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#Question id: 339


The edges of each base pair are exposed in the major and minor grooves, creating a pattern of hydrogen-bond donors and acceptors and of hydrophobic groups (allowing for Vander Waals interactions) that identifies the base pair. The edge of an C : G base pair displays the following chemical groups in the following order in the major groove:

#Unit 1. Molecules and their Interaction Relevant to Biology
  1. A hydrogen-bond acceptor (at N7 of guanine), a hydrogen-bond acceptor (the carbonyl on C6 of guanine), a hydrogen-bond donor (the exocyclic amino group on C3 of cytosine), and a small nonpolar hydrogen (the hydrogen at C5 of cytosine)

  2. A hydrogen-bond acceptor (at N7 of guanine), a hydrogen-bond acceptor (the carbonyl on C6 of guanine), a hydrogen-bond donor (the exocyclic amino group on C4 of cytosine), and a small nonpolar hydrogen (the hydrogen at C5 of cytosine)

  3. A small nonpolar hydrogen (the hydrogen at C5 of cytosine), a hydrogen-bond donor (the exocyclic amino group on C4 of cytosine), a hydrogen-bond acceptor (the carbonyl on C6 of guanine) and a hydrogen-bond acceptor (at N7 of guanine)

  4. A small nonpolar hydrogen (the hydrogen at C5 of cytosine), a hydrogen-bond acceptor (the exocyclic amino group on C4 of cytosine), a hydrogen-bond acceptor (the carbonyl on C6 of guanine) and a hydrogen-bond donor (at N7 of guanine)

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#Question id: 5735

#Unit 8. Inheritance Biology

Aneuploids may arise through nondisjunction, the failure of homologous chromosomes or sister chromatids to separate in meiosis. Following some statement are correct.

A)  Meiosis I non-disjunction leads to produce all gamete become aneuploidy

B) Meiosis II non-disjunction leads to produce half of gametes n+1 other half normal haploid

C) Meiosis I non-disjunction leads to produce half of gametes n+1 other half n-1

D) Either meiosis I or meiosis II non-disjunction leads to produce all aneuploidy gamete

TLS Online TPP Program

#Question id: 5736

#Unit 8. Inheritance Biology

Two zygote in which one A is 44+XYY and another B is 44+XXX karyotype is best described by

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#Question id: 5726

#Unit 8. Inheritance Biology

Following diagram represents the sequence of genes in a normal chromosome of a plant species

CORRECT combination for chromosomal mutation using

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#Question id: 5723

#Unit 8. Inheritance Biology

An individual heterozygous for a reciprocal translocation possesses the following chromosomes

I-  nonviability                                  II – Viability        

III- only translocated chromosome     IV- only normal chromosome

V- one translocated another normal

Which of the following above chromosome constituent and feature in their gamete result from?

A- alternate       B- adjacent-1     C- adjacent-2

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#Question id: 23966

#Unit 8. Inheritance Biology

Deletion in short arm of chromosome number 4, that leads to cause syndrome;

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#Question id: 5734

#Unit 8. Inheritance Biology

Following is heterozygote for inversion?

Following some statement given for above abnormality?

A) There is a less probability of a crossover occurring within inversion region and producing an enviable cross over meiotic product.

B) Chromosomes that have engaged in crossing-over separate in the normal fashion, without the creation of a bridge

C) The crossover produces chromatids that contain duplication and a deletion for different parts of the chromosome results non viable gamete

D) Double cross over within inversion loop will be results all viable gamete of parental and recombinant