As a team of explorers travels southward from the North Pole into Canada, they encounter a dense forest of pines and other coniferousevergreens. They leave the forest before the winter, which is long and cold. Which of these biomes has the team visited?
o A. temperate grasslands
B. Northwest coniferous forest
C. tropical dry forest
O
D. boreal forest (taiga)

Answer:

The answer is (d) Three genes, each with two alleles that act additively.

Explanation:

1/ The genotype of P is AABBCC and aabbcc, for example. Then, F1 should be AaBbCc.

The genotype of individuals with n heterozygous pair is 2^n. Thus, in this case, the number of genotype in F2 should be 2^3 * 2^3 = 8 * 8 = 64. We can get this conclusion by analyzing the number of 6 cm tails in F2: 1/64 with genotype aabbcc, and the number of 30 cm tails: 1/64 with genotype AABBCC. These two genotypes is as same as the ancestor in this experiment.

The genotype of an F2 piglet with an 18 cm tail is AaBbCc. If these genes show dominance, the tail lenght of F2 will be 30 cm. And there are 7 possible phenotypes. Thus, we can conclude that the genes act additively.

2/ 18 cm tail F2: AaBbCc, and 6 cm tail: aabbcc.

The offspring genotypes are:

• 18 cm AaBbCc, 14 cm AaBbcc, 14 cm AabbCc, 10 cm Aabbcc
• 14 cm aaBbCc, 10 cm aaBbcc, 10 cm aabbCc, 6 cm aabbcc

The frequency of piglets with 14 cm tail length should be 3/8 = 37.5%.

Answer and Explanation:

1) These results are consistent with option (d)Three genes, each with two alleles that act additively .

Quantitative heritability: Refers to the transmission of a phenotypic trait in which expression depends on the additive effect of a series of genes.

Polygenic heritability occurs when a trait is due to the action of more than one gene that can also have more than two alleles. This can cause many different combinations that are the reason for genotypic graduation.

Quantitative traits are those that can be measure, such as longitude, weight, eggs laid per female, among others. These characters do not group individuals by precise and clear categories. Instead, they group individuals in many different categories that depend on how the genes were intercrossed and distributed during meiosis. The result depends on how each allele of each gene contributed to the final phenotype and genotype.

In the exposed example, there are 7 different phenotypes  (6, 10, 14, 18, 22, 26, 30) and the majority of F2 individuals have 18-lengthed tails. This might be the heterozygotic phenotype for all the genes.The rest of the phenotypes are possible combinations of genes and their alleles. There are six possible combinations (apart from the 18 lenght form). This leads to 3 genes and two alleles in each gene.

So, until now we have:

• A pig with a 6 cm tail was mated to a pig with a 30 cm tail. All F1 offspring had a tail of length 18 cm. This is:

Parental)     6cm    x    30cm

F1)                       18 cm  

• An F2 generation was made by crossing F1 individuals. This resulted in many piglets whose tails ranged in 4-cm intervals from 6 to 30 cm (6, 10, 14, 18, 22, 26, 30). This is:

Parental)       18 cm   x   18 cm

F2)               6, 10, 14, 18, 22, 26, 30

• The majority of pigs had 18 cm tails, this means that 18 cm phenotype is a heterozygote for each gene

• 1/64 had 6 cm tails, and 1/64 had 30 cm tails, this means that these two phenotypes are the extreme traits, that is the recessive homozygote and the dominant homozygote.

• There are 7 phenotypes, one of them is the recessive form, the other is the dominant form and the majority is the heterozygotic form for every intervening gene. There are three genes with two alleles each:

Gene 1: allele A and a

Gene 2: allele B and b

Gene 3: Allele C and c

Phenotypes:

aabbcc: homozygotic recessive form

AABBCC: homozygotic dominant form

AaBbCc: heterozygotic form for every intervening gene

If the recessive form is 6cm length, and each dominant allele contributes in 4 cm to each phenotype, we get:  

• 6 cm length = aabbcc (1/64)

• 10cm length = Aabbcc  (A contributes 4cm)

• 14cm length = AaBbcc  (A and B contribute 4 cm each)

• 18cm length = AaBbCc (The majority) (A, B and C contribute 4cm each)

• 22cm length = AABbCc

• 26cm length = AABBCc

• 30cm length = AABBCC (1/64)

2) An F2 piglet with an 18 cm tail is heterozygous at each of the tail-length loci, and is mated to an F2 piglet with a 6 cm tail.  

Parental)       AaBbCc     x     aabbcc

Gametes)   ABC  ABc  AbC  Abc  aBC  aBc  abC  abc

                  abc   abc   abc   abc   abc   abc  abc  abc

Punnet square)  

            ABC        ABc       AbC       Abc      aBC       aBc        abC      abc

abc    AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc  

abc    AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc

abc   AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc

abc    AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc          

abc    AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc          

abc    AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc          

abc    AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc      

abc    AaBbCc AaBbcc AabbCc Aabbcc aaBbCc aaBbcc aabbCc aabbcc          

F3 genotype and phenotype)

8/64 AaBbCc = 18 cm

8/64 AaBbcc = 14cm

8/64 AabbCc = 14cm

8/64 Aabbcc = 10cm

8/64 aaBbCc = 14cm

8/64 aaBbcc = 10cm

8/64 aabbCc = 10cm

8/64 aabbcc = 6cm  

Each dominant allele contributes 4cm to the recessive homozygote form for each phenotype.