Three Mikan

So, you’ve read the article on dihybrid crosses, you’ve red the article on monohybrid crosses, and now you want to move on to polyhybrid crosses (but take it slow). Well, far be it from me to keep you from learning. Let’s see if you’ve got it…

In the hypothetical MPSN world, goldfish only have three traits: size, color, and teeth. The size gene comes in two flavors, big (B) and small (b). The color and teeth genes are similar, coming in gold(G) or black(g) and teeth(T) or no teeth(t), respectively. Let’s cross a homozygous dominant male with a homozygous recessive female. What are the phenotypes of the F1 offspring (the children)? And if we cross the F1 offspring with each other, what are the phenotypes and ratios of the F2 offspring (the grandchildren)?

Can you handle it? Now don’t scroll down until you think you have answered all the questions. Ready? GO!

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Step 1: Gather the Known Information

• Mom = (bbggtt)
• Dad = (BBGGTT)

Step 2: Find Unknown Information

The first thing we’ll need to do, now that we know our parental phenotypes is to figure out what types of gametes that they can produce. Since we know that mom ONLY HAS (b), (g), andt (t), then (bgt) is the only gamete she can produce. Likewise for dad: only his gametes will each be dominant (BGT).

If we cross these, we end up with 100% (BbGgTt) offspring in the F1 (children) generation. Thankfully since mom and dad only have dominant or recessive alleles to give, we can deduce that the offspring will each be heterozygous WITHOUT USING A PUNNETT SQUARE! Woot!

HOWEVER, to answer the second question, we must employ the help of a punnett square. Here goes:

So, what are the F1 gametes? Since they are all (BbGgTt), then they will all have the same set of gametes. (But those gametes can combine differently, so don’t assume that all the offspring of the F2 generation (grandchildren) will be the same like in the F1 generation.) Using what we know about permutations, we can calculate that there will be 8 different types of gametes. These F1 gametes are: (BGT), (BGt), (Bgt), (bgt), (bGt), (bgT), (BgT), (bGT). Now, to find out the phenotypes, we have to cross them. 123GO.

OY. That took me 20 minutes to type up… such a pain… But anyway, here’s what we found:

• Big, Gold, and with Teeth      = 27
• Big, Gold, and toothless        = 9
• Big, black, and with Teeth     = 9
• Big, black, and toothless       = 3
• small, Gold and with Teeth    = 9
• small, Gold, and toothless     = 3
• small, black, and with Teeth  = 3
• small, black, and toothless    = 1

As you can see, both for populations of Big F2 offspring (grandchildren) and small F2 offspring (grandchildren), we can see a difinite 9:3:3:1 ratio. [As you travel up the corporate ladder into bigger polyhybrid crosses, be on the look out for that 9:3:3:1 ratio]

Step 3: Check Your Work.

The original problem asked 2 questions: “What would the F1 phenotype be?” and “What would the F2 phenotypes and ratios be?” Have we answered both of those questions? Yes? Cool. Good work.

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Best of Luck,

Grey

Yak

• Now that you guys have a fairly good understanding of the difference between dominance and recessiveness, I’ll no longer be referring to traits as dominant or recessive. I’ll simply refer to one trait as (X) and another as (x) and I’ll expect you to know the difference. You’re growing up in your biological education, so it’s time to remove some training wheels. If you fall of the bike, get a band-aid, review the last few posts, and move on!

So since you’re growing up, come right over here. Sit with me, pour yourself a glass of scotch, and I’ll tell you all about my Uncle Vlad. (NOTE: MPSN DOES NOT CONDONE UNDERAGE DRINKING. THANK YOU.) Uncle Vlad is a yak breeder in the imaginary nation of Romania, the beta version. Uncle Vlad is known all throughout the countryside for his skill in breeding healthy, white-haired (h) yak that produce a sweet bright blue milk (m)… regardless of gender. Now I’m not sure if you know anything about yak breeding, but getting a combination like that is VERY rare. You see, brown hair (H) and bitter green milk (M) are the usual traits.

He once told me of his first yak; he won her at a village fair.  You see, there was an old monk (the kids liked to call him Mr. Greg) who had this crazy theory about the inner workings of having children. To the fair, he brought a herd of yak, and he told everyone there that if they could explain why the two parent yak had yaklets that looked the way they did, the winner could choose a yak from the herd to take home as his or her own. Uncle Vlad couldn’t resist, even though he had read all of crazy old Greg’s papers.

He first noticed that the two yak that had been called the parents were both brown, but the mom produced blue milk and the dad produced green milk. He also noticed that there were only two main types of yaklets: brown with blue milk and brown with green milk. Old Greg saw him concentrating on the yak, so he decided to help out a little. Beside the dad, he placed two large combs and two small buckets. Beside the mom, he placed one large and one small comb and one large bucket. Although, Uncle Vlad’s memory escapes him as to HOW he did it, he does recall winning the yak and eventually going to medical school.

By now, you’re probably certain that this story is utter mess, but it’s an entertaining way to say: LETS EXPLORE THE GENETICS OF THE BETA-ROMANIAN YAK. Mad at me? Oh well.

So let’s pretend that the parent yak went to Vegas for their honeymoon after the wedding and soon thereafter gave birth to non-identical triplets. (They don’t all have look the same because they don’t have identical DNA.) They all have brown hair, and two have green milk. If the parents decide to go on three more honeymoons and have triplets each time, how many kids total should we theoretically expect to produce blue milk?

I know, I know. This is a long article. “Get to the point.” I have heard your cries, so I’ll condense the technical side. Here goes.

Step 1: Collect given information.

I did this on purpose. Let’s go through the article and pick up the stuff that we already know before we start calculatin’.

• Brown hair = (H), White hair = (h), Green milk = (M), Blue milk = (m)
• Mom’s genotype = (HHmm) just in case you didn’t get that from the bucket & comb hint
• Dad’s genotype = (HhM?) The (?) is because that allele could either be a (t) or a (T); we don’t know yet.
• Kid’s phenotypes = brown hair for all,  green milk for 2, blue milk for 1

Step 2: Figure out genotypes.

In order to answer the question, we’ll need to know the genotypes of the parents. We already know mom’s but we’re missing one crucial allele from dad. Let’s look at the kids in order to figure it out. We already know the genotypes for the hair gene, so lets ignore it and move on to the milk gene.

There are kids that produce blue milk and kids that produce green milk. In order for this to occur, there must be kids who are (mm) AND kids who are either (MM) or (Mm). Regardless of genotype, the kids have to get alleles from the parents – one from each. Since mom can only give (m) alleles, we know that there are no kids that have the (MM) genotype for the milk gene. [Dad has the (M) allele, but he can only give ONE to the child.] So, we know that there are kids with (mm) and kids with (Mm).

We’ve figured out the genotypes of the kids, but that doesn’t help us figure out how to answer the question… or does it? Remember our point? We have to figure out dad’s genotype. He’s (HhM?). So what is the identity of the (?) allele? Well, just like we used mom to figure out that there are no (MM) kids, we’re gonna use the (mm) kid to figure out that dad’s mysterious allele is an (m). The kid gets one (m) from mom (mm) who ONLY has (m) to give, but he had to get the other one from somewhere. That must mean that dad is (Mm) for the milk gene.

Step 3: Test by Crossing

Now that we know that mom’s genotype is (HHmm) and that dad’s is (HhMm), we can figure out what we can expect as far as kids go. Since mom only has one type of allele for each gene, she can only produce one type of gamete (Hm). There are no other combinations. Dad, however, has many different combinations. In fact, we can figure out exactly how many by using basic probability rules. He has 2 genes represented, and both of those genes have 2 alleles represented. Therefore, (2×2)=4, and he can produce these gametes: (HM), (Hm), (hM), and (hm). As a cross, we can see the progeny below:

Step 4: Interpret results

The question asked us how many children, out of 12, would we expect to produce blue milk. (We therefore exclude the actual kids that have already been born.) Since 2/4 of the progeny had the genotype (mm) for the milk gene, we can conclude that 2/4 of the 12 children, aka 6, should produce blue milk.

So to recap:

1. Collect given info
2. Decipher genotypes of all individuals involved
3. Cross with Punnett square
4. Interpret results in light of the question asked.

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Best of Luck,

Grey

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Ok, so we’ve already done a cross, but in that situation, I wanted to focus more on the fact that we were dealing with a situation focused on only one gene. In this case, we’re still going to deal with only one gene, but now, we’re going to set out with the purpose of figuring out the parent’s genotype from using the offspring. If you’re just getting started with genetics, don’t worry. Soon, you’ll be able to do monohybrid crosses in your head. Scouts Honor.

So, here’s the problem:

Mrs. Payette breeds parrots and teaches them to talk. Unfortunately, learning to talk requires the ability to hear, and one of her male parrots, Dazzy, is deaf. He’s such a beautiful bird though, so she doesn’t want to keep him from breeding unless it’s there is absolutely no way that he’ll have hearing parrotlets. In this case, deafness is a dominant phenotype and the ability to hear is recessive. She wants us to figure out whether or not letting him breed is worth it.

Let’s identify the big problem. What is Dazzy’s genotype? We were told that dazzy shows a dominant phenotype, deafness, among a group of parrots who are all NOT DEAF, meaning that they are all recessive. So, we can assume that the rest of Mrs. Payette’s parrots are (dd) and that Dazzy is …. no wait, we don’t know what Dazzy is. His genotype is either (Dd) or (DD). Bingo.

There are no parrotlets yet, so there’s no way to figure this out. So let’s tell Mrs. Payette that we need behbehs.

(Elapsed time: 2 months)

SO, Dazzy and Cindy had parrotlets… 10 to be exact. Four of them can sleep through the squawking of all their comrades, so Mrs. Payette are pretty sure that they are deaf.

Because there are deaf parrotlets, we know that Dazzy’s genotype is (Dd). Did you figure out why? Well let’s make sure you know.

As you can see, if Dazzy were (DD), then ALL of his parrotlets would be deaf (Dd). Instead, 4/10 (almost half) are deaf, which makes sense if you look at the Punnett square. (We expected 50%. But hey nothing in life works out perfectly every time.)

Rules for Crosses:

1. FIND THE PROBLEM. Teachers are tricky; make sure to identify the actual question so that you don’t do a lot of work for nothing.
2. Identify the given information.
3. Try to identify all the genotypes of the individuals involved. If you can’t identify them all (which you probably won’t, since that’s the problem), then use the given information to find the not-given information.
4. Check your work with a Punnett Square.
5. Check to make sure that you answered the question completely – yet another way to easily lose points.

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As Always, Best of Luck,

Grey

In zebras, purple eyes are dominant to yellow ones. (I don’t know if this is true, but just go with it.)  A purple-eyed zebra-man has little zebralets with yellow-eyed zebra babe: 487 purple-eyed 520 yellow-eyed. What are the genotypes of the man and his kids? What are the probabilities of their possible children?

We’ll call the purple allele (P) because it’s dominant, and we’ll call the yellow allele (p) because it’s the recessive version of the gene. Now that we know how we’re gonna write the genotype of the parents, let’s figure those genotypes out.

We know that the female genotype is (pp) because we were told that, in this example, yellow is both recessive and the color of the female’s eyes.

Dad’s genotype is going to be a little harder to decide – but not much. Remember that with dominant alleles, both homozygotes and heterozygotes will show the dominant phenotype. If you’re not really good with the genetic jargan, that means that Dad will have purple eyes whether his genotype is (PP) or (Pp). That’s where we come in; we have to figure out which one he is.

This is usually where most people have problems. “Uhhhh… what now?” What a great question. Let’s figure out the children’s genotypes – since we already figured out everybody else’s. There are both purple and yellow eyed kids. That means that there are either (Pp) or (PP) AND there are (pp) represented.

REMEMBER that the children can only get alleles from mom and dad. SO if mom and dad don’t have them, the children can’t. Since Mom’s genotype is (pp), then she can only give (p) alleles to her kids. Since we know that there is going to be at least one (p) in all the kids, we can deduce that NONE of the kids will be (PP) and that the purple-eyed kids will have (Pp) as their genotype.

HEY! Wait a sec, we missed something. OH YEAH, we can also figure out Dad’s genotype now. Remember what I said earlier about kids only being able to get alleles from the parents? Let’s look at those yellow eyed kids again. They have a genotype of (pp) and we already know that one of those (p)’s came from Mom… so what about the OTHER one? It had to come from Dad, who has purple eyes. THEREFORE, his genotype must be (Pp).

FINALLY, we did ALL THAT WORK. Phew! But let’s check the question to make sure that we answered the question. I made that mistake a lot when I took genetics- so check to make sure that you’re done. There were two parts: genotype and probability. We know that Dad’s genotype is (Pp) and that his children are (Pp) and (pp). BUT we haven’t figured out the probabilities yet. Sorry.

It’s not that bad. Quit whining. Let’s just do this. It’s all down hill from here. Watch.

Mom: (pp)

Dad: (Pp)

If you didn’t get it that easily, check out my article on understanding Punnett Squares. But as you can see on the graph, 2/4 of the children are (Pp) and the other 2/4 are (pp). THEREFORE, we know that 0.5 of the children are (Pp) and the other 0.5 are (pp). And that makes sense seeing as 487+520 are 1007, and both numbers are just about half. NOW, we’re done.

Let’s recap the rules that we learned.

1. Name the alleles with the letter of the dominant. That way, it’s easier to remember which one is dominant.
2. The DOMINANT is capitalized and the recessive is not.
3. Use the genotypes of what’s given to find out the genotypes of what’s not.
4. If statistics are given, check the given data with the stats to make sure it makes sense.

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Best of Luck,

Grey