Download - 09 TYS Unit3.4 cell division and genetics

8/14/2019 09 TYS Unit3.4 cell division and genetics

http://slidepdf.com/reader/full/09-tys-unit34-cell-division-and-genetics 1/3

7/13/09

1

TYS Unit 3.4 A & B !

Genetics!

Unit 3.4A Question 1 !

a) R;!" Variety P and Q have stripes and spotson their bodies respectively that serveas a camouage among the water plantsin the river; !

" However, Variety R will stand out andwill be easily spotted by predators; !

Unit 3.4A Question 1 !Bi) Stripes w ith spots on them;!ii) Variety S has the be st camouage am ong

the water plants an d hence is the mostadapted compared to P,Q,R and S;!

" Hence, their survival rate increases,Hence, reproductive rate increases;!

Advantageou s alleles are passed on tooffspring. Hence, they sur vive better thu scausing an increase in numbe rs of S;!

Unit 3.4A Question 1c !

• Variety R may be completely wiped out asit is most predated upon; !

• Numbers of variety P and Q will decrease; !• As they mate with S, more offsprings will

inherit the characteristics of S and willsurvive better with the best camouage; !

Unit 3.4A Question 2 !(a) R: fertilization; "S: zygote; !

(b) (i) 27; "(ii) 54; !

(c) Lamb K was produced without requiringthe fusion of male and female gametes; !Lamb K is genetically identical to Sheep

J as the genetic material was obtainedfrom a cell in Sheep J; !

(d) Lamb K will be female as it is geneticallyidentical to Sheep J which is female; !

There would be 2 X chromosomes in thenucleus placed into the ovum, and henceLamb K will develop into a female; !

(e) Can ascertain success of procedure, aslamb will not resemble the other sheep asit has not inherited any genetic materialfrom it OR Sometimes donor of ovum maynot have a healthy uterus in which thezygotes can implant successfully; !



7/13/09

2

Unit 3.4B Question 1a !• Mouse 2" Mouse 3!" X" Y" " X" X!• 50%!

3.4B Qn1b & c !• (b)(i)" Mouse 2" " (ii)" Mouse 3!• Let B represent dominant allele for black coat !• Let b represent recessive allele for white coat !• P Phenotype: "Black (5) " Black !• P Genotype: " Bb" " x" Bb!• Gametes:"" B" b" " B" b!

• F1 Genotype:" BB" Bb" " Bb" bb!• F1 Phenotype:"Black" Black" " Black" White!• Ratio:" " Black"" :" White !" " " " " 3" :" 1!

3.4B Qn1d !• Let W represent dominant allele for short whiskers!• Let w represent rece ssive allele for long whisker s!• P Phenotype:"Short (3) " Long (4)!• P Genotype:" Ww" x" ww!• Gametes:"" W" w" " w" w!

• F1 genotype:" Ww" Ww" " ww" ww!• F1 Phenotype:"Short"Short"" Long" Long!• Ratio:" " 1 Short" :" 1 Long!" " " " " !

3.4B Qn 1d !• Let T represent dominant allele for long tail !• Let t represent recessive allele for short tail " !• Parents’Phenotype: "Short (3) " Long (4) !• Parents’Genotype:" tt "" x" Tt!• Gametes:"" t " t " " T" t !

• Offspring Genotype: "Tt" tt " " Tt" tt !• Offspring Phenotype: Long" Short "" Long" Short !• Ratio:" " 1 Short" :" 1 Long!" " " " " !

Unit 3.4B Question2 !Ai) Red Blood Cell !

ii) Haemoglobin !

b) Oxygen carrying capacity of blood is reduced asHaemoglobin S (HbS) is less efcient at carryingoxygen compared to normal haemoglobin. . Also,less oxygen diffuses into the cells due to adecrease in the surface area of the crescentshaped sickled cells as compared to the biconcaveRBC.!

Note: RBCs un dergo lysis and aggregate, increasingthe viscosity the blood. !

Unit 3.4B Question2 !

c) Mutation in the gene that codes forHaemoglobin. (HbS instead of HbA) !

d) Bb (sickle cell trait) !

e) Bb x Bb (25% chance of offsprings, bb) !



7/13/09

3

Unit 3.4B Question 3 !(a) Oval leaves!Bi) Let S represent allele for spear -shape d leaves!

Let s represent allele for roun d leaves!• P Phenotype:"Spear (D) " Round (E)!• P Genotype:" SS "" x" ss!• Gametes:"" S" S" " s" s!

• F1Genotype:" Ss " Ss " " Ss " Ss!• F1Phenotype:" Oval" Oval" " Oval" Oval!• Genotype of offspring is heterozygous.!

ii) BothS and s alleles express t hemselves (incompletedominance) leading to a phenotype that isinterme diate between t he two (oval leaves).!

Unit 3.4B Qn 5a !Father Mother

• Parents’ Phenotype: Normal " " Normal " !•

Parents’ Genotype:"

Aa"

x"

Aa!

• Gametes:"" " A" a" " A" a!

• Offspring Genotype :" AA" Aa" " Aa" aa !• Offspring Phenotype : Normal Normal Normal Disease!• Ratio: " " " 3 Normal: 1 Disease!

Unit 3.4B Question 5b !i) The gene is a small segment of DNA, a

hereditary factor borne on a particularlocus in a chromoso me that controls aparticular characteristic by coding for aprotein.!

ii)A mutation is a sudden change in a gene,chromosome structure, or chromosomenumber.!

c) Gp AB require both gp A and gp B/IA &IB alleles;!

No IB/group B in population;!

3.4A Essay Question 1a !• For discontinuous variation, the particular

characteristic of the organism falls into denitecategories, with no continuous pattern in apopulation. Examples: ABO blood group (4categories) !

• This is due to the characteristic being determinedby a single gene which expression is not affectedby the environment. !

• For continuous variation, the characteristic wouldfall into a range of values / intermediates.Examples: Height, weight !

• This is due to the characteristic being determined

by many genes where their expression is affectedby the environment. !

3.4B Essay Question 1ai) !• Alleles I A & I B are co-dominant. A combination of

both gives rise to a person w ith blood type AB. !• Phenotype: " Type A " " Type B !• Parents: " " I AI O "" x" I BI O!• Gametes: "" I A" I O" " I B" I O!

• Offspring: " I AI B "I AI O "" I BI O "I OI O!• Phenotype: " AB" A" " B" O!

Essay Question 1aii !• It is possible to have a child with the recessive

phenotype if both parents have a copy of therecessive allele (I o), ie they are heterozygous. !

• Phenotype: " Type B " " Type B !• Parents: " " I BI O "" x" I BI O!• Gametes: "" I B" I O" " I B" I O!

• Offspring: " I BI B " I BI O "" I BI O "I OI O!• Phenotype: " B" B" " B" O!• RATIO: " " 3 B Blood group : 1 O Blood group !