Question 1

ADENINE  + WITH THYMINE. 2

Accepted Answer

HYDROGEN BONDS.

Question 2

GUANINE + CYTOSINE

Accepted Answer

3 HYDROGEN BONDS. COMPLEMENTARY.

Question 3

THE QUANTITIES OF GC

Accepted Answer

AND AT ARE ALWAYS  SAME. BUT THE RATIO OF AT TO GC VARIES FROM SPECIES.

Question 4

EACH COMPLETE TURN

Accepted Answer

OF THIS HELIX /ARE 10 BASE PAIRS.

Question 5

DNA PASS GENETIC INFO

Accepted Answer

FROM GENERATION TO GENERATION FROM CELL TO CELL.

Question 6

INFINITE SEQUENCE OF

Accepted Answer

DNA BASES ALONG THE MOLECULE. MAKES GENETIC DIVERSITY.

Question 7

DNA STABLE LIKE TABLE

Accepted Answer

PASS GEN WITHOUT CHANGE.

Question 8

2 STRANDS JOINED BY

Accepted Answer

HYDROGEN BONDS.

Question 9

CAN SEPARATE DURING

Accepted Answer

DNA REPLICATION AND PROTEIN SYNTHESIS.

Question 10

LARGE  CARRY A LOT

Accepted Answer

OF GENETIC INFO.

Question 11

GENETIC INFO PROTECTED

Accepted Answer

BY CORRUPTION FROM CHEMICALS AND PHYSICAL FORCES.

Question 12

(HSW ) EVIDENCE DNA

Accepted Answer

IS HEREDITARY MATERIAL:

Question 13

DNA HALVED IN SPERM

Accepted Answer

CELLS QUANTITY.

Question 14

MICE : BACTERIUM THAT

Accepted Answer

CAUSED PNEUMONIA. 2 FORMS SAFE STRAIN R S STRAIN.

Question 15

MICE INJECTED WITH LIVING

Accepted Answer

R STRAIN AND DEAD S STRAIN. BOTH GROUPS REMAINED HEALTHY.

Question 16

BUT WHEN INJECTED

Accepted Answer

WITH BOTH THEY GOT PNEUMONIA.

Question 17

EXPERIMENT

Accepted Answer

WAS REPEATED.

Question 18

DNA PASSED FROM

Accepted Answer

ONE BACTERIUM TO ANOTHER –BY VIRUSES.

Question 19

ELECTRON MICROSCOPES

Accepted Answer

/ IMAGES OF DNA GOING VIRUSE TO BACTERIAL CELLS.

Question 20

GENE SECTIONS OF DNA

Accepted Answer

THAT CONTAIN THE CODED INFO FOR MAKING POLYPEPTIDES./PROTEIN/ ENZYMES.

Question 21

GENE IS SEQUENCE OF DNA

Accepted Answer

BASES THAT DETERMINE THE POLYPEPTIDE.

Question 22

A POLYPEPTIDE IS A

Accepted Answer

SEQUENCES OF AMINO ACIDS.

Question 23

AS CODE HAS 3 BASES

Accepted Answer

CALLED TRIPLET CODE. 64 POS CODES ONLY 20 AMINO A.

Question 24

SOME AMINO HAVE MORE

Accepted Answer

THAN ONE CODE.

Question 25

IN EUKARYOTES . NON

Accepted Answer

CODING DNA . INTRON.

Question 26

FEATURES OF THE

Accepted Answer

TRIPLET CODE:

Question 27

AMINO ACID MORE

Accepted Answer

THAN ONE CODE DEGENERACY OF THE CODE.

Question 28

TRIPLET CODE READ IN

Accepted Answer

ONE DIRECTION ON DNA STRAND.

Question 29

THE START SEQUENCE

Accepted Answer

ALWAYS METHIONINE CAN BE REMOVED.

Question 30

3 TRIPLET CODES MARKS

Accepted Answer

END OF POLYPEPTIDE CHAIN .‘STOP CODES’.

Question 31

DNA AND CHROMOSOMES

Accepted Answer

PROKARYOTES CELL

Question 32

PROKARYOTES /  DNA

Accepted Answer

MOLS SMALLER / CIRCULAR / NOT ASSOCIATED WITH PROTEIN /NO CHROMS.

Question 33

DNA  LINEAR AND LARGER

Accepted Answer

/ASSOCIATED WITH PROTEINS/ FORM CHROMOSOMES.

Question 34

CHROMOSOME ONLY

Accepted Answer

VISIBLE WHEN CELL IS DIVIDING.

Question 35

2 THREADS HELD

Accepted Answer

BY CNETROMERE.

Question 36

THE DNA IN CHROMOSOMES

Accepted Answer

/ HELD IN POSITION BY PROTEINS.

Question 37

2M  DNA  COILED

Accepted Answer

AND FOLDED.

Question 38

HELIX WOUND ROUND

Accepted Answer

PROTEINS TO FIX IT IN POSITION,

Question 39

A LOT OF DNA PACKED IN

Accepted Answer

SINGLE CHRMOSOME.

Question 40

THE CHROMOSOME

Accepted Answer

CONTAINS A SINGLE MOL IF DNA.

Question 41

SINGLE DNA MANY

Accepted Answer

GENES LENGTH.

Question 42

ALWAYS EQUAL/ CHROMOSOME

Accepted Answer

OCCUR IN PAIRS (HOMOLOGOUS PAIRS).

Question 43

HOMOLOGOUS CHROMOSOMES

Accepted Answer

SEXUALLY EGG AND SPERM, ONE CHROMOSOME EACH.

Question 44

MOTHERS CHROMOSOMES

Accepted Answer

IN EGG (MATERNAL CHROMS), FATHER (PATERNAL CHROMS.)

Question 45

IN MEIOSIS THE HALVING

Accepted Answer

OF NUMBER OF CHROMOSOMES ENSURE EACH DAUGHTER CELL HAS ONE CHROMOSOMES FORM EACH HOMOLOGOUS PAIR.

Question 46

SO EACH CELL RECEIVES ONE

Accepted Answer

SET OF INFO FOR EACH CHARACTERISTIC.

Question 47

HAPLOID CELL COMBINEDIPLOID

Accepted Answer

STATE AND HOMOLOGOUS CHROMOSOMES RESTORED.

Question 48

INDIVIDUAL INHERITS ONE

Accepted Answer

ALLELE FROM EACH PARENT , MAY BE SAME OR DIFF.

Question 49

WHEN ALLELE WILL CODE FOR

Accepted Answer

A DIFF POLYPEPTIDE. / DIFF SEQUENCE OF AMINO ACID /HENCE A DIFF PROTEIN.

Question 50

PROTEIN MAY NOT FUNCTION

Accepted Answer

/ NOT COMPLEMENTARY/ENZYME NO FUNCTION.

Question 51

MITOSIS: PRODUCES 2 DAUGHTER

Accepted Answer

NUCLEI WITH THE SAME NUMBER OF CHROM AS PARENT.

Question 52

MEIOSIS: PRODUCES 4 DAUGHTER

Accepted Answer

NUCLEI /EACH WITH HALF THE NUMBER OF CHROM AS PARENT.

Question 53

MEIOSIS GOOD / MAINTAIN

Accepted Answer

CONSTANT NUMBER OF CHROMOSOMES.

Question 54

PROCESS OF

Accepted Answer

MEIOSIS  1

Question 55

THE HOMOLOGOUS CHORMOSOMES

Accepted Answer

PAIR UP/ CHROMATIDS WRAP AROUND EACH OTHER.

Question 56

END HOMOLOGOUS PAIRS /SEPARATED /ONE

Accepted Answer

CHROMOSOMES FROM EACH PAIR GOING/ INTO ONE OF THE 2 DAUGHTER CELLS.

Question 57

THE SECOND MEIOTIC DIVISION /

Accepted Answer

THE CHROMATIDS MOVE APART.

Question 58

END 4 CELLS HAVE BEEN FORMED

Accepted Answer

. 23 CHROMS.

Question 59

MIOSIS ALLOWS /ORGANISM TO

Accepted Answer

ADAPT AND SURVIVE/ IN CHANGING WORLD.

Question 60

BRINGS ABOUT

Accepted Answer

GENETIC VARIATION BY:

Question 61

INDEPENDENT SEGREGATION OF

Accepted Answer

HOMOLOGOUS CHROMOSOMES, 2. RECOMBINATION OF HOMOLOGOUS CHROMOSOMES BY CROSSING OVER.

Question 62

LOCUS – THE POSITION OF A

Accepted Answer

GENES ON A CHROMOSOMES OR DNA MOL.

Question 63

ALLELE – ONE OF THE

Accepted Answer

DIFF FORMS OF A GENE.

Question 64

INDEPENDENT SEGREGATION

Accepted Answer

OF HOMOLOGOUS CHROMOSOMES

Question	Answer
RUNGS DNA LADDER SAME LENGTH/ ON OF ECAH TYPE BASE.
ADENINE + WITH THYMINE. 2	HYDROGEN BONDS.
GUANINE + CYTOSINE	3 HYDROGEN BONDS. COMPLEMENTARY.
THE QUANTITIES OF GC	AND AT ARE ALWAYS SAME. BUT THE RATIO OF AT TO GC VARIES FROM SPECIES.
EACH COMPLETE TURN	OF THIS HELIX /ARE 10 BASE PAIRS.
DNA PASS GENETIC INFO	FROM GENERATION TO GENERATION FROM CELL TO CELL.
INFINITE SEQUENCE OF	DNA BASES ALONG THE MOLECULE. MAKES GENETIC DIVERSITY.
DNA STABLE LIKE TABLE	PASS GEN WITHOUT CHANGE.
2 STRANDS JOINED BY	HYDROGEN BONDS.
CAN SEPARATE DURING	DNA REPLICATION AND PROTEIN SYNTHESIS.
LARGE CARRY A LOT	OF GENETIC INFO.
GENETIC INFO PROTECTED	BY CORRUPTION FROM CHEMICALS AND PHYSICAL FORCES.
(HSW ) EVIDENCE DNA	IS HEREDITARY MATERIAL:
DNA HALVED IN SPERM	CELLS QUANTITY.
MICE : BACTERIUM THAT	CAUSED PNEUMONIA. 2 FORMS SAFE STRAIN R S STRAIN.
MICE INJECTED WITH LIVING	R STRAIN AND DEAD S STRAIN. BOTH GROUPS REMAINED HEALTHY.
BUT WHEN INJECTED	WITH BOTH THEY GOT PNEUMONIA.
EXPERIMENT	WAS REPEATED.
DNA PASSED FROM	ONE BACTERIUM TO ANOTHER –BY VIRUSES.
ELECTRON MICROSCOPES	/ IMAGES OF DNA GOING VIRUSE TO BACTERIAL CELLS.
GENE SECTIONS OF DNA	THAT CONTAIN THE CODED INFO FOR MAKING POLYPEPTIDES./PROTEIN/ ENZYMES.
GENE IS SEQUENCE OF DNA	BASES THAT DETERMINE THE POLYPEPTIDE.
A POLYPEPTIDE IS A	SEQUENCES OF AMINO ACIDS.
AS CODE HAS 3 BASES	CALLED TRIPLET CODE. 64 POS CODES ONLY 20 AMINO A.
SOME AMINO HAVE MORE	THAN ONE CODE.
IN EUKARYOTES . NON	CODING DNA . INTRON.
FEATURES OF THE	TRIPLET CODE:
AMINO ACID MORE	THAN ONE CODE DEGENERACY OF THE CODE.
TRIPLET CODE READ IN	ONE DIRECTION ON DNA STRAND.
THE START SEQUENCE	ALWAYS METHIONINE CAN BE REMOVED.
3 TRIPLET CODES MARKS	END OF POLYPEPTIDE CHAIN .‘STOP CODES’.
DNA AND CHROMOSOMES	PROKARYOTES CELL
PROKARYOTES / DNA	MOLS SMALLER / CIRCULAR / NOT ASSOCIATED WITH PROTEIN /NO CHROMS.
DNA LINEAR AND LARGER	/ASSOCIATED WITH PROTEINS/ FORM CHROMOSOMES.
CHROMOSOME ONLY	VISIBLE WHEN CELL IS DIVIDING.
2 THREADS HELD	BY CNETROMERE.
THE DNA IN CHROMOSOMES	/ HELD IN POSITION BY PROTEINS.
2M DNA COILED	AND FOLDED.
HELIX WOUND ROUND	PROTEINS TO FIX IT IN POSITION,
A LOT OF DNA PACKED IN	SINGLE CHRMOSOME.
THE CHROMOSOME	CONTAINS A SINGLE MOL IF DNA.
SINGLE DNA MANY	GENES LENGTH.
ALWAYS EQUAL/ CHROMOSOME	OCCUR IN PAIRS (HOMOLOGOUS PAIRS).
HOMOLOGOUS CHROMOSOMES	SEXUALLY EGG AND SPERM, ONE CHROMOSOME EACH.
MOTHERS CHROMOSOMES	IN EGG (MATERNAL CHROMS), FATHER (PATERNAL CHROMS.)
IN MEIOSIS THE HALVING	OF NUMBER OF CHROMOSOMES ENSURE EACH DAUGHTER CELL HAS ONE CHROMOSOMES FORM EACH HOMOLOGOUS PAIR.
SO EACH CELL RECEIVES ONE	SET OF INFO FOR EACH CHARACTERISTIC.
HAPLOID CELL COMBINEDIPLOID	STATE AND HOMOLOGOUS CHROMOSOMES RESTORED.
INDIVIDUAL INHERITS ONE	ALLELE FROM EACH PARENT , MAY BE SAME OR DIFF.
WHEN ALLELE WILL CODE FOR	A DIFF POLYPEPTIDE. / DIFF SEQUENCE OF AMINO ACID /HENCE A DIFF PROTEIN.
PROTEIN MAY NOT FUNCTION	/ NOT COMPLEMENTARY/ENZYME NO FUNCTION.
MITOSIS: PRODUCES 2 DAUGHTER	NUCLEI WITH THE SAME NUMBER OF CHROM AS PARENT.
MEIOSIS: PRODUCES 4 DAUGHTER	NUCLEI /EACH WITH HALF THE NUMBER OF CHROM AS PARENT.
MEIOSIS GOOD / MAINTAIN	CONSTANT NUMBER OF CHROMOSOMES.
PROCESS OF	MEIOSIS 1
THE HOMOLOGOUS CHORMOSOMES	PAIR UP/ CHROMATIDS WRAP AROUND EACH OTHER.
END HOMOLOGOUS PAIRS /SEPARATED /ONE	CHROMOSOMES FROM EACH PAIR GOING/ INTO ONE OF THE 2 DAUGHTER CELLS.
THE SECOND MEIOTIC DIVISION /	THE CHROMATIDS MOVE APART.
END 4 CELLS HAVE BEEN FORMED	. 23 CHROMS.
MIOSIS ALLOWS /ORGANISM TO	ADAPT AND SURVIVE/ IN CHANGING WORLD.
BRINGS ABOUT	GENETIC VARIATION BY:
INDEPENDENT SEGREGATION OF	HOMOLOGOUS CHROMOSOMES, 2. RECOMBINATION OF HOMOLOGOUS CHROMOSOMES BY CROSSING OVER.
LOCUS – THE POSITION OF A	GENES ON A CHROMOSOMES OR DNA MOL.
ALLELE – ONE OF THE	DIFF FORMS OF A GENE.
INDEPENDENT SEGREGATION	OF HOMOLOGOUS CHROMOSOMES
EACH CHROMOSOME LINES	UP/RANDOM HOMOLOGOUS PARTNER.
SO CHROMOSOMES THAT GOES	INTO THE DAUGHTER CELL AT MEIOSIS 1 IS RANDOM.
MOVEMENT	DEPENDS/ON LINE UP.
VARIETY FORM NEW	GENETIC COMBINATIONS :
MEMBER OF A HOMOLOGOUS	PAIR OF CHROMOSOMES / SAME GENES / SAME CHARACTERISTICS
THE ALLELES MAY DIFFER	(BROWN .BLUE EYE.). PRODUCED NEW GENETIC COMBINATIONS.
STAGE 2 : AT END OF MEIOSIS 1 :	THE HOMOLOGOUS CHROMOSOMES HAVE SEGREGATED INTO 2 SEPARATE CELLS.
STAGE 3: AT END OF MEIOSIS 2 –	THE CHROMOSOME HAVE SEGREGATED INTO CHROMATIDS PRODUCING 4 GAMETES THE GAMETES ARE
THE GAMETES DIFFERNET	/ DIFF CHRMOSOMES.
RANDOM FUSING CREATE	VARIETY IN OFFSPRING.
GENETIC RECOMBINATION	BY CROSS OVER:
AFTER EACH CHROMOSOME	LINES UP ALONGSIDE ITS HOMOLOGOUS PARTNER.
THE CHROMATIDS EACH PAIR	TWISTED AROUND ONE ANOTHER.
TWISTING CREATES	TENSIONS /BREAKING POINT.
BROKEN PORTIONS REJOIN	WITH THE CHROMATIDS OF /HOMOLOGOUS PARTNER. SAME PRORTIONS.
NEW GENETIC	VARIETY.
NO RECOMBINATION BY CROSS OVER 2	CELLS PRODUCED . RECOMBINATION / 4 CELLS ARE PRODUCED.

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