Newer methods of staining chromosomes, such as chromosome painting, has advanced
ID: 102632 • Letter: N
Question
Newer methods of staining chromosomes, such as chromosome painting, has advanced genetic research tremendously. This is because it allows for: a. Specificity b. Wide array of color variety c. Distinguish bands from one another d. Allows structure abnormalities to be detected If two genes are completely unlinked, what is their frequency of recombination? a. 0% b. 75% c. 50% d. 25% At what stage of meiosis does crossing over occur? a. metaphase I b anaphase I c telophase I d prophase I During a dihybrid cross involving 2 linked genes, 14 % of the resulting gametes showed a recombinant genotype. These two linked genes are ______ mapping units apart. a. 28 b. 1.4 c. 56 d. 14 When analyzing the progeny of a test-cross. The non-recombinants phenotype will have the ____ progeny: the ______ recombinants will have the least a. least, single b. most double c. most, single d. least, double In a crossing-over event where three genes are being analyzed, the coefficient of equal to 0.01. What assumption can be made a. The interference between one crossing over event is affecting another b. There is no interference going on c. The genes are not located on the same gene d. The genes are located less than 20cm from each otherExplanation / Answer
20. Chromosome painting refers to the hybridization of fluorescently labeled chromosome-specific probe to cytological preparations. With the help of this technique the individual chromosome in the metaphase and anaphase can be visualized and both the structural and numerical chromosomal aberrations can be detected with high sensitivity.
Until now the chromosome banding analyses was used in detection of aberrations in cytogenetic screening tests with the use material like Giemsa. But with the use of FISH( fluorescence in situ hybridization) the it has become easy to detect the reciprocal translocations.
The amplified DNA fragments are labeled with appropriate fluorescent dyes and attached to the metaphase chromosomes.
The fluorescent labeled DNAs will attach to the analogous chromosome.
If a part of painted chromosome undergoes an exchange with non-painted chromosome it is possible to detect the aberration.
Therefore the option is D.
21. The genes found on different chromosomes or are very far from one another on the same chromosome independent assortment will take place and the genes are said to be unlinked.
On the other hand if the genes are linked they are placed very close to each other on the same chromosome.
The Independent assortment if viewed with regard to the recombination then in the test cross the recombinants will always make the 50 % of the progeny that means 25% of each recombinant type in the total progeny.
Therefore the option is C.
22. Meiosis 1 is also called as the reductional division as the chromosome number becomes half in this phase. It is divided into 4 stages, Prophase I, Metaphase I, Anaphase I, TelophaseI.
Prophase I- The longest phase in which the chromosome condense and crossing over takes place between the non-sister chromatids. It is distinguished into 5 stages:
Leptotene- The chromosome condense and coil.
Zygotene- There is synapsis formation between the homologous chromosome pair.
Pachytene- Crossing Over takes place and it is marked by chiasma formation.
Diplotene-The synaptonemal complex breaks down.
Diakinesis- The final stage in which the homologous chromosome move apart and terminalization occurs.
Therefore the option is D.
23. The Recombination frequency is proportional to the distance between the genes upto some point or in proper range.
1% of recombination frequency corresponds to map unit and one map unit is also known as one centiMorgan or 1 cM.
Therefore if the RF between two genes say gene A and gene B is 14% the distance will be mu or cM.
Therefore the option is D.
24. In analyzing the progeny of test cross, If we compare the parental with the double cross overs it will be found that the non-recombinants are most frequent and the double crossovers are least frequent.
Therefore the option is B.
25. If cc is then interference is 0.
If cc is in positive value the interference is in negative value
And if cc is 0 then interference is 1.
Therefore the option is B.