Answer:
1/3 of black F2 progeny will be homozygous.
Explanation:
Here, BB = black coat color = true breeding dominant black strain
bb = white coat color = true breeding recessive white strain
When true breeding black and white guinea pigs are mated to give F1 progeny:
BB X bb = Bb ( all black guinea pigs )
When two of the F1 guinea pigs are mated to give F2 progeny:
Bb X Bb = BB, Bb, Bb, bb
F2 progeny has 3/4 progeny as black (BB and Bb) . Out of them 1/3 are homozygous and 2/3 are heterozygous.
Hence, 1/3 of black F2 progeny will be homozygous.
Answer:
i would think vegitarians because we an eventually run out of meat but we can keep growing vegtables
Explanation:
<span>Somatic mosquito cells have 6 chromosomes. Its gamete has 3 chromosomes.
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Mosquitoes have 6 number of chromosome but when mosquitoes<span> make sperm or egg </span>cells<span>, meiosis reduces the </span>chromosome<span> number to 3.
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*the chromosome number in somatic body cells of a mosquito --- 6.
<span>the haploid (n) number is 3.
the diploid (2n) number is 6.</span>
Answer:
Short answer is primers are partially complementary.
Explanation:
Forward primer: 5'-AGTCTACTCGTAACCGGTTACC-3'
Reverse primer: 5'-TAAGGCATCATGGTAACCGGTT-3'
When we write reverse primer 5' to 3' we can easily see that
3'-TTGGCCAATGG---5' is complementary to the forward primers'
5'---AACCGGTTACC-3' sequence. So instead of binding to the template DNA these primers might bind each other resulting with reduction of efficiency of DNA amplification.
