1) Insertions and deletions of single nucleotides: A) cause frameshift mutations
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Question
1) Insertions and deletions of single nucleotides:
A) cause frameshift mutations.
B) cause cancer.
C) shorten chromosomes.
D) cause missense mutations.
2) In genetics, the dash symbol (–) is a “wild card” that stands for either the dominant allele or the recessive allele; for example, R– means the individual has either the genotype RR or Rr. In Duroc-Jersey pigs, genotypes R– S– have a red coat color, R– ss and rr S– have sandy-colored coats, and rr ss pigs are white. R and S show independent assortment. What is the ratio of red : sandy : white among progeny of the cross Rr Ss × Rr Ss?
A) 9:6:1
B) 10:3:3
C) 12:3:1
D) 9:3:1
3) Imagine that blending inheritance was true, and black and white rabbits mated as in the example in the book. If the offspring show only half the intensity of black pigment after one generation, how many generations would be required for them to show 1/16th the intensity of black pigment?
A) 4
B) 5
C) 7
D) 6
4) In the F2 generation of a homozygous round (AA) × homozygous wrinkled (aa) cross in peas, two round seeds are chosen at random. What is the probability that one is AA and the other Aa? The key here is the stipulation that the chosen seeds are round (i.e., AA or Aa).
A) 2(2/3)(1/3)
B) (1/3)^2
C) (2/3)^2
Explanation / Answer
1. In insertion there is addition of a piece of DNA which changes the number of DNA bases in the gene. As a result the resulting protein from the gene may not function properly.
In deletion the number of DNA bases will change by removal of a piece of DNA. Deletions are small when few base pairs within the gene are deleted or sometimes the deletion is large if the entire gene or several neighboring genes are deleted. This results in the production of dysfunctional protein.
The insertions and deletions change the reading frame of a gene. This is called Frameshift mutation. This mutation actually shifts the frame of the three bases coding for amino acids. Thus the code is changed and the protein produced generally is nonfunctional. Therefore insertions, deletions and duplications can all cause the frameshift mutation.
The option is A.
2. The Law of Independent Assortment is the basic principle of genetics developed by Gregor Mendel. This principle states that the allele for a trait separate independently during the formation of gametes.
Mendel formulated this principle by performing dihybrid cross in pea plants which were true-breeding for two traits.
In the cross given:-
RrSs * RrSs
The genotype and the phenotype of the F2 individuals will be as under-
RRSS, RRSs, RrSS and RrSs - 9/16 individuals will have phenotype red coat colour
RRyy and Rryy - 3/16
rrYY and rrYy - 3/16
rryy - 1/16 individual will have white coloured coat.
Therefore the ratio is 9:6:1 and correct option is A.
3. In 19th century prior to the development of genetics the theory of blending inheritance was most studied and followed. It is a hypothetical model and not a scientific theory. According to this model the inherited traits are determined randomly between the range of homologues traits found in parents.
This theory has the drawback as it offers the uniform blending of the characters found in parents in the offspring. For example height of a person having one tall and one short parent was thought to be intermediate between the two parents. This theory also fails to explain how the traits which are thought to be disappeared down the lineage reasserted itself in some generations. If blending inheritance was true there will be no variation and individual will appear as clones.
Following the blending inheritance theory the pigment will be 1/ 16th of original black pigment after 4 generations. Therefore the option is A.
4. The cross between homozygous round seeds and homozygous wrinkled seeds in pea plant;
AA * aa
In F1 generation all offspring will have genotype Aa and thus round seeds.
In F2 generation when F 1 individuals are self-pollinated,
Aa * Aa
Three possible genotypes can result- AA, Aa and aa from the above cross. The AA and Aa will produce round seed phenotype and aa will produce wrinkled seed phenotype. Thus the probability of wrinkled seed is ¼ or .25%.
Of the round seed phenotype two Aa genotype is there and one AA genotype is there. Therefore the probability of round seed phenotype will be 2/3 and 1/3. In the example two seeds are randomly selected therefore the probability of them to be round seeded is 2(2/3)(1/3).
Therefore the option is A.
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