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Let's address each of these genotypes step-by-step, assuming independent assortment of the four genes. In a tetrahybrid cross involving the genotypes AaBbCcDd, each allele has an equal probability of being passed on. We will calculate the probabilities for each specific genotype in the [tex]$F_2$[/tex] generation.
### (a) aabbccdd
This genotype is homozygous recessive for all alleles (aa bb cc dd).
The probability of obtaining aa from Aa x Aa is:
[tex]\[ P(aa) = \frac{1}{4} \][/tex]
The probability of obtaining bb from Bb x Bb is:
[tex]\[ P(bb) = \frac{1}{4} \][/tex]
The probability of obtaining cc from Cc x Cc is:
[tex]\[ P(cc) = \frac{1}{4} \][/tex]
The probability of obtaining dd from Dd x Dd is:
[tex]\[ P(dd) = \frac{1}{4} \][/tex]
Since the genes assort independently, we multiply the probabilities:
[tex]\[ P(aabbccdd) = \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{256} \][/tex]
So, the probability is:
[tex]\[ P(aabbccdd) = 0.00390625 \][/tex]
### (b) AaBbCcDd
This genotype is heterozygous for all alleles (Aa Bb Cc Dd).
The probability of obtaining Aa from Aa × Aa is:
[tex]\[ P(Aa) = \frac{1}{2} \][/tex]
The probability of obtaining Bb from Bb × Bb is:
[tex]\[ P(Bb) = \frac{1}{2} \][/tex]
The probability of obtaining Cc from Cc × Cc is:
[tex]\[ P(Cc) = \frac{1}{2} \][/tex]
The probability of obtaining Dd from Dd × Dd is:
[tex]\[ P(Dd) = \frac{1}{2} \][/tex]
Multiplying the probabilities gives:
[tex]\[ P(AaBbCcDd) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{16} \][/tex]
So, the probability is:
[tex]\[ P(AaBbCcDd) = 0.0625 \][/tex]
### (c) AABBCCDD
This genotype is homozygous dominant for all alleles (AA BB CC DD).
The probability of obtaining AA from Aa x Aa is:
[tex]\[ P(AA) = \frac{1}{4} \][/tex]
The probability of obtaining BB from Bb x Bb is:
[tex]\[ P(BB) = \frac{1}{4} \][/tex]
The probability of obtaining CC from Cc x Cc is:
[tex]\[ P(CC) = \frac{1}{4} \][/tex]
The probability of obtaining DD from Dd x Dd is:
[tex]\[ P(DD) = \frac{1}{4} \][/tex]
Multiplying these probabilities gives:
[tex]\[ P(AABBCCDD) = \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{256} \][/tex]
So, the probability is:
[tex]\[ P(AABBCCDD) = 0.00390625 \][/tex]
### (d) AaBBccDd
This genotype is heterozygous for Aa and Dd, homozygous dominant for BB, and homozygous recessive for cc.
The probability of obtaining Aa from Aa x Aa is:
[tex]\[ P(Aa) = \frac{1}{2} \][/tex]
The probability of obtaining BB from Bb x Bb is:
[tex]\[ P(BB) = \frac{1}{4} \][/tex]
The probability of obtaining cc from Cc x Cc is:
[tex]\[ P(cc) = \frac{1}{4} \][/tex]
The probability of obtaining Dd from Dd x Dd is:
[tex]\[ P(Dd) = \frac{1}{2} \][/tex]
Multiplying these probabilities gives:
[tex]\[ P(AaBBccDd) = \left(\frac{1}{2}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{2}\right) = \frac{1}{64} \][/tex]
So, the probability is:
[tex]\[ P(AaBBccDd) = 0.015625 \][/tex]
### (e) AaBBCCdd
This genotype is heterozygous for Aa, homozygous dominant for BB and CC, and homozygous recessive for dd.
The probability of obtaining Aa from Aa x Aa is:
[tex]\[ P(Aa) = \frac{1}{2} \][/tex]
The probability of obtaining BB from Bb x Bb is:
[tex]\[ P(BB) = \frac{1}{4} \][/tex]
The probability of obtaining CC from Cc x Cc is:
[tex]\[ P(CC) = \frac{1}{4} \][/tex]
The probability of obtaining dd from Dd x Dd is:
[tex]\[ P(dd) = \frac{1}{4} \][/tex]
Multiplying these probabilities gives:
[tex]\[ P(AaBBCCdd) = \left(\frac{1}{2}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{128} \][/tex]
So, the probability is:
[tex]\[ P(AaBBCCdd) = 0.0078125 \][/tex]
### (a) aabbccdd
This genotype is homozygous recessive for all alleles (aa bb cc dd).
The probability of obtaining aa from Aa x Aa is:
[tex]\[ P(aa) = \frac{1}{4} \][/tex]
The probability of obtaining bb from Bb x Bb is:
[tex]\[ P(bb) = \frac{1}{4} \][/tex]
The probability of obtaining cc from Cc x Cc is:
[tex]\[ P(cc) = \frac{1}{4} \][/tex]
The probability of obtaining dd from Dd x Dd is:
[tex]\[ P(dd) = \frac{1}{4} \][/tex]
Since the genes assort independently, we multiply the probabilities:
[tex]\[ P(aabbccdd) = \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{256} \][/tex]
So, the probability is:
[tex]\[ P(aabbccdd) = 0.00390625 \][/tex]
### (b) AaBbCcDd
This genotype is heterozygous for all alleles (Aa Bb Cc Dd).
The probability of obtaining Aa from Aa × Aa is:
[tex]\[ P(Aa) = \frac{1}{2} \][/tex]
The probability of obtaining Bb from Bb × Bb is:
[tex]\[ P(Bb) = \frac{1}{2} \][/tex]
The probability of obtaining Cc from Cc × Cc is:
[tex]\[ P(Cc) = \frac{1}{2} \][/tex]
The probability of obtaining Dd from Dd × Dd is:
[tex]\[ P(Dd) = \frac{1}{2} \][/tex]
Multiplying the probabilities gives:
[tex]\[ P(AaBbCcDd) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{16} \][/tex]
So, the probability is:
[tex]\[ P(AaBbCcDd) = 0.0625 \][/tex]
### (c) AABBCCDD
This genotype is homozygous dominant for all alleles (AA BB CC DD).
The probability of obtaining AA from Aa x Aa is:
[tex]\[ P(AA) = \frac{1}{4} \][/tex]
The probability of obtaining BB from Bb x Bb is:
[tex]\[ P(BB) = \frac{1}{4} \][/tex]
The probability of obtaining CC from Cc x Cc is:
[tex]\[ P(CC) = \frac{1}{4} \][/tex]
The probability of obtaining DD from Dd x Dd is:
[tex]\[ P(DD) = \frac{1}{4} \][/tex]
Multiplying these probabilities gives:
[tex]\[ P(AABBCCDD) = \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{256} \][/tex]
So, the probability is:
[tex]\[ P(AABBCCDD) = 0.00390625 \][/tex]
### (d) AaBBccDd
This genotype is heterozygous for Aa and Dd, homozygous dominant for BB, and homozygous recessive for cc.
The probability of obtaining Aa from Aa x Aa is:
[tex]\[ P(Aa) = \frac{1}{2} \][/tex]
The probability of obtaining BB from Bb x Bb is:
[tex]\[ P(BB) = \frac{1}{4} \][/tex]
The probability of obtaining cc from Cc x Cc is:
[tex]\[ P(cc) = \frac{1}{4} \][/tex]
The probability of obtaining Dd from Dd x Dd is:
[tex]\[ P(Dd) = \frac{1}{2} \][/tex]
Multiplying these probabilities gives:
[tex]\[ P(AaBBccDd) = \left(\frac{1}{2}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{2}\right) = \frac{1}{64} \][/tex]
So, the probability is:
[tex]\[ P(AaBBccDd) = 0.015625 \][/tex]
### (e) AaBBCCdd
This genotype is heterozygous for Aa, homozygous dominant for BB and CC, and homozygous recessive for dd.
The probability of obtaining Aa from Aa x Aa is:
[tex]\[ P(Aa) = \frac{1}{2} \][/tex]
The probability of obtaining BB from Bb x Bb is:
[tex]\[ P(BB) = \frac{1}{4} \][/tex]
The probability of obtaining CC from Cc x Cc is:
[tex]\[ P(CC) = \frac{1}{4} \][/tex]
The probability of obtaining dd from Dd x Dd is:
[tex]\[ P(dd) = \frac{1}{4} \][/tex]
Multiplying these probabilities gives:
[tex]\[ P(AaBBCCdd) = \left(\frac{1}{2}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{1}{128} \][/tex]
So, the probability is:
[tex]\[ P(AaBBCCdd) = 0.0078125 \][/tex]
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