1. In a series of crosses with peas that have round seeds and yellow cotyledons
ID: 51246 • Letter: 1
Question
1. In a series of crosses with peas that have round seeds and yellow cotyledons with pea plants that that have wrinkled seeds and green cotyledons, the following was seen:
Parents progeny
round, yellow x round, yellow ¾ round, yellow
¼ wrinkled, yellow
wrinkled yellow x round, yellow 6/16 wrinkled, yellow
2/16 wrinkled, green
6/16 round, yellow
2/16 round green
Determine the genotypes of the parents.
2. What is the probability of having 3 girls in a family of 8 children?
3. What is the probability of having at least 3 girls in a family of 8 children?
Please explain your answers! Thanks
Explanation / Answer
1. In a series of crosses with peas that have round seeds and yellow cotyledons with pea plants that that have wrinkled seeds and green cotyledons, the following was seen:
Parents Progeny
1)round, yellow x round, yellow ¾ round, yellow ¼ wrinkled, yellow
2)wrinkled yellow x round, yellow 6/16 wrinkled, yellow
2/16 wrinkled, green
6/16 round, yellow
2/16 round green
In first case where the cross is between Round, yellow x round, yellow we see that none of the progeny has Green phenotype so the gametes should be (in Bold)
RY
rY
RY
RRYY(Round Yellow)
RrYY(Round Yellow)
rY
RrYY(Round Yellow)
rrYY(Wrinkled Yellow)
Which on crossing give us the desired Progeny phenotype
So the parents should be:
RrYY X RrYY
In the second case we see 6/16 wrinkled, yellow
2/16 wrinkled, green
6/16 round, yellow
2/16 round, green
Phenotype which should have the following gametes(In Bold)
RY
Ry
rY
ry
rY
RrYY (Round Yellow)
RrYy (Round Yellow)
rrYY(Wrinkled, Yellow)
rrYy(Wrinkled, Yellow)
ry
RrYy (Round Yellow)
Rryy (Round green)
rrYy(Wrinkled, Yellow)
rryy(Wrinkled, Green)
rY
RrYY(Round Yellow)
RrYy ((Round Yellow)
rrYY(Wrinkled, Yellow)
rrYy(Wrinkled, Yellow)
ry
RrYy(Round Yellow)
Rryy(Round green)
rrYy(Wrinkled, Yellow)
rryy(Wrinkled, Green
This would give the Parent genotype as
wrinkled yellow x round, yellow rrYy XRrYy
2. Probability of having 3 girls in a family of 8 children.
We know the male is XY and female is XX, If we have a look at the probability of having a male or female it would be XY * XX, which would be XX,XX,XY,XY i.e. Probability of having a male (P(M))=Probability of having a female(P(F))=1/2.
Now out of a litter of 8 children, the probability of having 3 females would be:
=8C3 * P(F)3 *P(M)5
=8!/3!(8-3)! *( ½)3 *(1/2)5
=56*(1/8)*(1/32)
= 7/16
=0.43
3. Probability of having at least 3 girls in a family of 8 children.
We know that each child can be a male or female. So the total number of events is equal to 28=256
The question says atleast 3 girls, so the number of girls can be 3,4,5,6,7 or 8
Hence,8C3+8C4+8C5+8C6+8C7+8C8
=8!/3!5!+8!/4!4!+8!/5!3!+8!/6!2!+8!/7!1!+8!/8!0!
=56+70+56+28/3+8+1
=191+28/3
=601/768
=0.782
1. In a series of crosses with peas that have round seeds and yellow cotyledons with pea plants that that have wrinkled seeds and green cotyledons, the following was seen:
Parents Progeny
1)round, yellow x round, yellow ¾ round, yellow ¼ wrinkled, yellow
2)wrinkled yellow x round, yellow 6/16 wrinkled, yellow
2/16 wrinkled, green
6/16 round, yellow
2/16 round green
In first case where the cross is between Round, yellow x round, yellow we see that none of the progeny has Green phenotype so the gametes should be (in Bold)
RY
rY
RY
RRYY(Round Yellow)
RrYY(Round Yellow)
rY
RrYY(Round Yellow)
rrYY(Wrinkled Yellow)
Which on crossing give us the desired Progeny phenotype
So the parents should be:
RrYY X RrYY
In the second case we see 6/16 wrinkled, yellow
2/16 wrinkled, green
6/16 round, yellow
2/16 round, green
Phenotype which should have the following gametes(In Bold)
RY
Ry
rY
ry
rY
RrYY (Round Yellow)
RrYy (Round Yellow)
rrYY(Wrinkled, Yellow)
rrYy(Wrinkled, Yellow)
ry
RrYy (Round Yellow)
Rryy (Round green)
rrYy(Wrinkled, Yellow)
rryy(Wrinkled, Green)
rY
RrYY(Round Yellow)
RrYy ((Round Yellow)
rrYY(Wrinkled, Yellow)
rrYy(Wrinkled, Yellow)
ry
RrYy(Round Yellow)
Rryy(Round green)
rrYy(Wrinkled, Yellow)
rryy(Wrinkled, Green
This would give the Parent genotype as
wrinkled yellow x round, yellow rrYy XRrYy
2. Probability of having 3 girls in a family of 8 children.
We know the male is XY and female is XX, If we have a look at the probability of having a male or female it would be XY * XX, which would be XX,XX,XY,XY i.e. Probability of having a male (P(M))=Probability of having a female(P(F))=1/2.
Now out of a litter of 8 children, the probability of having 3 females would be:
=8C3 * P(F)3 *P(M)5
=8!/3!(8-3)! *( ½)3 *(1/2)5
=56*(1/8)*(1/32)
= 7/16
=0.43
3. Probability of having at least 3 girls in a family of 8 children.
We know that each child can be a male or female. So the total number of events is equal to 28=256
The question says atleast 3 girls, so the number of girls can be 3,4,5,6,7 or 8
Hence,8C3+8C4+8C5+8C6+8C7+8C8
=8!/3!5!+8!/4!4!+8!/5!3!+8!/6!2!+8!/7!1!+8!/8!0!
=56+70+56+28/3+8+1
=191+28/3
=601/768
=0.782
RY
rY
RY
RRYY(Round Yellow)
RrYY(Round Yellow)
rY
RrYY(Round Yellow)
rrYY(Wrinkled Yellow)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.