Question

Chapter 4: Extensions of Mendelian Inheritance Help Seve & You cross a rose comb chicken with a peacomb chicken and get all walnut comb progeny. You then cross two of the walnut combed chickens from the F1 generation and get 9 walnut combs: 3 rose combus:3pea combs : 1 single comb. Which of the following would best explain these results? Underline means any allele could be present. For example R could be RR or Rr Skoped Rose comb Pea comb Walnut comb Single corb Multiple Choice O walnut-RP Segler, rose R DO < Prey 4 of 62 Next >

Answer 1

Answer to the question is option A. Walnut=R_P_ ; single= rrpp; Rose = R_pp; pea = rrP_.

This is an example of quantative traits. Quantative means the quantity of genes or number of genes present. Here, in chicken comb case we are getting a typical Mendelian ratio as obtained in dihybrid cross in F2 generation. We get,

9= walnut comb

3= rose comb

3 = pea comb

1 = single comb

On observing these progenies we can conclude that single comb is recessive trait and expressed in Recessive condition i.e. alleles. Walnut comb has maximum progeny hence would be the dominant trait.

In F1 generation as given in question, that on crossing a rose comb chicken with pea comb chicken we get all walnut comb chicken. As we recall the dihybrid cross of pea plants performed by Mendel in his experiments took a yellow round(RRYY) and green wrinkled (rryy). On crossing we get a hybrid as RrYy ( yellow and Round). Here 2 genes are responsible for 2 traits i.e. seed shape and seed color. The same kind is implied here in chicken comb but here the 2 genes are responsible for a single character i.e. the comb shape.

This type of traits are quantative trait in which a single trait is responsible because of differe genes pairs as here R,r and Pp.

So, take rose comb allele = R;

Pea comb allele= P;

In F1, when rose comb is crossed with pea comb chicken we get all walnut type. This is possible only when rose comb have genotype = RRpp and pea comb = rrPP.

RRpp X rrPP

We will get, RrPp. All progenies are RrPp, walnut comb; this shows R is Dominant over r and P dominant over p. Hence, walnut comb is Dominant trait and in a population it can be expressed in genotypes as RrPp, RRPP, RrPP and RRPp.

In F2 we self crossed the F1 progeny that is RrPp, we get progenies in ratio =9:3:3:1.

Cross RrPp X RrPp

Possible Gametes from both the parents are :

RP, Rp, rP, rp.

	RP	Rp	rP	rp
RP	RRPP	RRPp	RrPP	RrPp
Rp	RRPp	RRpp	RrPp	Rrpp
rP	RrPP	RrPp	rrPP	rrPp
rp	RrPp	Rrpp	rrPp	rrpp

All colored Red are Walnut comb =9

All colored yellow are Rose comb =3

All colored green are pea comb = 3

One with no color or say white is single comb =1

This shows that when both pairs of gene are dominant it will have walnut comb, when both are completely recessive it will be single comb, when rose (R) is dominant and pp(recessive) present along will have rose comb, and when P is Dominant and rr(recessive) will have pea comb.

So, walnut comb exist when genotypes can be --- RRPP, RrPP, RRPp, RrPp.

Rose comb = RRpp or RRpp.

Pea comb = rrPP or rrPp.

Single comb = rrpp.

To derive at the answer we can use options also as assuming that as the answer. Good luck!!!

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