Tag: trihybrid cross

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Questions Related to trihybrid cross

Multiple choice trihybrid cross classical genetics botany

In humans, pointed eyebrows (B) are dominant over smooth eyebrows (b). Seemas father has pointed eyebrows, but she and her mother have smooth. What is the genotype of the father?

  1. BB

  2. Bb

  3. bb

  4. BbBb
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since pointed eyebrow (B) is dominant over smooth eyebrows (b). So, Seema can have smooth eyebrows only if her father is heterozygous for the dominant character. If her father has the genotype Bb and her mother is bb (smooth eyebrows) then her genotype will be either Bb (pointed) or bb(smooth eyebrows). If however, her father is homozygous for the trait the gamete being transferred from him will have the allele B and hence the progeny will only show the genotype Bb for pointed eyebrows. So, the correct option is 'Bb'.

Multiple choice trihybrid cross classical genetics botany

On selfing a plant of ${F} _{1}$ generation with genotype "AABbCC", the genotypic ratio in ${F} _{2}$-generation will be

  1. $1:2:1$
  2. $1:1$
  3. $9:3:3:1$
  4. $27:9:9:9:3:3:3:1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The possible gametes from the given genotype AABbCC are ABC and AbC.

The various possibilities of their their selfing would be :
1.  ABC × ABC = AABBCC
2. ABC × AbC = AABbCC
3. AbC × ABC = AABbCC
4. AbC × AbC = AAbbCC
It is evident from the above crosses that the genotypic ratio of F2 generation will be 1:2:1.
So, the correct option is '1:2:1'.

Multiple choice trihybrid cross classical genetics botany

In $TtggRr \times TtGgRr$, the percentage of recessive individuals will be

  1. 12

  2. 6

  3. 25

  4. 3
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a trihybrid cross, there are 32 possible combinations of alleles in the Punnett square out of which only one of the 32 combinations is recessive

i.e.3.125%
So, the correct option is '3'.

Multiple choice trihybrid cross classical genetics botany

In trihybrid, when all gene show complete linkage then what is the ratio of test cross?

  1. 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1

  2. 1 : 1

  3. 1 : 1 : 1 : 1

  4. 9 : 3 : 3 : 1
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A trihybrid cross is between two individuals that are heterozygous for three different traits. For example, RrYyCc x RrYyCc is a trihybrid cross where, 

RR = round, Rr = round, rr = wrinkled
YY = green, Yy = green, yy = yellow

CC = smooth, Cc = smooth, cc = constricted
The complete linkage is when genes are so closely associated that they are always trans­mitted together, and that does not undergo crossing over. 
So, in this case, the ratio of test cross for trihybrid with complete linkage will be 1:1. 
Thus, the correct answer is '1:1'.

Multiple choice trihybrid cross classical genetics botany

Self fertilising trihybrid plants form

  1. Eight different gametes and 64 different zygotes

  2. Four different gametes and sixteen different zygotes

  3. Eight different gametes and sixteen different zygotes

  4. Eight different gametes and thirty two different zygotes
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A total number of types of gamete produced by an organism is 2$^n$, where n is the number of heterozygous genes present. As we know that trihybrid plant is heterozygous for three genes, these total possible gametes by it = 2$^3$ = 8. During fertilization, any of the 8 types of sperms can fuse with any of the 8 types of egg to produce 64 possible combinations of zygotes. This makes option A correct but C and D wrong. Four different types of gametes are produced by an organism who is heterozygous for two genes; 2$^2$ = 4 and resultant four gametes produce total 4 x 4 = 16 zygotic combinations. Thus, option B is wrong.

Multiple choice trihybrid cross classical genetics botany

How many different kinds of gametes will be produced by a plant having the genotype AaBbCC?

  1. Two

  2. Three

  3. Four

  4. Nine
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

AaBbCC will produce 4 types of gametes which are as follows- ABC, AbC, aBC, abC. The number of gametes formed is decided by the number of heterozygous alleles present in the given genotype. 2^n is the formula used to find it out, where n=number of heterozygous alleles present in the genotype. Say for example, in the above genotype Aa & Bb are the 2 heterozygous alleles, so here n=2. Putting the values in the formula , we get 2^2=4. Hence 4 types of gametes are formed.

So the correct option is 'four'.

Multiple choice trihybrid cross classical genetics botany

If a quantitative character is influenced by the additive effect of four genes how many phenotypic categories are expected in the progeny of a tetrahybrid cross

  1. $5$
  2. $7$
  3. $9$
  4. $11$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For quantitative inheritance by four genes, we have a total of 8 alleles that will determine the phenotypes of the individual. Those offspring with all 8 dominant alleles ( Let's assume it as AABBCCDD ) will show the highest additive effect. Gradually as we keep decreasing the number of dominant alleles, additive effect will become less prominent. Hence, offspring with 7,6,5,4,3,2 and 1 dominant alleles will show different phenotypes. Finally, those individuals with no dominant alleles i.e. all recessive ( aabbccdd ) will show the least additive effect of characters. If we count all of them we have a total of 09 types of phenotypes possible. 

So, the correct option is '9'.

Multiple choice trihybrid cross classical genetics botany

When a hybrid (AaBbCcDd) is selfed then the genotypes AABbCCDd, AaBBCcDd, AaBbCcDd, aabbccdd would be in a proportion of?

  1. $2 : 4 : 8 : 21$
  2. $4 : 8 : 16 : 1$
  3. $4 : 8 : 16 : 27$
  4. $8 : 4 : 16 : 81$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The tetrahybrid cross of AaBbCcDd can be easily calculated by assuming their independent assortment.

The probability of both the alleles same is 1/4
The probability of both alleles different is 1/2
Hence when we calculate this for all the conditions AABbCCDd, AaBBCcDd, AaBbCcDd, aabbccdd is 4:8:16:1

So the correct answer is ' 4  : 8 : 16 : 1'.