# biology sample part 1 by 7Jg4vq

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• pg 1
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Name ___________________________________

Date ____________________________________

Problem 1
a.      Dominant allele is __Black(B)___________
b.      Symbols for the alleles: Yellow _____y_____ Black ____B______
c.      Explanation:
The black color is due to dominant allele B. The phenotypic expression of recessive allele y is only
possible in the absence of allele B.As the cross between a black and yellow produces mostly black
puppies, the black color is dominant.

d.      Percentage of yellow puppies: ____25%_________
Work:

Mother

B                 y

B       BB                  By

Father           By                  yy                                               i
y

¼ x 100=25%
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Problem 2 (Refer to on-line answer sheet for full form)

Copper & White---Brown and White           Copper---Brown & White         Grandparents

__ __ cW __ __        __ _CW_ __ __           __ _cc_ __ __         __ _CW_ __ __

Brown --------------------------------Brown & White           Parents

__ _Cc_ __ __                          __ _CW_ __ __

_____________________________________

Brown      Brown & White        Brown       Copper & White       Babies

__ _CC_ __ __      __ CW__ __ __       __ _Cc_ __ __         __ _CW_ __ __
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Problem 3
a.) Ellen's genotype: ___Tt_______ Edward's genotype: ____Tt_____
b.) Chance for a normal child: __________
Work:                                          Ellen
T                t
Edward       T     TT                 Tt
t     Tt                 tt

Tay Sachs disease will occur only when the genotype is tt. Therefore the probability is 50% as the
allele of combination to be formed as tt rather than safe Tt or TT.

BONUS: The recessive allele of the Tay Sach’s Disease always remains in the gene pool of humans in
heterozygous forms eg.Tt. The allele expresses itself when in homozygous recessive form i.e.tt. Thus
even the individuals carrying the expressive form die, the allele is conserved in recessive form.
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Problem 4

Mr. and Mrs. Smith get: ________baby 1 and 4____________________
Mr. and Mrs. Abernathy get: ____baby 2 and 3______________________
For Mr and Mrs Smith

IA              i

IB   IAIB            IB i

i   IA i            ii

All the four blood groups A,b,AB,and Oare possible.

For Mr and Mrs Abernathy,

IA               IB

i    IAi             IB i

i   IA i            IB i

Only A and B blood groups are possible.

Therefore the child 1 and 4 cannot belong to Mr. and mrs Abernathy.
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Problem 5

Could Man #1 be this baby's father? ___No______
Why or why not?

The man cannot be the father of the girl because the X chromosome of the girl comes from
her father should be defective. The father has normal vision which means that the X
chromosome is not defective.

Blood group analysis does not prove this fact in his case.

Could Man #2 be this baby's father? _____no____
Why or why not?

IA                     IB

IA    IAIA              IAIB

IA i              IB i
i

The blood group off none of the possible offspring comes to be O. Hence the man cannot be the father of the
child.

Could Man #3 be this baby's father? ___yes______
Why or why not?
6

IA             i

IB       IAIB                  IBi

i    IA i                  ii

Since the blood group of the offspring can match with the above Punnet analysis the man can be the
possible father. Morever, the X chromosome is affected as he is color blind which is transmitted to the girl.

Problem 6

a.) Ethyl's Genotype: ____Eett_______ Egbert's Genotype: __EeTt_________
b.) Fraction of children with same phenotype as Ethyl: ____12.5%_________
Work:                ET         eT          et tE
ET       EETT       EeTT    ETet    EETt

eT       EeTT       eeTT    eeTt    EeTt
et       EeTt       eeTt    eett    Eett
Et       EETt       EeTt    Eett    EEtt

2/16 x100=12.5%

Problem 7

a.) Female fly's genotype: ___Ww_____ Male fly's genotype: ___WW______
b.) Phenotypic Ratio: ______3:1__________________
Work:
7
W                 w female

male           WW                   Ww             W

W     WW                   Ww

Problem 8

a.) Genotype of F1 flies: _Cncn__________ Phenotype: ____Normal________
b.) Phenotypic Ratio: ______3:1______________
Work:

CnCn x cncn         parents

F1= Cncn

F2=                 Cn    cn     Cn       cn
Cn     CnCn   Cncn   CnCn    Cncn

cn     Cncn   cncn   Cncn    cncn

Cn     CnCn   CNcn   CnCn    CNcn

cn
Cncn   cncn   Cncn    cncn

Problem 9

a.) Parental flower colors: ___Red_________ and ____white________
b.) F1 flower color: ___Pink___(CRCW)_____________
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c.) F2 Phenotypic Ratio: _____1:2:1__________RED;PINK:WHITE_____________________
Work:
CR          CW
CR      CRCR           CRCW
CW      CRCW           CWCW

d.) After 10 generations of selection, phenotypic ratio will be:
_____1:2:1_____________________. The ratio remains the same because none of the alleles is
dominant.

Problem 10

a.) Gerbil colors: BBdd: __Tan__________ bbDD: ___grey________
b.) BbDd color: ___black____________
c.) Fraction of tan gerbils in F2: ____25%________
Work:
:                 BD      Bd     bD      bd
BD         BBDD   BBDd   BbDD    BbDd

Bd         BBDd   BBdd   BbDd    Bbdd

BbDD   BDbd   bbDD    bbDd
bD
BdDd   Bdbd   bbDd    bbdd
bd

fraction of tan gerbils:4/16x100=25%

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