Question

Please help me answer this question. It matters a lot that I get the answer correct. Please do not answer if you're not 100% sure, thank you very much for understanding.

1. S. superis is a rare flowering plant that can produce blooms in two colours, purple (P) and teal (p), and two shapes, four petals (F) and seven petals (f).  Purple is dominant to teal, and four is dominant to seven petals. S. superis is a diploid organism, 2n  = 4, that can both self-fertilize and mate with other plants of the same species (out-cross).  You find one of these rare plants in your garden and it has purple flowers with four petals.  Assuming this lonely plant must self-fertilize what potential offspring phenotypes could you see?

Select one:

a. Purple with seven petals

b. Purple with four petals

c. Teal with seven petals

d. All of the phenotypes listed are possible depending on the genotype of the plant.

and ,

Your friend is working with S. superis plants in the lab and has set up a cross between a plant with four purple petals and a plant with seven teal petals. Of the 200 offspring from the cross all have purple petals but about 1/2 have four petals and 1/2 have seven petals.

If you were to write out the cross that your friend set up what would it look like genotypically?

Select one:

a. PPFF X ppff

b. PpFf X ppff

c. PpFF X ppff

d. PPFf X ppff

e. PpFf X PpFf

2.

Merfolk have three pairs of autosomes and one sex chromosome; males are X0 and females are XX.

In merfolk the genes controlling tail fin shape and scale colour are on different autosomes. Solid tail fin (Ff) is dominant to forked tail fin (ff), and the F allele is recessive lethal. There are two alleles of the scale colour gene – C^B(blue) and C^R (red)– C^B and C^R show incomplete dominance with heterozygotes being purple.

The silent trait is X-linked and dominant (X^S) to normal voice (X⁺).

Atolla is a male merfolk, he has a forked tail fin, blue scales, and is voiceless (silent). Brachy is a female merfolk with a solid tail fin, purple scales and a normal voice.

a.

What is the probability that a female child of a mating between Brachy and Atolla will be silent (it doesn't matter what tail shape or scale colour the child has)?

Select one:

a. 1

b. 0

c. 1/2

d. 3/4

e. 1/4

b.

What is the probability that a mating between Brachy and Atolla would produce a child with a forked tail and purple scales (doesn’t matter whether they are silent or not)?

Select one:

a. 1/2

b. 2/3

c. 1/3

d. 1/4

e. 3/4

c.

You mate a female merfolk with blue scales who is heterozygote for silent and a carrier of forked tail, to a silent male with blue scales who is also a carrier of forked tail. Which of the following are TRUE statements (you can choose more than one)

For this question ignore the phenotypes that are unmentioned. For example "a blue female" is a blue female who can be any colour, and either be silent or normal.

Select one or more:

a. This mating can produce silent or normal voiced female offspring

b. The probability of an offspring from this mating having a forked tail and normal voice is 1/6

c. This mating can produce silent or normal voiced offspring

d. The probability of a female offspring having a forked tail and blue scales is 1/3

Answer 1

Ans 1) given information are :

Flower color --- purple = P ------ dominant

Teal = p ------- recessive

Shape ----- four petals = F ------ dominant

Seven petals = f ------- recessive

And the plant is able to self fertilize as well as cross fertilize.

A purple color and four petal flower plant is in garden and is self fertilizing so the types of progeny it will produce.

The flower is purple colored so can be PP or Pp and has foure petals so can be FF or FF.

So possible Genotype of the flower are = PPFF, PpFf, PpFF and PPFf. If the flower with anyone of these Genotypes is fertilized is capable of producing all the types of progenies given in the option.

So, option d is correct answer.

Ans 1b) my friend crossed 4 petal purple color flower plant with 7petal teal colored flower plant and got ALL PURPLE colored flower , 1/2 have 4 petals and 1/2 have 7 petals.

For this the possible Genotypes of the plant will be

4 petal purple plant = PPFF, PpFf, PPFf,PpFF

7 petal teal plant = ppff.

And to get the progenies as above with all purple colored flower the Genotype of 4 petal purple flower will be PP as if it will be heterozygous then the progenies would have teal flowers by crossing with the P2 7petal teal flower plant. So, 4 petal purple flower will be PP.

The progenies have both 4 and 7 petals this means the 4 petal purple flower is heterozygous Ff. As only in this case when 4 petal is crossed with 7 petal FF will produce both number petals.

So, the Genotype of the parents are PPFf X ppff

So, Option d is correct answer.

Ans 2) the given information about merfolk are

Male = X0

Female = XX

1) tail == solid tail = FF ---- dominant

Forked tail fin = ff

F = recessive lethal == if FF it is lethal, can't survive

2) Scale color == CB = Blue

CR = Red

CBCR = purple. --- incomplete dominance

3) Voice == trait is X-linked

XS == Silent ------ dominant

X+ == Normal voice ----- recessive

Atolla = Male merfolk

If has forked tail fin = ff; Blue scale color = CBCB; and is silent = XS.

Genotype = ff, CBCB, XS.

Brachy = female merfolk

It has solid tail = FF; purple scale color = CBCR; and has normal voice = X+X+ (recessive trait)

Ans 2 a) when Atolla and Brachy are crossed, the Probability of getting a FEMALE child Silent is

Atolla = XS X Brachy = X+X+

	X+	X+
XS	XSX+	XSX+

Both females are silent as silent is Dominant over normal voice.

Probability = 2/2 or 1.

Answer is Option a.

Ans 2 b) probability of getting a child with forked tail and purple scale color without considering the voice trait.

Atolla = forked fin= ff ; Blue scale = CBCB == ffCBCB

Brachy = solid tail = Ff; purple scale= CBCR == FfCBCR

By using forked line method

Answer is Option d.

Ans 2 c) given information are :

Femlae merfolk

Blue scale = CBCB

Silent = heterozygote = XSX+

Forked tail = carrier = Ff

Genotype = Ff CBCB XSX+

Make merfolk

Blue scale = CBCB.

Silent = XS

Forked tail = Carrier = Ff

Genotype = Ff CBCB XS

Statement 1 -- mating can produce silent or normal voice offspring.

Cross XS with XSX+

	XS	X+
XS	XSXS	XSX+

Conclusion = no, mating can't produce normal voice femlae offspring.

Statement 2 -- forked tail and normal voice Probability is 1/6 from this cross.

Ff XSX+ female crossed with FfXS male

Conclusion = statement is wrong. Probability is 1/4 of forked tail and normal voice offspring.

Statement 3-- this mating can produce normal voice and silent offspring.

Yes, it can produce both kinds of offsprings. As shown in image above of cross of only voice trait.

Statement 4 -- probability of female offspring having forked tail and blue scales is 1/3.

F = femlae ; M= male ; (1*1, 2*1, 1*2, 2*1 ,etc likewise means crossed female with the numbered male Gamete)

Conclusion = false, the Probability of female offspring to have forked tail and blue scales is 1/2.

Answer is Option 3.

Thank you....???