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## Homework Statement

A researcher studied six independently assorting genes

in a plant. Each gene has a dominant and a recessive allele:

R black stem, r red stem; D tall plant, d dwarf plant; C full

pods, c constricted pods; O round fruit, o oval fruit; H hairless

leaves, h hairy leaves; W purple fl ower, w white fl ower. From

the cross (P1) Rr Dd cc Oo Hh Ww (P2) Rr dd Cc oo Hh ww,

(a) How many kinds of gametes can be formed by P1?

(b) How many genotypes are possible among the progeny of

this cross?

(c) How many phenotypes are possible among the progeny?

(d) What is the probability of obtaining the Rr Dd cc Oo hh ww

genotype in the progeny?

(e) What is the probability of obtaining a black, dwarf, constricted,

oval, hairy, purple phenotype in the progeny?

**2. The attempt at a solution**

(a)2

(a)

^{n}= no. of gametes where n is the no. of heterozygous loci.

Ans:

__Rr__

__Dd__cc

__Oo__

__Hh__

__Ww__

2

^{5}=32 gametes.

**(b)**I'm confused as how to solve this.

According to Griffith (sum no. 3c) no. of genotypes = 3

^{n}, n-is the no. of genes (and not heterozygous genes).

Is this formula correct?

I tried to check. Like, in monohybrid cross, 3

^{1}=3 and dihybrid cross 3

^{2}=9. But there the loci were also heterozygous. While crossing two individuals AABb and AaBB (both heterozygous for one lous), AABB, AaBB, AABb and AaBb were the progenies. Here unlike the formula (3

^{2}=9) 4 different genotypes are obtained.

So here I'm confused as what the formulae should be.

**I don't know what the formulae is. For straight monohybrid and dihybrid crosses though 3**

(c)

(c)

^{n}works, there were 3 and 9 phenotypes respectively.

**(d)**Can this be found out without a Punnett square?

**(e)**Seems related to d.

Any help will be much appreciated.

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