<em>A</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em>
YY
<em>E</em><em>x</em><em>p</em><em>l</em><em>a</em><em>n</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em>:</em>
Y-purple while y-white
YY is the genotype of a pure breed purple flowered pea plant
Yy is the genotype of a heterozygous pea plant flower (purple is dominant over white)
yy is the genotype of a white flowered pea plant that is generally called as recessive
Observation is broken up into three parts. The first is the date. Then you have the activity and last you have a brief description of what is being observed.
<span>The sequence of alternation of generation is; gametes->zygote->sporophyte->spores->gametophyte->gametes.
The attached diagram shows clearly this looped cycle. Alternation of generation
occurs in a more advanced land plant that
has distinct
haploid and diploid phases in their life
cycle. The diploid phase usually involves
the sporophyte while the haploid phase involves the gametophyte</span>
The answers would be:
Genotype Phenotype
Tt Tall stemmed
tt Short stemmed
Genotypic ratio : 2:2 or 1:1
Phenotypic ratio: 2:2 or 1:1
<u />
<u>You can read on to see how this was done:</u>
Tall stems (T) are dominant to short stems (t).
First figure out the genotypes of the parents. We have a short-stemmed plant and a heterozygous long-stemmed plant cross.
For short stem to occur, you need 2 pairs of short alleles. So the first parent would have a genotype of tt.
Heterozygous long-stemmed means that the parent has one of each allele. So the genotype of the second parent would be, Tt.
Now we can make our Punnett Square.
tt x Tt
<u> t t </u>
<u>T | Tt | Tt</u>
<u>t | tt | tt</u>
Let's list down the genotypes and phenotypic results.
Genotype no. Phenotype
Tt 2 Tall stemmed
tt 2 Short stemmed
So from that we can answer the other questions:
Genotypic ratio : 2:2 or 1:1
Phenotypic ratio: 2:2 or 1:1
