Contact Form

Name

Email *

Message *

Friday, 23 July 2021

Molecular basis of inheritance-I MCQs ll NEET ll NCERT ll Biology

Molecular basis of inheritance-I  MCQs with explanation

  Watch lectures on my you tube channel: https://bit.ly/3Bzgc2k

 

1. Genetic code consists of

(a) adenine and guanine

(c) cytosine and guanine

(d) all the above.

Answer and Explanation:

1. (d): The genetic information is transferred from DNA to m-RNA to protein. The proteins are made up of some 20 amino acids whose sequence is hidden in the sequence of nucleotides of mRNA. Hence, genetic code consists of all 20 amino acids. Thus genetic code is the relationship of amino acids sequence in a polypetide and nucleotide/base sequence in mRNA antisense strand and DNA.

2. Haploids are able to express both recessive and dominant alleles/mutations because there are

(a) many alleles for each gene

(b) two alleles for each gene

(c) only one allele for each gene in the individual

(d) only one allele in a gene.

Answer and Explanation:

2. (c): Haploids are able to express both recessive and dominant alleles/ mutations because there are only one allele for each gene in the individual. Hybrid is a organism containing two different alleles or individual containing both dominant and recessive genes of an allelic pairs.

3. A family of five daughters only is expecting sixth issue. The chance of its beings a son is

(a) zero

(b) 25%

(c) 50%

(d) 100%.

Answer and Explanation:

3. (c): A family of five daughter only is expecting sixth issue. The chance of its beings a son is 50%. Human have 22 pairs chromosomes which are XX in females and XY in males. So if we cross the parents there is 1 : 1 chance for boy and girl.

4. Mutations used in agriculture are commonly

(a) induced

(b) spontaneous

(c) lethal

(d) recessive and lethal.

Answer and Explanation:

4. (a): Mutations used in agriculture are commonly induced. Induced mutations can be produced artificially by certain mutagenic agents called as mutagens. Ethyl methyl sulphonate (EMS) is the most extensively used chemical mutagen in different crop plants nowadays.

5. DNA replication is

(a) conservative and discontinuous

(b) semiconservative and semidiscontinuous

(c) semiconservative and discontinuous

(d) conservative.

Answer and Explanation:

5. (b): DNA replication is semi-conservative and semi-discontinuous. Meselson and Stahl, 1958 by using 14N and 15N confirmed that the replication of DNA in E- coli is semi conservative. Carin, 1963 by using 3H in thymidine also confirmed semi-conservative mode of DNA replication in this bacterium.

6. Diploid chromosome number in human is

(a) 46

(b) 44

(c) 48

(d) 42.

Answer and Explanation:

6. (a): Diploid chromosome number in human is 46. All animals including man have a characteristics number of chromosomes in their body cells called diploid (or 2n) number. These occur as homologous pairs, one member of each pair have been acquired from the gamete of one of the two parents of the individual whose cells are being examined. It is actually 44 + XX or 44 + XY.

7. In the genetic dictionary, there are 64 codons as

(a) 64 amino acids are to be coded

(b) 64 types of tRNAs are present

(c) there are 44 nonsense codons and 20 sense codons

(d) genetic code is triplet.

Answer and Explanation:

7. (d): Codon is a sequence of three nucleotides coding for one aminoacid in a polypeptide chain. There are four different types of nucleotides – A, T, G and C and since a codon is a triplet thus 64 (4 x 4 x 4) distinct triplets of purines or pyrmidine bases determine the 20 amino acids. But out of these 64 codons, only 61 codons code for individual amino acids during protein synthesis. The three codons that do not code for amino acids are the termination or nonsense codons. These are UAA, UAG and UGA. So the remaining 61 codons code for the 20 amino acids.

8. An octamer of 4 histones complexed with DNA forms

(a) endosome

(b) nucleosome

(c) mesosome

(d) centromere.

Answer and Explanation:

8. (b): An octamer of 4 histones complexed with DNA forms nucleosome. The association of histones with DNA is very characteristic. It involves the formation of linear array of spherial structures called nucleosomes. This structure contains four pairs of histones (H2A, H2B, H3 and H4) in a ball; around which is wrapped a stretch of about 150 base pairs of DNA.

9. Bateson used the terms coupling and repulsion for linkage and crossing over. Name the correct parental of coupling type along with its cross over or repulsion

(a) coupling AABB, aabb; Repulsion AABB, aabb

(b) coupling AAbb, aaBB; Repulsion AaBb, aabb

(c) coupling aaBB, aabb; Repulsion A ABB, aabb

(d) coupling AABB, aabb : Repulsion AAbb, aaBB.

Answer and Explanation:

9. (d): Bateson and Punnet explained that when two dominants enter from the same parent-they try to remain together, called coupling. When two dominants enter from different parents they try to remain seperate called repulsion. Bateson and Punnett (1906) used the term coupling and repulsion in sweet pea (Lathyrus odoratus) for linkage and crossing over. The correct parental of coupling type along with its cross over or repulsion is coupling AABB, aabb : Repulsion AAbb; aaBB.

10. The process of transfer of genetic information from DNA to RNA/formation of RNA from DNA

(a) transversion

(b) transcription

(c) translation

(d) translocation.

Answer and Explanation:

10. (b): The process in living cells in which the genetic information of DNA is transferred to a molecule of messenger RNA (mRNA) is the first step in protein synthesis (see also genetic code). Transcription takes place in the cell nucleus or nuclear region and is regulated by transcription factors.

11. Escherichia coli fully labelled with 15N is allowed to grow in 14N medium. The two strands of DNA molecule of the first generation bacteria have

(a) different density and do not resemble parent DNA

(b) different density but resemble parent DNA

(c) same density and resemble parent DNA

(d) same density but do not resemble parent DNA.

Answer and Explanation:

11. (a): E-coli fully labelled with 15N is allowed to grow in 14N medium. The two strands of DNA molecule of the first generation bacteria have different density and do not resemble parent DNA. Meselson and Stahl, 1958 by using 14N and 15N confirmed that the replication of DNA in E-coli is semi-conservative in nature.

12. Khorana first deciphered the triplet codons of

(a) serine and isoleucine

(b) cysteine and valine

(c) tyrosine and tryptophan

(d) phenylalanine and methinonine.

Answer and Explanation:

12. (b): Khorana synthesised a chain of alternate nucleotide GUGUGUGUGU. He found that it stimulated synthesis of a peptide having alternate valine-cysteine- valine-cysteine.

13. Experimental material in the study of DNA replication has been

(a) Escherichia coli

(b) Neurospora crassa

(c) Pneumococcus

(d) Drosophila melanogaster.

Answer and Explanation:

13. (a): Experimental material in the study of DNA replication has been Escherichia coli. E-coli fully labelled with 15N is allowed to grow in 14N medium. The two strands of DNA molecule of the first generation bacteria have different density and do not resemble parent DNA. Meselson and Stahl, 1958 by using 14N and 15N confirmed that the replication of DNA in E-coli is semi-conservative in nature.

14. Out of 8 ascospore formed in Neurospora the arrangment is 2a : 4a : 2a showing

(a) no crossing over

(b) some meiosis

(c) second generation division

(d) first generation division.

Answer and Explanation:

14. (c): Out of 8 ascospore formed in Neurospora the arrangement is 2a : 4a : 2a showing second generation division. In Neurospora, products of meiosis remain linearly arranged and undergo one mitosis. Crossing over occurs in four strand stage.

15. When a certain character is inherited only through female parent, it probably represents

(a) multiple plastid inheritance

(b) cytoplasmic inheritance

(c) incomplete dominance

(d) mendelian nuclear inheritance.

Answer and Explanation:

15. (b): When a certain character is inherited only through female parent, it probably represents cytoplasmic inheritance. Plasma genes present in the maternal cytoplasm are transmitted to offspring hence this type of inheritance is termed as cytoplasmic or maternal inheritance.

16. Nucleotide arrangement in DNA can be seen by

(a) X-ray crystallography

(b) electron microscope

(c) ultracentrifuge

(d) light microscope.

Answer and Explanation:

16. (a): Nucleotide arrangement in DNA can be seen by X-ray crystallography. Watson and Crick, 1953 proposed the double helical model for DNA. They were awarded Nobel prize in 1962. This model was developed by them on the basis of several previous observations including the of-helix of Pauling, 1951 and X-ray reflection studies of Franklin and Gosling, 1953.

Watch lectures on my you tube channel: https://bit.ly/3Bzgc2k 

17. A DNA with unequal nitrogen bases would most probably be

(a) single stranded

(b) double stranded

(c) triple stranded

(d) four stranded.

Answer and Explanation:

17. (a): A DNA with unequal nitrogen bases would most probably be single stranded. Nitrogenous bases are unequal in number in single stranded DNA, because they do not posses complementary base pairs.

18. The process of translation is

(a) ribosome synthesis

(b) protein synthesis

(c) DNA synthesis

(d) RNA synthesis.

Answer and Explanation:

18. (b): The process of translation is protein synthesis. Emil Fischer, a German chemist established that the proteins are polymers of amino acids. There are some twenty amino acids involved in protein synthesis. In translation, the message coded by DNA on m-RNA is translated into a specific protein.

19. During DNA replication, the strands separate by

(a) DNA polymerase

(b) topoisomerase

(c) unwindase/helicase

(d) gyrase.

Answer and Explanation:

19. (c): During DNA replication, the strands separate by unwindase/helicase. The molecule is unwound by DNA unwinding proteins called helicases. The helicases II and III get attached to logging strand and protein to the leading strand. The formation of bands is avoided by single stranded DNA binding proteins (SSB).

20. Because most of the aminncids are represented by more than one codon, the genetic code is

(a) overlapping

(b) wobbling

(c) degenerate

(d) generate.

Answer and Explanation:

20. (c): Because most of the amino acids are represented by more than one codon, the genetic code is degenerate. Certain amino acids are identified by more than one codons. This phenomenon is called as degeneracy e.g. only AUG codes for methionine and UGG tryptophan.

21. Who proved that DNA is basic genetic material?

(a) Griffith

(b) Watson

(c) Boveri and Sutton

(d) Hershey and Chase.

Answer and Explanation:

21. (d): Hershey and Chase proved that DNA is a basic genetic material. Hershey and Chase, 1952, by using s32 and s35 with a T-2 type phage concluded that DNA is the genetic material.

22. The transforming principle of Pneumococcus as found out by Avery, MacLeod and McCarty was

(a) mRNA

(b) DNA

(c) protein

(d) polysaccharide.

Answer and Explanation:

22. (b): The transforming principle of Pneumococcus as found out by Avery, MacLeod and McCarty was DNA. In 1944, Avery, MacLeod and McCarty repeated Griffith’s experiment successfully. They separated the proteins, carbohydrates and DNA of S III strains and separately mixed them in the pure cultures of R II. Only DNA could bring about transformation of R II type into S III and not the proteins or the carbohydrates.

23. Initiation codon of protein synthesis (in eukaryotes) is

(a) GUA

(b) GCA

(c) CCA

(d) AUG.

Answer and Explanation:

23. (d): A gene or a cistron transcribes to form mRNA which consists of several coding and non coding regions. The coding region always starts with the codon AUG. It is called initiation codon. It is found in both eukaryotes and prokaryotes. It codes for amino acid methionine.

24. Nucleosome core is made of

(a) HI, H2A, H2B and H3

(b) HI, H2A, H2B, H4

(c) HI, H2A, H2B, H3 and H4

(d) H2A, H2B, H3 and H4.

Answer and Explanation:

24. (d): Nucleosome core is made up of H2A, H-,B, H, and H4. It is about 7-10 nm in diameter, consisting of histones around which a DNA strand, about 120 base pair long is wrapped in chromosomes.

25. Sex is determined in human beings

(a) by ovum

(b) at time of fertilization

(c) 40 days after fertilization

(d) seventh to eight week when genitals differentiate in foetus

Answer and Explanation:

25. (b): Sex is determined in human beings at the time of fertilization. Sex of human body is determined by the karyotype of the zygote or fertilized egg. Sex of the baby depends upon the sperm which fertilizes the ovum.

26. In protein synthesis, the polymerization of amino acids involves three steps. Which one of the following is not involved in the polymerisation of protein?

(a) termination

(b) initiation

(c) elongation

(d) transcription.

Answer and Explanation:

26. (d): Transcription is the mechanism of copying the message of DNA on RNA with the help of enzyme RNA polymerase. It is meant for taking the coded information from DNA to the site where it is required for protein synthesis.

Translation or protein synthesis is a complicated process involving several steps such as – activation of amino acid, transfer of amino acid to /-RNA, initiation of polypeptide synthesis, elongation of polypeptide chain and, termination of polypeptide chain. This entire process occurs in the cytoplasm over the ribosomes. Transcription and translation are two entirely different processes that are separated by time and space.

27. There are special proteins that help to open up DNA double helix in front of the replication fork. These proteins are

(a) DNA ligase

(b) DNA topoisomerase I

(c) DNA gyrase

(d) DNA polymerase I.

Answer and Explanation:

27. (b): DNA is a double helical molecule and it opens to form a replication fork for its replication. The two strands of DNA are joined with the help of H-bonds between the strands. Topoisomerases are specialized to cause nicks or breaks in the double helix and helps separate the DNA stands. Helicase unwinds the DNA helix from that nick caused by the topoisomerase and this seperates the two strands.

DNA gyrase introduces negative supercoils in DNA strands of prokayotes.

DNA polymerase adds nucleotides units to the 3′ end of a DNA chain. DNA ligase joins the ends of DNA.

28. Initiation codon in eukaryotes is

(a) GAU

(b) AGU

(c) AUG

(d) UAG.

Answer and Explanation:

28. (c): The base triplets found in mRNA that helps in protein synthesis is called as codon. Initiation codon is triplet of ribonucleotides in messenger RNA (mRNA) that is present at the start of the mRNA coding sequence and initiates polypeptede formation or translation. The triplet is AUG and codes for the amino acid methionine: In bacteria the start codon is either AUG coding forN-formyl methionine or GUG coding for valine.

29. ‘Lac operon’ in E. coli, is induced by

(a) ‘I’ gene

(b) promoter gene

(c) P-galactosidase

(d) lactose.

Answer:

(c) P-galactosidase

30. Anticodon is an unpaired triplet of bases in an exposed position of

(a) <-RNA

(b) w-RNA

(c) r-RNA

(d) both ‘b’ and ‘c’.

Answer and Explanation:

30.

(a): Anticodon is the sequence of three nucleotides in a transfer RNA molecule that pairs with a complementary sequence of three nucleotides (codon) on a molecule of messenger RNA. /-RNA has clove like shape or L shape (three dimensional). It has G at 5′ end CCA at 3′ end. CCA at 3′ end is meant for attaching to a specific amino acid (AA-binding site). On the opposite side lies an anticodon that is complementary to a specific codon of mRNA. The two are called recognition sites.

31. The point, at which polytene chromosome appear to be attached together, is called

(a) centromere

(b) chromomere

(c) chromocentre

(d) centriole.

Answer and Explanation:

31. (c): Chromocentre is junction point of the chromosomes in the polytene salivary glands of Drosophila larvae. Unlike the situation in other cells, the giant chromosomes of these cells persist through interphase, the two homologous copies of each chromosome are attached together throughout their lengths and all the chromosomes are joined together by a chromocentre.

32. In split genes, the coding sequences are called

(a) exons

(b) cistrons

(c) introns

(d) operons.

Answer and Explanation:

32. (a): Split gene are those genes that consist of continuous sequence of nucleotide (coding sequence) interrupted by intervening sequences. Most eukaryotic genes are split as are genes of some animal viruses. The continous coding sequences are called exons and the intervening non-coding sequences are called introns. These introns are not represented in mRNA transcribed from the gene and are not utilized for the synthesis of proteins.

33. The polytene chromosomes were discovered for the first time in

(a) Drosophila

(b) Musca domestica

(c) Chironomus

(d) Musca nebula.

Answer and Explanation:

33. (c): In salivary glands cells of Chironomus tantans giant chromosomes were observed by E.GBalbini for the first time in 1881. The availability of these chromosomes greatly helped in the study of Drosophila cytogenetics in which they were discovered later.

34. Each chromosome at the anaphase stage of a bone marrow cell in our body has

(a) two chromatids

(b) several chromatids

(c) no chromatids

(d) only one chromatid.

Answer and Explanation:

34. (d): A bone marrow cell is a somatic cell and therefore mitosis takes place in it. During mitosis anaphase chromosomes split at centromere. Sister chromatids seperate from each other, so that the two sister chromatids are separate structures and can now be called as chromosomes.

35. If the DNA codons are ATG ATG ATG and a cytosine base is inserted at the beginning, then which of the following will result?

(a) CAT GAT GATG

(b) a non-sense mutation

(c) C ATG ATG ATG

(d) CATGATGATG

Answer and Explanation:

35. (a): Nonsense mutation is a mutation which interconverts a nonsense to or from a sense-coding triplet, resulting in an abnormally foreshortened or elongated polypeptide chain. But in this example cytosine is added at the beginning so CAT GAT GATG. will result.

Watch lectures on my you tube channel: https://bit.ly/3Bzgc2k 

36. Barr body in mammals represents

(a) all the heterochromatin in male and female ceils

(b) the Y-chromosome in somatic cells of male

(c) all the heterochromatin in female cells

(d) one of the two X-chromosomes in somatic cells of females.

Answer and Explanation:

36. (d): Barr body is a condensed heterochromatic copy of the X chromosome, visible by staining the interphase nucleus of somatic cells of the homogametic sex, for example the human female. Named after M. Barr, these condensed chromosomes offer an easy way of determining the true genetic sex of individuals with intermediate secondary sexual characteristics. The total number of Barr bodies is always one less than the total number of X chromosomes present in the cell or the organism.

37. If the sequence of bases in DNA is ATTCGATG, then the sequence of bases in its transcript will be

(a) GUAGCUUA

(b) AUUCGAUG

(c) CAUCGAAU

(d) UAAGCUAC.

Answer:

(d) UAAGCUAC.

38. An environmental agent, which triggers transcription from an operon, is a

(a) depressor

(b) controlling element

(c) regulator

(d) inducer.

Answer:

(d) inducer

39. The lac operon is an example of

(a) repressible operon

(b) overlapping genes

(c) arabinose operon

(d) inducible operon.

Answer and Explanation:

39. (d): Lac operon (lactose operon). Genetic system of E.coli is responsible for the uptake and initial catabolism of lactose. The lac operon consists of three structural genes (lac Z, lac Y, lac A), lac Z codes for p- galactosidase which hydrolyses lactose to glucose and galactose. (lacY codes for lac permease, a membrane- bound protein constituent of the lactose transport system, lac A codes of thioglactoside transacetylase, an enzyme of uncertain metabolic function.

Bacteria growing in a lactose-free medium contain about five molecules of P-glactosidase per cell. The appearance of P-galactosidase is coordinated with the production of permease and transacetylase. On removing lactose from the medium enzyme synthesis stops.

Such enzymes whose synthesis can be induced by adding the substrate are known as inducible enzymes and the genetic systems responsible for the synthesis of such an enzyme are known as inducible operons.

40. The wild type E.coli cells are growing in normal medium with glucose. They are transferred to a medium containing only lactose as sugar. Which of the following changes take place?

(a) the lac operon is induced

(b) E.coli cells stop dividing

(c) the lac operon is repressed

(d) all operons are induced.

Answer and Explanation:

40. (a): When E.coli bacteria are transfered to medium containing lactose, then the lac opeon is indueed. The lac opeaon consists of 3 structural gene (lac Z, lac Y and lac A). It involves the synthesis of P-galactosidase enzyme in E.coli, which hydrolyses lactose into glucose and galactose.

41. Radio-tracer technique shows that DNA is in

(a) multi-helix stage

(b) single-helix stage

(c) double-helix stage

(d) none of these.

Answer and Explanation:

41. (c): Autoradiography is the study of labelled precursors like 3H by knowing the movement of radioactivity with the help of photographic films and emulsions at short intervals. 14C and 3H are incorporated in bases like thymidine, uridine and amino acids to study the structure of DNA and proteins. Radio tracer technique shows that DNA is in double helical form.

42. The maximum formation of /n-RNA occurs in

(a) ribosome

(b) nucleoplasm

© cytoplasm

(d) nucleolus.

Answer and Explanation:

42. (d): Nucleolus is a plasmosome body that is formed around the nucleolus organizer and is located in the secondary constriction on that chromosome. It is made up of RNA and proteins. The associated nucleolar chromatin contains DNA. It forms mRNA that has low molecular weight. Ribosomes are mainly concerned with proteins synthesis. They are sites for synthesis of rRNA and tRNA is synthesized in the cytoplasm.

43. Which of the following serves as a terminal codon?

(a) UAG

(b) AGA

(c) AUG

(d) GCG.

Answer and Explanation:

43. (a): In eukaryotes, the termination of polypeptide chain during translation is brought about by the terminating codons. These are UAA, UAG and UGA and these are called as amber, ochre and opal respectively. These codons are also called as nonsense codons as they donot code for any amino acid. In phages and bacteria there may be many initiating and terminating codons and thus as many polypeptides are synthesized. AUG is the initiation codon. It codes for methionine amino acid.

44. Identify the one, which causes mutation

(a) cosmic rays

(b) hromosomes

(c) crossing over

(d) X-rays.

Answer and Explanation:

44. (d): Mutations can be artificially induced with the help of mutagenic agents which can be broadly classified into two groups, physical mutagens and chemical mutagens. Physical mutagens are mainly radiations, although change in pH value (acidity) or temperature shocks may also induce mutations.

Among ionizing radiations, more commonly X-rays, gamma rays, beta rays and neutrons are used for inducing mutations. X-rays are produced in a X-ray machine when energy charged particles like cathode rays (electrons) impinge on a suitable target like tungston.

45. Lampbrush Chromosomes are seen in which typical stage?

(a) mitotic metaphase

(b) meiotic prophase

(c) mitotic anaphase

(d) mitotic prophase.

Answer and Explanation:

45. (b): Lampbrush chromosomes (Ruckert, 1892) are large sized diplotene chromosome bivalents with a length of 400 – 1000 nm each and a total length of 5900 in Triturus (Salamander = Newt). A lampbrush chromosome is made of two homologous chromosomes held at several places by chiasmata. Each chromosome has an axis with alternate chromosomes held at several places by chiasmata.

Each chromosome has an axis with alternate chromomeres and interchromomeric areas. Many of the chromomeres give out lateral loops of various sizes which are thin in the region of origin and thick in the area where they are wound back into chromomeres. Loops possess a number of copies of the same gene and are meant for rapid transcription and production of materials like yolk. Hence, lampbrush chromosomes occur in oocytes.

46. Which of the following step of translation does not consume a high energy phosphate bond?

(a) peptidyl transferase reaction

(b) aminoacyl /-RNA binding to ,4-site

(c) translocation

(d) amino acid activation.

Answer and Explanation:

46. (a): Protein synthesis or translation consists of ribosomes, amino acids, mRNA, tRNAs and aminoacyl tRNA synthetases. The ribosomes have two binding sites namely aminoacyl site or A-site and peptide site or P- site. The starting amino acid methionine lies at the P- site of the ribosome.

The next incoming tRNA is called amino acyl tRNA, it is bound to A-site. A peptide bond is formed between COOH group of the t-RNA at P-site and NH2 group of aminoacyl t-RNA. This is facilitated by the enzyme peptidyl transferase and does not require high energy phosphate bonds.

Watch lectures on my you tube channel: https://bit.ly/3Bzgc2k 

47. The RNA that picks up specific amino acid from amino acid pool in the cytoplasm to ribosome during protein synthesis is called

(a) r-RNA

(b) RNA

(c) m-RNA

(d) t-RNA.

Answer and Explanation:

47. (d): Transfer RNA or tRNA help in transfer of amino acids to ribosomes mRNA complex to form the polypeptide chain. It has four key regions a carrier and recognition end, enzyme site and ribosome site. This recognition end has three anticodons with the help of which aminoacids are identified. r-RNA forms 67% of 70s ribosomes and 50% of 80s ribosomes.

mRNA carries the coded information from DNA for the synthesis of proteins.

48. DNA synthesis can be specifically measured by estimating the incorporation of radio-labelled

(a) thymidine

(b) deoxyribose sugar

(c) uracil

(d) adenine.

Answer and Explanation:

48. (a): Autoradiography is the study of labelled precursors like 3H by knowing the movement of radioactivity with the help of photographic films and emulsions at short intervals.

Radioactive material like tritiated thymidine which is formed by replacing normal hydrogen of thymidine with H3 (heavy isotope of hydrogen). Thymidine only is used for this purpose because RNA will not be labelled by this.

49. Genetic identity of a human male is determined by

(a) sex-chromosome

(b) cell organelles

(c) autosome

(d) nucleolus.

Answer and Explanation:

49. (a): Sex chromosomes are those chromosomes whose presence, absence or particular form determines the sex of the individual in unisexual or dioecious organisms, e.g., XX – XY. XY method (XX – XY). It occurs in mammals and many insects with females having homomorphic XX sex chromosomes and males possessing heteromorphic XY – chromosomes. In human being the Y-chromosome is straight and acrocentric (centromere subterminal near one end).

It is about 2 µm in length and is thus the shortest of all chromosomes. X- chromosome is 5.0 – 5.5 µm in length. It is metacentric (centromere in the middle). Despite differences in morphology, XY chromosomes synapse during zygotene. They have two parts, homologous and differential. Homologous regions of the two take part in synapsis.

50. Genes located at the same locus of chromosomes are called

(a) multiple alleles

(b) polygenes

(c) oncogenes

(d) none of these.

Answer and Explanation:

50. (a): Genes located on the same locus of chromosomes are called multiple alleles. They are produced due to repeated mutations of the same gene but in different directions. Multiple alleles being located on the same locus do not show crossing over.

  Watch lectures on my you tube channel: https://bit.ly/3Bzgc2k

No comments:

Post a Comment

APOMIXIS AND POLYEMBRYONY l Sexual Reproduction in Flowering Plants - NCERT Based

  A POMIXIS AND P OLYEMBRYONY   ·          Although seeds, in general are the products of fertilisation, a few flowering plants have e...