
Match the genetic cross of parents on left (list 1) with the resulting genotypes on right (list2) of the offspring most likely to be produced from that cross.
List 1 List 2 a. BB $\times$ bb 1. $100\%$ bb b. Bb $\times$ Bb 2. $25\%$ BB, $50\%$ Bb, $25\%$ bb c. BB $\times$ BB 3. $100\%$ BB d. bb $\times$ bb 4. $100\%$ Bb
A. a-2, b-1, c-3, d-4
B. a-4, b-2, c-3, d-1
C. a-1, b-3, c-2, d-4
D. a-1, b-2, c-4, d-3
| List 1 | List 2 | ||
| a. | BB $\times$ bb | 1. | $100\%$ bb |
| b. | Bb $\times$ Bb | 2. | $25\%$ BB, $50\%$ Bb, $25\%$ bb |
| c. | BB $\times$ BB | 3. | $100\%$ BB |
| d. | bb $\times$ bb | 4. | $100\%$ Bb |
Answer
585.9k+ views
Hint: The inheritance is the process by which the characters are passed on from parents to their offspring; it is the basis of heredity. Gregor Mendel was the first scientist who applied statistical analysis and mathematical logic in biology. He experimented with garden peas to find out the inheritance pattern of the genes.
Complete answer:
Mendel experimented with garden pea with seven contrasting characters to find out the inheritance pattern of various genes. Out of which he also studied the inheritance of one gene.
In this case, we have to find the genotype of the offspring due to one gene. Thus, we will find it by crossing the parents with Punnett square method. The following results will be observed:
In case 1, where BB x bb, the resulting genotypes will be Bb, Bb, Bb and Bb. Therefore, all progeny will be $100\% $ Bb due to the presence of dominant alleles. Thus, the correct option for (a) is 4.
In case 2, where Bb$ \times $Bb, the genotypes observed are BB, Bb, Bb and bb. Here, the resulting genotype will be $25\% $ BB, $50\% $ Bb and $25\% $bb. Thus, the correct option for (b) is 2.
In the 3rd case, where BB$ \times $BB , the offspring produced will have the genotypes of BB that is $100\% $ BB as only the dominant allele is present. Thus, a suitable option for (c) is 3.
In the last case, in which bb$ \times $bb is crossed, the resulting genotypes will all be bb ( $100\% $bb), because of homozygous recessive alleles. Thus, the correct alternative for (d) is 1.
Hence, The correct answer is option (B): a-4, b-2, c-3, d-1.
Note: Genes are the units of inheritance. The genes contain the information which is required to express a particular trait in an organism. Genes which code for a pair of contrasting characters are called alleles. The alleles can be similar as in case of homozygotes or can be dissimilar in case of heterozygotes.
Complete answer:
Mendel experimented with garden pea with seven contrasting characters to find out the inheritance pattern of various genes. Out of which he also studied the inheritance of one gene.
In this case, we have to find the genotype of the offspring due to one gene. Thus, we will find it by crossing the parents with Punnett square method. The following results will be observed:
In case 1, where BB x bb, the resulting genotypes will be Bb, Bb, Bb and Bb. Therefore, all progeny will be $100\% $ Bb due to the presence of dominant alleles. Thus, the correct option for (a) is 4.
In case 2, where Bb$ \times $Bb, the genotypes observed are BB, Bb, Bb and bb. Here, the resulting genotype will be $25\% $ BB, $50\% $ Bb and $25\% $bb. Thus, the correct option for (b) is 2.
In the 3rd case, where BB$ \times $BB , the offspring produced will have the genotypes of BB that is $100\% $ BB as only the dominant allele is present. Thus, a suitable option for (c) is 3.
In the last case, in which bb$ \times $bb is crossed, the resulting genotypes will all be bb ( $100\% $bb), because of homozygous recessive alleles. Thus, the correct alternative for (d) is 1.
Hence, The correct answer is option (B): a-4, b-2, c-3, d-1.
Note: Genes are the units of inheritance. The genes contain the information which is required to express a particular trait in an organism. Genes which code for a pair of contrasting characters are called alleles. The alleles can be similar as in case of homozygotes or can be dissimilar in case of heterozygotes.
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