due to single X- chromosome.
4. Affected mother (XaXa) gives all sons affected (XaY) and daughter as earner (XaX).
Female progenies develop disease when the father is affected (XaY) and mother is earner (XaX) for the allele.


*Multiple options can be correct
QUESTION: 13

Which of the following can be said about continuous traits? 

Solution:

Mendel’s laws do not apply to these traits as several genes are responsible for a single trait and it is impossible to track each on them on the basis of Mendel’s principles. These genes may or may not necessarily be present on the same chromosome. The genotype as well as the environment influence these traits. 

*Multiple options can be correct
QUESTION: 14

Which of the following is true for a sex- influenced trait? 

Solution:

The expression of a phenotype is limited to one sex in a sex limited trait but is not true for  a sex-influenced trait.

*Multiple options can be correct
QUESTION: 15

Which of the following is correct for a species that has a chromosome number 2n = 20?

Solution:

The species is a diploid and the gametes would be haploid and contain n number of chromosomes i.e. 10. The somatic cells would contain 2n number of chromosomes i.e 20. Only 10 different types of chromosomes are present in the species. 

*Multiple options can be correct
QUESTION: 16

Which of the following are true for Karyotyping?

Solution:

The counting and identification of chromosomes in the karyotype of an individual can diagnose aneuploidies, diseases caused by an alteration in the number of chromosomes in relation to the normal number in the species. It does not detect diseases caused due to mutations.

*Multiple options can be correct
QUESTION: 17

Which of the following is  not true for Klinefelter’s Syndrome?

Solution:

Males are affected in Klinefelter’s syndrome. They have 2 X and 1 Y chromosome and are sterile. 

*Answer can only contain numeric values
QUESTION: 18

Two heterozygous individuals having the genotype AaBb are crossed with each other. The probability of having a baby with genotype AaBb is _______ (Answer upto two decimal points)


Solution:

When AaBb and AaBb are crossed with each other, the probability of getting a genotype Aa will be ½. Same way the probability of getting genotype Bb will be ½.
Using the product rule, to get an individual with both Aa and Bb, we will multiply the individual probabilities. This gives us ¼ as the probability of getting a baby with genotype AaBb.

*Answer can only contain numeric values
QUESTION: 19

The probability of having a hemophilic baby if the mother is the carrier and the father is unaffected by this X-linked recessive disease will be _______ (Answer upto two decimal points)


Solution:

The mother will have genotype XX’ and father will be XY. The genotypes of their babies could be XX, XY, XX’ and X’Y. Only the male child will be affected. Therefore, the probability is ¼ or 25%.

*Answer can only contain numeric values
QUESTION: 20

In a flock of sheep, 9% of the population has black wool and 91% has white wool. The percent of  population is heterozygous for this trait if black wool is recessive and the population is in Hardy Weinberg Equilibrium will be ________% (Answer in integer)


Solution:

q2 = 0.09, then q = 0.3
p = 1-q = 0.7
heterozygous population = 2pq = 2*0.7*0.3 = 42%

*Answer can only contain numeric values
QUESTION: 21

Albinism is a recessive condition in humans. If 85% of a population is albino, then the allele frequency of the dominant allele will be (Answer upto three decimal points) __________


Solution:

If q2 = 0.85
Then q = “0.85 = 0.921
p = 1 – q = 1 – 0.921 = 0.079

Related tests