Answer:
Explanation:
To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.
The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.
So, en the exposed example:
- J and K are autosomal genes
- J and K are separated by 60 M.U.
- 60 M.U. means that there is 60% of recombination.
Cross) J K / j k x j k / j k
Gametes) JK Parental jk, jk, jk, jk
jk Parental
Jk Recombinant
jK Recombinant
One map unit equals 1% of recombination frequency. This means that every 100 meiotic products, one of them is a recombinant one.
1 M.U. -------------- 1% recombination
60 M.U. ------------ 60% recombination
30% Jk + 30% jK
100 M.U. - 60 M.U. = 40 M.U.
40M.U.--------------40 % Parental (Not recombinant)
20% JK + 20% jk
Punnet Square) JK jk Jk jK
jk JK/jk jk/jk Jk/jk jK/jk
J K / j k = 20%
j k / j k = 20%
J k / j k = 30%
j K / j k = 30%
As a mechanism of evolution, natural selection can be most closely equated with UNEQUAL REPRODUCTIVE SUCCESS.
Mechanism of evolution refers to micro-evolution that involves mutation, gene flow and natural selection. Unequal reproductive success is also related to other processes of micro evolution such as natural selection. The organisms that can reproduce unequally and survive even in adverse condition, increase in numbers.
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<span> Wow, this is a big question, but here goes. Freckles q: 1) Since freckles is the dominant phenotype, this means that Olivia must have the recessive phenotype, which is only given by the genotype ff. FF and Ff would both give freckles because, again, freckles is dominant. So any genotype with F_ would give freckles. 2)Same answer as #1. 3) Since they only have f alleles to pass on, both the dad and the mom will pass on one f each to 100% of their children. So all will have genotype ff and no freckles.
Hairline q: Since the parents are heterozygous, this means that their genotype has one of each allele to give Hh. (Even though you said they are quadruplets, I'm going to assume they are fraternal (different DNA) b/c otherwise they'd always have the same genotype if they are identical). 1) Okay, so when you do the Punnett square for Hh x Hh, you'll get 1HH:2Hh:1hh. Since widow's peak is dominant, any genotype with H_ gives the widow's peak. So, there are 3 options from the Punnett square, so you'll have 3 of the kids with widow's peak. Theres a 3/4 chance that a child will have widow's peak from these parents. 2) Since straight hairline is recessive, you need hh to get this phenotype and there is only one option. So, there will only be one child with the straight hairline. (1/4 chance). 3) Homozygous dominant means that you have two of the dominant allele, so HH. Since there is only 1 option, there is only one child with this genotype.
Tongue q: Since rolling the tongue is a dominant trait and the parents both can't roll their tongues, they must have a homozygous recessive genotype for this to happen (remember in dominance, any genotype with a dominant allele will give the dominant phenotype), so they have tt. 1) Since again they can only pass on t alleles, the kids will all have tt, so no one can roll their tongues. 2) None of them are hybrids because there was no variety in the genotypes or anything. Both parents had tt, so they were same in genotype. 3) She will have tt because of the above stated reasons.
Dimples q: Since all four kids have dimples, the dominant phenotype, they must all have the genotype D_ (either Dd or DD). 1) Since Marcus is a hybrid, this means that he had parents that were DD x dd to give him the genotype Dd. Since he has a recessive allele d in his genotype, Olivia must have all dominant alleles to make sure that each child has at least one dominant D. So, she must have the genotype DD. 2) Since she is DD, the dominant alleles will make her have dimples.
Earlobe q: 1) Since the parents are EE both, a cross of EE x EE will give EE genotype children. So, all children have EE, this means they all have free earlobes. So the ratio is 100% free to 0% attached earlobes. 2)Homozygous means they have two of the same alleles. Since all of them have EE, 100% of them are homozygous.
PTC q: Marcus has genotype bb and Olivia has genotype Bb because she is heterozygous. 1) The cross of bb x Bb gives 1Bb:1bb, So, 1/2 can taste the paper, so 50% can taste. 2) Since Violet can't taste the paper, she must be recessive and have the genotype bb. Since both of the boys can taste, they must have the genotypes Bb. Since 1/2 is already Bb, Claudia must be bb to help create the 50% that can't taste in the kids. 3)So, 2 people out of the family can taste the paper. Even though Olivia has Bb and should be able to taste, she can't. So, only Jonas and Nathan can taste the paper.
Pheww... done.. Hope this helps! :)<span>
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