Question

(More Genetics Problems and Difficult Genetics Problems). MORE GENETICS PROBLEMS 1. When tall plants mate with dwarf plants, only tall plants occur in the offspring. Assign symbols and show the phenotypes and genotypes in the parental (P) and first filial generations for a cross between tall plants and dwarf plants that produces only tall plants. 2. Show the expected outcome when the F1 plants in problem #1 are crossed. Be sure to give both phenotype and genotype ratios for these second filial (F2) generation plants. 3. Assume that in aardvarks gray fur dominates over brown fur. If, in a certain litter, half of the offspring show gray fur and half brown fur, indicate what the parents' genotypes and phenotypes probably were. 4. In humans, six fingers (F) is the dominant trait over five fingers (). This illustrates the often overlooked fact that dominance does not necessarily mean commonness of occurrence. If both parents are heterozygous for six fingers, what is the probability of their having six-fingered children? Five-fingered children? 5. In one of his experiments, Mendel crossed plants with round, yellow seeds with plants having wrinkled, green seeds. The resulting plants all had round, yellow seeds. When these seeds were used, the plants growing from them were crossed among themselves, and the offspring showed 4 phenotypes. 315 of the seeds were round and yellow, 101 were wrinkled and yellow, 108 were round and green, and 32 were wrinkled and green. Use appropriate symbols and method to explain these results. 6. In the Japanese four o'clock, flower color results from genes showing incomplete dominance or blending. For example, a cross between a homozygous red and a homozygous white will always result in a pink flower. What is the probable outcome (offspring phenotypes) of a cross between two pink flowers? Between a red and a pink flower? .. .. . . . . ......aunt ler (h) and the trotting gait(Tdominates

Answer 1

Q1) Here in question-1 tall plants are crossed with dwarf plants, tall character is dominant over dwarfness, as a single character depends on two alleles one allele from father and one allele from mother. Here tallness can be symbolised as “TT” and dwarfness can be symbolised as “tt”, they produced tall plants that means in the parental generation or in “P” generation the phenotype of plants are one is tall and its genotype is “TT” and the other parent is phenotypically dwarf and its genotype is “tt”, they are both homozygous for the characters, because when they are crossed they produced only tall plants.

Here when “TT” crossed with “tt”

“TT” produced a single type of gamete that is “T” and “tt” also produced a single type of gamete that is “t”

It can be seen from the below punnette square

	T	T
t	Tt	Tt
t	Tt	Tt

The offspring produced from a cross between two individuals from parental generation is called first filial generation or F1 generation

Here the phenotype of F1 generation is tallness and the genotype of F1 generation is “Tt”.

Phenotype is the external appearance or physical appearance and genotype is the internal nature of the individual or here the plants.

Q2) In question-2, F1 plants of question-1 are crossed that means “Tt” crossed with “Tt”

They produced two types of gametes they are “T” and “t”

When they are crossed the result will be as shown in the below punnette square

	T	t
T	TT	Tt
t	Tt	tt

Here in F2 generation there will be 3 tall plants and 1 dwarf plant.

The phenotypic ratio is 3:1, because external appearance of 3 plants are tall and one plant is dwarf.

The genotypic ratio is 1:2:1 because here when the genetic makeup of the plants seen, it is seen one plant is pure tall or homozygous for the character tallness because it gets the tall gene from both the parents, 2 plants are heterozygous tall because they get one tall gene from one parent and one dwarf gene from the other parent, and one is pure dwarf or homozygous dwarf in nature which gets one dwarf gene from each parent.

Q3) In question-3 it is said that in case of aardvarks grey fur is dominant over brown fur, suppose grey fur is symbolised by the alleles “GG” and brown fur by “gg”, here it is given that in a certain litter half of the offspring show grey fur and half show brown fur, this means the phenotypic as well as the genotypic ratios are 1:1, this is seen in case of test crosses.

A test cross is a cross when the heterozygote individual crossed with a homozygous recessive individual. Here the parent genotype are “Gg” and “gg” when they are crossed they produced half grey and half brown individuals.

“Gg” produced two types of gametes that is “G” and”g”, “gg” produced only one type of gamete that is “g”.

It can be seen in the below punnette square.

	G	g
g	Gg	gg
g	Gg	gg

So 50% of offspring are grey and 50% are brown.

So parent’s genotype are “Gg” and “gg” and phenotype are grey and brown.

Here both phenotypic and genotypic ratios are same that is 1:1.

Q4) Here in question-4 it is said that in humans six finger (F) character is dominant over five finger character (f) . In the question both the parent are heterozygous that is both have genotype “Ff” it is asked about the probability of having six fingered children and five fingered children.

When “Ff” crossed with ”Ff” the result can be seen in the below punnette square.

	F	f
F	FF	Ff
f	Ff	ff

In this cross there are three events taking place they are

1. Two F gametes unite (to form FF genotype)
2. F gamete from mother unites with f gamete of father (giving Ff genotype)
3. F gamete from mother unites with F gamete from father (giving rise to Ff genotype).

Here sum rule of probability can be applied because events are mutually exclusive (cannot happen at the same time), each individual event has a probability of ¼, so probability of having a offspring with dominant phenotype here that is having a child with six fingers will be [ (probability of getting F from mother and F from father) + (probability of ” F “from father and” f” from mother)+ (probability of getting “f” from father and “F” from mother)] = [ ¼ + ¼ +1/4] = ¾

When ¾ is multiplied with 100 the percentage will be 75%.

And the probability of having a five fingered child which is a recessive character will be ¼ or 25% because in only one case it can happen that is when the child gets a “f” or recessive gene from his father as well as a “f” gene from his mother.

From the punnette square also the probability of having six fingered children and five fingered children can be accessed, that is the probability of having a six fingered children is 75% and probability of having five fingered children is 25%.