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In a testcross involving F(1) dihybrid f...

In a testcross involving `F_(1)` dihybrid flies, more parental-type offsping were produces than the recombinant-type offspring. This indicates

A

Both of the characters are controlled by more than-one gene

B

The two genes are located on two different chromosomes.

C

Chromosomes failed to separate during meiosis

D

The two genes are linked and present on the same chromosome

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Two genes 'A' and 'B' are linked. In a dihybrid cross involving these two genes, the F_(1) hetrozygote is crossed with homozygous recessive parental type (aa bb). What would be the ratio of offspring in the next generation?

Two genes R and Y are located very close on the chromosomal linkage map of maize plant. When RRYY and rryy genotypes are hybridized the F_(2) segregation will show 1. segregation in the expected 9:3:3:1 ratio 2. segregation in 3:1 ratio 3. higher number of parental types 4. higher number of the recombinant types

Knowledge Check

  • Let us take an electrical conductor in which the electrical energy supplied is entirely converted into heat . If , for the conductor , the terminal potential difference = V , the curent through it = I and its resistance = R , then the electrical energy cnsumed in its t is , W=I^(2)t (from Ohm'slaw R=(V)/(I) ) . So , if thr electrical and the heat energies both are expressed in joule , the heat developed in tiem t is H=I^(2)Rt . However , If H is expressed in the conventional unit calorie , then from the law , W = JH , we may write H=(I^(2)Rt)/(J) , where , J = mechanical equivalent of heat =4.2J."cal"^(-1) . The resistance R of a conducting wire depends on its material , its length l and its area of cross section a .The resistivity of the meterial of the conductor is , rho=(RA)/(l) . When more than one heat -producing conductors are kept in series in a circuit , the same current passes through each of them , but as their resistance are f=different in general , the terminal potential differences are also unequal . On the other hand , each conductor has the same terminal potential difference in a parallel combination , however , the curents through them are different . The terminal potential difference and the currents through two conducting wires are both in the ratio 2 : 1 The ratio of the rates of heat evolved in them is

    A
    `1:1`
    B
    `2:1`
    C
    `4:1`
    D
    `8:1`
  • Let us take an electrical conductor in which the electrical energy supplied is entirely converted into heat . If , for the conductor , the terminal potential difference = V , the curent through it = I and its resistance = R , then the electrical energy cnsumed in its t is , W=I^(2)t (from Ohm'slaw R=(V)/(I) ) . So , if thr electrical and the heat energies both are expressed in joule , the heat developed in tiem t is H=I^(2)Rt . However , If H is expressed in the conventional unit calorie , then from the law , W = JH , we may write H=(I^(2)Rt)/(J) , where , J = mechanical equivalent of heat =4.2J."cal"^(-1) . The resistance R of a conducting wire depends on its material , its length l and its area of cross section a .The resistivity of the meterial of the conductor is , rho=(RA)/(l) . When more than one heat -producing conductors are kept in series in a circuit , the same current passes through each of them , but as their resistance are f=different in general , the terminal potential differences are also unequal . On the other hand , each conductor has the same terminal potential difference in a parallel combination , however , the curents through them are different . Heat is produced at hte rate of 8"cal".s^(-1) in a uniform wire , when its terminal potential difference is 10 V What would be the rate in another wire of the same material , of the same potential difference ?

    A
    `32"cal".s^(-1)`
    B
    `16"cal".s^(-1)`
    C
    `4"cal".s^(-1)`
    D
    `2"cal".s^(-1)`
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