B.Sc. Microbiology/Biotech II Cell biology and Genetics Unit 5 microbial genetics
09 TYS Unit3.4 cell division and genetics
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7/13/09 1 TYS Unit 3.4 A & B Genetics Unit 3.4A Question 1 a) R; " V ariety P and Q have stripes an d spots on their bodies respectively that serve as a camouflage among the wat er plants in the river; " However, Variety R will stand out and will be easily spotted by predators; Unit 3.4A Question 1 Bi) Stripes with spots on them; ii) Variety S has the best camou flage among the water plants and hence is the most adapted compared to P ,Q,R and S; " Hence, their survival rate increases, Hence, reproductive rate increases; Advantageous alleles are passed on to offspring. Hence, they survive better thus causing an increase in numbe rs of S; Unit 3.4A Question 1c • Variety R may be completely wiped out as it is most predated upon; • Numbers of variety P and Q will decrease; • As they mate with S, more offsprings will inherit the charac teristics of S an d w ill sur vive bett er w ith the best camouflage; Unit 3.4A Question 2 (a) R: fertilization; " S: zygote; (b) (i) 27; " (ii) 54; (c) Lamb K was produced w ithout requiring the fusion of male and female gametes; Lamb K is genetically identical to Sheep J as the genetic mat erial was obt ained from a cell in Sheep J; (d) Lamb K will be female as it is genetically identical to Sheep J which is female; There would be 2 X chromosomes in the nucleus pla ced into the ovum, a nd hence Lamb K will develop into a female; (e) Can ascertain success of procedure, as lamb will not resemble the other sheep as it has not inherited any genetic material from it OR Sometimes donor of ovum may not have a healthy uterus in which the zygotes can implant successfully;
Transcript of 09 TYS Unit3.4 cell division and genetics
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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; !
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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) !
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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 !
